part a. solve ve of the following eight problems... find the characteristic of ... part b. solve ve...

Mathematics Department Qualifying Exam: Algebra Fall 2012

Part A. Solve five of the following eight problems:

1. Let

R =

{[a b2b a

] ∣∣∣∣ a, b ∈ Z}

and S = {a+ b√

2 | a, b ∈ Z}

Define ϕ : R→ S by ϕ

([a b2b a

])= a+ b

√2. Prove that ϕ is a ring isomorphism.

You may assume that R and S are commutative rings and that ϕ is a well-defined function.

2. (a) Let G1 and G2 be groups and consider their direct product G1×G2. Prove that G1×G2

is Abelian if and only if both G1 and G2 are Abelian.

(b) The dihedral group Dn of order 2n (n ≥ 3) has a subgroup H of order n consisting ofrotations, and several subgroups of order 2. Prove that Dn cannot be isomorphic to thedirect product of H and any of the groups of order 2.

3. Let G be a cyclic group of order 8. Prove that G has exactly one element of order 2. Is therea non-Abelian group of order 8 with this property?

4. Let G be a group and a, b ∈ G. Assume that |a| = 12, |b| = 22 and 〈a〉 ∩ 〈b〉 6= {e}, provethat a6 = b11.

5. An (additive) Abelian group G is said to be divisible if for every element a ∈ G, and everyk ∈ Z \ {0}, there is an element x ∈ G such that kx = a.

Show that Q is divisible but that Z is not.

6. Recall that the center of a group G is the subset Z(G) = {a ∈ G | ax = xa for all x ∈ G}.

(a) Prove that Z(G) is a subgroup of G;

(b) Prove that Z(G) is normal in G.

7. Let v be a fixed vector in Rn. Show that the set of all matrices A ∈Mn(R) such that Av = 0is a left ideal of Mn(R).

8. (a) Find the characteristic of the ring Z4⊕4Z (which, using a different notation, is Z4×4Z).

(b) Let R be a commutative ring of characteristic 2. Prove that the set of all idempotentsin R is a subring of R.Remark: x ∈ R is an idempotent iff x2 = x.

Part B is on the back!!!

Part B. Solve five of the following eight problems :

1. Let T : R2 → R2 be a linear transformation given by T (x1, x2) = (x1 + x2,−3x1 − 3x2).

(a) Determine whether or not T is invertible.

(b) Determine whether or not the vector w = (2, 5) is in the range of T .

(c) Find the dimension of the kernel of T .

2. Suppose that B = P−1AP (where A, B and P are square matrices of the same size, and Pis invertible) and that x is an eigenvector of A corresponding to an eigenvalue λ. Show thatP−1x is an eigenvector of B corresponding also to λ.

3. The following 3× 3 matrix depends on c:

A =

1 1 2 43 c 2 80 0 2 2

.(a) For each c, find a basis for the column space of A.

(b) For each c, find a basis for the null space of A.

(c) For each c, find the complete solution to Ax =

1c0

.

4. Given A =

[0 1−2 −3

], find A12.

5. Suppose that A is an invertible matrix and similar to B. Show that B is invertible, and thatA−1 is similar to B−1.

6. Justify your answer to each part.

(a) Let A be a fixed 3 × 3 matrix, and let V be the set of all 3 × 3 matrices B such thatAB = BA. Is V a subspace of M3×3(R)?

(b) Let N be the set of all continuous functions on [−1, 1] so that f(x) < 0 for x ∈ [−1, 1].Is N a subspace of C[−1, 1]?

(c) Let V and V ′ be vector spaces, and let T : V → V ′ be a linear transformation. Is thekernel of T a subspace of V ?

7. Let V be a vector space, and u,v1,v2, ...,vn,w ∈ V . Suppose that u ∈ Span{v1,v2, ...,vn,w},but u /∈ Span{v1,v2, ...,vn}. Show that w ∈ Span{u,v1,v2, ...,vn}.

8. Let f(t) = 2, g(t) = −2t− 1, and h(t) = 7t2 − 2t+ 7. Consider the inner product

< p(t), q(t) >= p(−1)q(−1) + p(0)q(0) + p(1)q(1)

in the vector space P2 (note that f(t), g(t), h(t) ∈ P2). Use the Gram-Schmidt process todetermine an orthogonal basis for the subspace of P2 spanned by the polynomials f(t), g(t),and h(t).

Algebra Qualifying Exam, Solutions Fall 2012

Part A.

1. Let A =

[a b2b a

]and let B =

[c d

2d c

]. First we show that ϕ preserves the two operations.

We want to show:

ϕ(A+B) = ϕ(A) + ϕ(B) and ϕ(AB) = ϕ(A)ϕ(B).

for all A,B ∈ R.We have

ϕ(A+B) = ϕ

([a+ c b+ d

2(b+ d) a+ c

])= a+ c+ (b+ d)

√2

=(a+ b

√2)

+(c+ d

√2)

= ϕ(A) + ϕ(B)

and

ϕ(AB) = ϕ

([ac+ 2bd ad+ bc2ad+ 2bc ac+ 2bd

])= ac+ 2bd+ (ad+ bc)

√2

= (a+ b√

2)(c+ d√

2)

= ϕ(A)ϕ(B).

To show ϕ is one-to-one, note that ker(ϕ) = {A ∈ R | ϕ(A) = 0}. So, if A =

[a b2b a

]is such

that ϕ(A) = 0, then a+ b√

2 = 0. Hence, a = b = 0 because√

2 /∈ Q. It follows that A = 0,for all A ∈ ker(ϕ), and thus ker(ϕ) = {0}.

Finally, we show ϕ is onto. For y = a+ b√

2 ∈ S, we have y = ϕ(A) where A =

[a b2b a

]∈ R.

Thus the image of ϕ is all of S.

2. (a) Let (a1, a2), (b1, b2) be arbitrary elements in G1×G2. Thus a1, b1 ∈ G1 and a2, b2 ∈ G2.Then

(a1, a2)(b1, b2) = (a1b1, a2b2) and (b1, b2)(a1, a2) = (b1a1, b2a2)

It follows that (a1, a2)(b1, b2) = (b1, b2)(a1, a2) if and only if (a1b1, a2b2) = (b1a1, b2a2),which is true if and only if a1b1 = b1a1 and a2b2 = b2a2.

(b) Let R be a rotation in 360/n degrees of a regular n-gon. Then, the rotation subgroup oforder n of Dn is 〈R〉. Let H denote a subgroup of order 2 (generated by any reflection).Since both subgroups are cyclic, they are also both abelian. Hence, using part (a) weget that 〈R〉 ×H is also abelian.Since |〈R〉 ×H| = 2n and 〈R〉 ∩H = {e} then 〈R〉 ×H ∼= Dn, but this contradicts thefact that Dn is not abelian. �

3. Every cyclic group of order 8 is isomorphic to Z8. The orders of elements in Z8 are

|0| = 1 |1| = 8 |2| = 4 |3| = 8 |4| = 2 |5| = 8 |6| = 4 |7| = 8

It follows that there is exactly one element in Z8 having order 2.

The non-abelian groups of order 8 are D4 and Q8. We know that D4 contains three non-trivial reflections (all of them having order 2), hence this cannot be our example. Let us lookat the orders of elements in Q8

|1| = 1 |i| = | − i| = |j| = | − j| = |k| = | − k| = 4 | − 1| = 2

So, Q8 has the desired property.

4. First we note that 〈a〉∩〈b〉 is a subgroup of the groups 〈a〉 and 〈b〉, thus (Lagrange’s Theorem)|〈a〉 ∩ 〈b〉| divides |〈a〉| = 12 and |〈b〉| = 22. So, |〈a〉 ∩ 〈b〉| divides gcd(12, 22) = 2. Since〈a〉 ∩ 〈b〉 6= {e}, then |〈a〉 ∩ 〈b〉| 6= 1, and so |〈a〉 ∩ 〈b〉| = 2. Hence, there is exactly onenon-identity element in 〈a〉 ∩ 〈b〉, call it x, and |x| = 2.

Working similarly to problem 3 above, or using that the number of elements of order 2 in acyclic group equals φ(2) = 1, we get that the only element of order 2 in 〈a〉 must be x. Since|a6| = 2 (using |a| = 12) then a6 = x. We get that b11 = x in the same way. �

5. Let a ∈ Q, and k ∈ Z \ {0}. We want to find an element x ∈ Q such that kx = a.

Just solving for x in kx = a we get x =a

k, which is a rational (recall that k 6= 0). Done.

Now we want to find a ∈ Z, and k ∈ Z \ {0} such that there is no element x ∈ Z such thatkx = a.

The idea would be to force x to be a rational that is not an integer. So, let us take a = 3

and k = 2. This forces x =3

2/∈ Z.

6. Solution 1. (a) First note that e ∈ Z(G) since e · x = x · e, for all x ∈ G. Now letm,n ∈ Z(G), and x ∈ G. Then, using associativity a few times, we get

x(mn) = (xm)n = (mx)n = m(xn) = m(nx) = (mn)x

therefore mn ∈ Z(G).Let n ∈ Z(G), and x ∈ G. Since x and n commute we have x = nxn−1 and thusn−1x = xn−1, which means that n−1 commutes with x. Therefore n−1 ∈ Z(G). Itfollows that Z(G) ≤ G.

(b) To show that Z(G) is normal in G, let m ∈ Z(G) and x ∈ G. Then xm = mx andthus xmx−1 = m ∈ Z(G).

Solution 2. For every x ∈ G consider the function φx : G→ G defined by φx(g) = xgx−1.This function is an automorphism of G, as

(i) φx(gh) = x(gh)x−1 = xg(xx−1)hx−1 =(xgx−1

) (xhx−1

)= φx(g)φx(h).

(ii) Ker(φx) = {g ∈ G; xgx−1 = e} = {g ∈ G; xg = x} = {g ∈ G; g = e} = {e}.(iii) For any given g ∈ G, φx(x−1gx) = g.

recall that Aut(G) is a group under composition. We define the function φ : G→ Aut(G)by φ(x) = φx. This is a homomorphism, as

φxy(g) = (xy)g(xy)−1 = xygy−1x−1 = x(ygy−1

)x−1 = φx

(ygy−1

)= φx (φy(g)) = (φx ◦ φy) (g)

for all g ∈ G.Finally,

Ker(φ) = {x ∈ G; φx = e}= {x ∈ G; φx(g) = g, for all g ∈ G}= {x ∈ G; xgx−1 = g, for all g ∈ G}= Z(G)

Since Ker(φ) = Z(G) and the domain of φ is G then Z(G) E G.

7. Let I = {A ∈Mn(R); Av = 0}. Clearly I ⊆Mn(R), and O ∈ I (zero matrix, that is).Let A,B ∈ I, then

(A−B)v = Av −Bv = 0− 0 = 0

so, A−B ∈ I.Now let a ∈ I and B ∈Mn(R), then

(BA)v = B(Av) = B0 = 0

and thus BA ∈ I.

8. (a) We know that char(Z4) = 4, since 4a ≡ 0 (mod 4), for all a ∈ Z4 and that there is nopositive integer k < 4 such that ka ≡ 0 (mod 4), for all a ∈ Z4.On the other hand, char(4Z) = 0 since there is no positive integer n such that n(4l) = 0for all 4l ∈ 4Z.

Now let (a, 4l) ∈ Z4 ⊕ 4Z. Then n(a, 4l) = (na, 4nl) for any positive integer n. Theobservation above implies that there is no way to get a 0 in the second component of(na, 4nl), thus there is no positive integer n such that n(a, 4l) = (0, 0) ∈ Z4⊕4Z. Hence,char(Z4 ⊕ 4Z) = 0.

(b) Let R be a commutative ring with char(R) = 2. Then 2x = 0 for all x ∈ R (note thatthis is equivalent to x = −x for all x ∈ R).

Recall that x ∈ R is an idempotent if x2 = x. Let S = {x ∈ R| x2 = x} be the set ofall idempotents in R. Clearly S 6= ∅ since 0 ∈ S (for 02 = 0).

Let x, y ∈ S (thus x2 = x and y2 = y). Then

(x− y)2 = (x− y)(x− y) = x2 − xy − yx+ y2 = x2 − 2xy + y2

(since R is commutative). But this is equivalent to (x− y)2 = x2 + y2 = x+ y = x− y,since y = −y. Hence x− y ∈ S.

Moreover, (xy)2 = x2y2 (since R is commutative), which is equivalent to (xy)2 = xy.Therefore xy ∈ S and S is a subring of R, by the Subring Test.Note that if rings were considered to always have a one, then we would also have that1 ∈ S, as 12 = 1. �

Part B.

1. (a) We want to find the standard matrix of T (that is, the matrix of T relative to thestandard basis for R2). For that, we compute T (1, 0) = (1,−3) and T (0, 1) = (1,−3),

and therefore the standard matrix of T is A =

[1 1−3 −3

].

Then det A = 0, so the matrix A is singular (NOT invertible), which implies that thelinear transformation T is NOT invertible.

(b) Recall that the range of a function T is the set of all images T (x). So, a vector w = (2, 5)is in the range of T if the equation T (x) = w has solutions. This equation is equivalentto the system

x1 + x2 = 2

−3x1 − 3x2 = 5

where we wrote x = (x1, x2), and used the formula for T . We can easily see that theabove system is inconsistent (it has no solutions), thus w = (2, 5) is NOT the range ofT .

(c)

ker T ={

(x1, x2) ∈ R2 | T (x1, x2) = (0, 0)}

={

(x1, x2) ∈ R2 | (x1 + x2,−3x1 − 3x2) = (0, 0)}

={

(x1, x2) ∈ R2 | x1 + x2 = 0 and − 3x1 − 3x2 = 0}

={

(x1, x2) ∈ R2 | x1 + x2 = 0}

= {(x1,−x1); x1 ∈ R}= Span({(1,−1)})

Therefore dim(ker T ) = 1. �

2. If x is an eigenvector of A corresponding to λ, then Ax = λx, and x 6= 0. Since P is invertibleit cannot be the zero matrix, thus Px 6= 0. Then we have

B(P−1x) = ((P−1AP )P−1)x = (P−1A)x = P−1(Ax) = P−1(λx) = λ(P−1x)

which shows that P−1x is an eigenvector of B corresponding to λ. �

3. (a) Call the columns of this matrix C1, C2, C3, and C4.We notice that 2C1 + C3 = C4, and thus Span{C1, C2, C3, C4} = Span{C1, C2, C3}.Also, since C3 /∈ Span{C1, C2} then C3 will be in the basis of the column space we willfind. The only thing left to check is whether C1 and C2 are linearly independent. Butthis is easy, as their first component is the same. It follows that we have two cases:(i) When c = 3 we get that C1 = C2 and thus {C1, C3} is a basis for the column space.(ii) When c 6= 3 we get that C1 and C2 are linearly independent, and thus a basis isgiven by {C1, C2, C3}, or any three column vectors in R3, as these three columnsgenerate the whole space R3 (column vectors).

A different solution for part (a) follows:

(a) Elimination gives 1 1 2 40 c− 3 −4 −40 0 2 2

,which gives two cases: c = 3 and c 6= 3. If c 6= 3, then c− 3 is a pivot. Hence,1 1 2 4

0 c− 3 −4 −40 0 2 2

→1 0 0 2

0 1 0 00 0 1 1

,

which implies that a basis for the column space C(A) is the first three columns of A:1

30

,1c0

,2

22

.

If c = 3, then c− 3 = 0, so1 1 2 40 c− 3 −4 −40 0 2 2

→1 1 0 2

0 0 1 10 0 0 0

,so a basis for the column space is the first and third columns of A:

130

,2

22

.

(b) Since the column space has dimension 2 or 3, depending on the value of c, we have totake cases:(i) If c 6= 3, the column space has dimension 3, and thus the nullity must be equal to 1.Solving the homogeneous system gives the basis for the null space N(A) as

−20−11

.

(ii) Here the nullity should be 2, as the dimension of the column space is 2 as well.Solving the homogeneous system gives the basis for the null space N(A) as

−1100

,−20−11

.

(c) Then again, given the two cases that naturally arise in this problem, we must considertwo cases here as well. In either case we just look for a particular solution, to laterattach to it those vectors in the null space... or, if you prefer, just solve the system thatyou get when you take the cases.What is interesting here is that the particular solution is common to both cases:

xp =

0100

Hence, the complete solutions are:

• c 6= 3

x1x2x3x4

=

0100

+ t

−20−11

for any scalar t.

• c = 3

x1x2x3x4

=

0100

+ s

−1100

+ t

−20−11

for scalars s and t.

4. We need to first diagonalize A. The characteristic equation for A, gives

0 = |A− λI| = λ2 + 3λ+ 2 = (λ+ 1)(λ+ 2),

so the eigenvalues of A are -1 and -2.

For λ = −1: Solve (A+ I)v = 0 to obtain that a basis for the eigenspace is {v1} =

{[1−1

]}.

For λ = −2: Solve (A+2I)v = 0 to obtain that a basis for the eigenspace is {v2} =

{[1−2

]}.

We can form the invertible matrix P =[v1 v2

]=

[1 1−1 −2

], and with D =

[−1 00 −2

]we

get A = PDP−1. It follows that

A12 = PD12P−1 =

[1 1−1 −2

] [(−1)12 0

0 (−2)12

] [2 1−1 −1

]=

[−4094 −40958190 8191

].

5. If A and B are similar matrices, then B = PAP−1, for some invertible matrix P . SinceA,P and P−1 are invertible matrices, and since the product of invertible matrices is againan invertible matrix, we conclude that B is invertible. Moreover, we have

B−1 = (PAP−1)−1 = (P−1)−1A−1P−1 = PA−1P−1

which implies that A−1 and B−1 are similar matrices. �

6. (a) First of all, it is clear that the zero matrix is in V . Now take two matrices B, C in V ,thus we know that AB = BA and AC = CA. Is B + C in V ?

A(B + C) = AB +AC = BA+ CA = (B + C)A

So, B + C ∈ V . Similarly, for α ∈ R we get

A(αB) = αAB = αBA = (αB)A

So, V is a subspace of M3×3(R).

(b) Note that the zero function does not belong to this set, as it is not negative (it is alwayszero). So, N is not a subspace of C[−1, 1]

(c) Let 0V and 0V ′ be the zero vectors for V and V ′, respectively. T is linear implies thatT (0V ) = 0V ′ ; hence, 0V ∈ ker(T ). Therefore, ker(T ) is non-empty.

Let u and v be elements of ker(T ). Then, by the linearity of T ,

T (u + v) = T (u) + T (v) = 0V ′ + 0V ′ = 0V ′

Thus, u + v ∈ ker(T ).

Let u be an element of ker(T ) and let α be a scalar with the associated vector spaceV . Then, again by the linearity of T ,

T (αu) = αT (u) = α0V ′ = 0V ′

Thus, αu ∈ ker(T ).

Since ker(T ) is nonempty and is closed under vector addition and scalar multiplication,ker(T ) is a subspace of V .

7. Since u ∈ Span{v1,v2, ...,vn,w}, there exist constants c1, c2, ..., cn, d such that

u = c1v1 + c2v2 + · · ·+ cnvn + dw.

Solving for dw givesdw = u− (c1v1 + c2v2 + · · ·+ cnvn) .

If d = 0, then this would imply that

u = c1v1 + c2v2 + · · ·+ cnvn,

but this is impossible because u /∈ Span{v1,v2, ...,vn}. Therefore, we can write

w = d−1u− d−1c1v1 − d−1c2v2 − · · · − d−1cnvn.

Hence, w can be written as a linear combination of u,v1,v2, ...,vn, and thus w is an elementin Span{u,v1,v2, ...,vn}.

8. We need first to find a basis for the space Span{f(t), g(t), h(t)}. Since the equation α(2) +β(−2t − 1) + γ(7t2 − 2t + 7) = 0 has only the trivial solution α = β = γ = 0, the set{f(t), g(t), h(t)} is linearly independent, and therefore forms a basis for Span{f(t), g(t), h(t)}.So, we will apply the Gram-Schmidt process to the basis B = {f(t), g(t), h(t)}.Let

p1(t) = f(t) = 2

p2(t) = g(t)− < g, p1 >

< p1, p1 >p1(t)

p3(t) = h(t)− < h, p1 >

< p1, p1 >p1(t)− < h, p2 >

< p2, p2 >p2(t)

Then we have

< g, p1 > = (1)(2) + (−1)(2) + (−3)(2) = −6

< p1, p1 > = (2)(2) + (2)(2) + (2)(2) = 12

and thus p2(t) = −2t.

Moreover, we have

< h, p1 > = (16)(2) + (7)(2) + (12)(2) = 70

< h, p2 > = (16)(2) + (7)(0) + (12)(−2) = 8

< p2, p2 > = (2)(2) + (0)(0) + (−2)(−2) = 8

implying that p3(t) = 7t2 − 14

3. The set {p1(t), p2(t), p3(t)} is an orthogonal basis for the

space Span{f(t), g(t), h(t)}.

Mathematics Department Qualifying Exam : Algebra Spring 2012


1. Prove that the set

M =

{[m n2n m

]; m,n ∈ Z

}is isomorphic, as a ring, to the ring Z[2] = {m + n

√2 |m,n ∈ Z}.

2. (a) How many elements of order 7 does the group Z21 ⊕ Z35 have? Justify your answer.(b) How many cyclic subgroups of order 7 does the group Z21 ⊕ Z35 have? Justify youranswer.

Note: You may prefer to use the notation Z21×Z35 instead of Z21⊕Z35. You are free to doso.

3. (a) Find all group homomorphisms from Z42 to Z12 (give formulas for each). How manyhomomorphisms are there from Z42 onto Z12?(b) Are all the group homomorphisms in part (a) also ring homomorphisms?

4. Let G = GL(2,R) and H = {A ∈ G ; det(A) = 3k, k ∈ Z}. Prove that H is a subgroup ofG. Moreover, show that H E G.

5. Let n ∈ N. Prove that all ideals in Zn are principal.

6. Let R be a commutative ring with one. Prove that R is a field if and only if {0} and R arethe only (two-sided) ideals in R.

7. (a) Suppose that H is a normal subgroup of G with [G : H] = 24 and |H| = 11. If x ∈ Gand x11 = e, prove that x ∈ H.(b) Let H = {(1), (12)(34)}. Prove or disprove: H is normal in A4.

8. Let G be a group. Assume that G is isomorphic to all its non-trivial subgroups. Prove thatG is isomorphic to either Z or Zp, for some prime p.



Notation: (a) Mn(K) is the set of n× n matrices with entries in K.(b) Pn is the set of all polynomials (with real coefficients) with degree at most n (including thezero polynomial).

1. Let v be a fixed vector in Rn, where n > 1. Assume that M ∈ Mn(R) \ {0} is such thatMv = 0. Prove that there exists a matrix N ∈Mn(R) such that (MN)v 6= 0.

2. Define T : M2(C)→M2(C) by T (A) = A + AT .(a) Show that T is a linear transformation.(b) Show that the range of T is the set of all B in M2(C) with the property that BT = B.(c) Describe the kernel of T .

3. Recall that the trace of a square matrix is the sum of the entries on the main diagonal. LetW be the subset of M2(R) of matrices with trace equal to 0.(a) Show that W is a subspace of M2(R).

(b) Let S =

{[1 00 −1

],

[0 10 0

],

[0 01 0

]}⊆W . Show that S is a basis for W .

4. Let V be the space C[0, 1] of real-valued continuous functions defined on [0, 1], and considerthe inner product on V defined by

〈f, g〉 =

∫ 1

0

f(t)g(t) dt.

Let W be the subspace spanned by the polynomials p1(t) = 1, p2(t) = 2t−1, and p3(t) = 12t2.Use the Gram-Schmidt process to find an orthogonal basis for W .

5. Define T : P2 → R2 by T (p) =

[p(0)p(2)

].

(a) Show that T is a linear transformation.(b) Find the kernel of T , and a basis for it.(c) Find the matrix for T relative to the standard bases for P2 and R2 respectively.

6. Let A be a square matrix. Prove or disprove the following statements:(a) The matrices A and AT have the same eigenvalues, counting multiplicities.(b) If A is an n× n diagonalizable matrix, then each vector in Rn can be written as a linearcombination of eigenvectors of A.(c) If A is diagonalizable, then the columns of A are linearly independent.

7. Let T : P1 → P1 be a linear transformation such that T (a + bt) = (−2a + b) + (a + 2b)t.Find the eigenvalues of T . Then choose one of the eigenvalues and find a basis for thecorresponding eigenspace.

8. Show that B = {t − 1, t + 1} forms a basis for P1, and find the change-of-coordinates (i.e.change of basis) matrix from the standard basis C = {1, t} to the basis B.

Algebra Qualifying Exam, Solutions Spring 2012

Part A.

1. Define f : Z[2]→M by f(m+ n√2) =

[m n2n m

]. We have

f((m+ n√2) + (r + s

√2)) = f((m+ r) + (n+ s)

√2) =

[m+ r n+ s2(n+ s) m+ r

]=

[m n2n m

]+

[r s2s r

]= f(m+ n

√2) + f(r + s

√2)

and

f((m+ n√2)(r + s

√2)) = f((mr + 2ns) + (ms+ nr)

√2) =

[mr + 2ns ms+ nr2(ms+ nr) mr + 2ns

]=

[m n2n m

] [r s2s r

]= f(m+ n

√2)f(r + s

√2).

Hence, f is a ring homomorphism.

Since[m n2n m

]=

[0 00 0

]iff m = n = 0, we have ker f = {0} and so f is one-to-one. If[

m n2n m

]∈M then m,n ∈ Z, and then m+ n

√2 ∈ Z[2]. Moreover,

f(m+ n√2) =

[m n2n m

]and thus f is onto. Hence f is an isomorphism of rings. �

2. (a) Let (a, b) ∈ Z21 ⊕ Z35. Then |(a, b)| = lcm(|a|, |b|). So, when |(a, b)| = 7 we have twocases:Case 1: |a| = 7 and |b| = 1 or 7. Since Z21 has a unique cyclic group of order 7 and anycyclic group of order 7 has six generators, there are six choices for a. Similarly, there areseven choices for b (one for |b| = 1 and six for |b| = 7). This gives 42 choices for (a, b).Case 2: |a| = 1 and |b| = 7. Since Z35 has a unique cyclic group of order 7 and any cyclicgroup of order 7 has six generators, there are six choices for b. There is only one choice fora. So, this case yields six more possibilities for (a, b).

Thus Z21 ⊕ Z35 has 42 + 6 = 48 elements of order 7.

(b) Because each cyclic group of order 7 has six elements of order 7 and no two of thecyclic subgroups can have an element of order 7 in common, there must be 48/6 = 8 cyclicsubgroups of order 7. �

3. (a) We know that a homomorphism f : Z42 → Z12 is determined by its action on a generatorof Z42. Assume that f(1) = a, for some a ∈ Z12. Then |a| divides |Z12| = 12 and |Z42| = 42(why?), so |a| divides gcd(12, 42) = 6.Case 1: If |a| = 1, then a = 0.Case 2: If |a| = 2, then a = 6.Case 3: If |a| = 3, then a = 4 or a = 8.Case 4: If |a| = 6, then a = 2 or a = 10.Notice that if f(1) = a, then f(x) = f(x · 1) = xf(1) = xa, for all x ∈ Z, because f isoperation preserving. Therefore, the homomorphisms from Z42 to Z12 are:

f1(x) = 0 f2(x) = 6x f3(x) = 4x f4(x) = 8x f5(x) = 2x f6(x) = 10x

None of the above functions are onto, since none of the possible values for a is a generatorfor Z12. Thus there are no homomorphisms from Z42 onto Z12.(b) We want to check whether any of the homomorphisms between the (additive) groups Z42

and Z12 found in (a) are ring homomorphisms. If such an f is then

a = f(1) = f(1 · 1) = f(1) · f(1) = a2

Out of the six possible a’s listed above we get

02 = 0 62 = 0 6= 6 42 = 4 82 = 4 6= 8 22 = 4 6= 2 102 = 4 6= 10

and thus the only candidates to be ring homomorphisms are f1(x) = 0 and f3(x) = 4x.f1 is clearly a homomorphism of rings, and

f3(xy) = 4(xy) = 42(xy) = (4x)(4y) = f(x)f(y)

which means that f3 is also a homomorphism of rings. �

4. Since det(A) = 3k 6= 0, for all A ∈ H then H ⊆ G. Moreover, det(I) = 1 = 30, and thus theidentity matrix lives in H.Let A,B ∈ H then det(A) = 3k and det(B) = 3t, for some k, t ∈ Z. Then

det(AB−1) = det(A) det(B−1) = det(A) det(B)−1 = 3k · 3−t = 3k−t

Since k − t ∈ Z, then AB−1 ∈ H. It follows that H ≤ G.In order to show that H E G we let A ∈ H (with det(A) = 3k) and B ∈ G (with det(B) =n 6= 0) and we want to show that BAB−1 ∈ H. Note that

det(BAB−1) = det(B) det(A) det(B−1) = det(B) det(A) det(B)−1 = n · 3k · n−1 = 3k

which implies that BAB−1 ∈ H. �

5. We know that the subgroups of (Zn,+) look like < [d] >, where d|n. But

< [d] >= {[d]k; k ∈ Z} = {[dk]; k ∈ Z} = {[dk]; [k] ∈ Zn} = {[d][k]; [k] ∈ Zn} = ([d])Zn = ([d])

where ([d]) denotes the ideal generated by [d].Note that we are using that Zn is a commutative ring with one to get that ([d])Zn = ([d]).

6. (=⇒) If R is a field and I 6= {0} is an ideal in R, then there exists a ∈ I, a 6= 0. Since R is afield, every nonzero element is a unit, and therefore 1 = a−1a ∈ I, since I is an ideal. Hencefor all r ∈ R we have r = r · 1 ∈ I, and I = R.(⇐=) R is a commutative ring with one, so to show that R is a field we only need to showthat all nonzero elements in R are units. Let 0 6= a ∈ R and consider I = (a), the principalideal generated by a. Note that, since R is a commutative ring with one, then (a) = Ra.I 6= {0} since 0 6= a ∈ I. If {0} and R are the only ideals in R, then we must have I = R.But R is a ring with one, hence 1 ∈ I. In other words, there exists r ∈ R such that 1 = ra.Hence r = a−1 (using that R is a commutative ring), and thus a is a unit. �

7. (a) First note that (xH)11 = x11H = eH = H, by hypothesis. Hence |xH| divides 11.But since G/H is a group of order 24, we must also have that |xH| divides 24. Sincegcd(11, 24) = 1, it must be the case that |xH| = 1. That is, xH = H. It follows that x ∈ H,as required.(b) As (123)H(123)−1 = (123)H(132) = {e, (14)(23)} 6= H, then H is not normal in A4. �

8. If G is isomorphic to all its non-trivial cyclic subgroups then it must be cyclic, as it wouldbe isomorphic to all its non-trivial cyclic subgroups. Hence,- If G is infinite we are done, as G ∼= Z.- If G is finite then G ∼= Zn, for some n. If n were not prime, then there would be a primeq dividing n. Consider H =< [q] >. This is a subgroup of Zn of order n/q. We get acontradiction because if G were isomorphic to H then n = n/q and that is false. So, n mustbe prime, and G is isomorphic to Zp, for some prime p. �

Part B.

1. Let v ∈ Rn and M ∈Mn(R) \ {0} be such that Mv = 0. Since M 6= 0 then there is a vectorw such that Mw 6= 0. Consider N to be any matrix mapping v to w.Why does such a matrix exist? One way to think about it is to create two matrices: A thatmaps v to, let us say, (1, 0, 0 · · · , 0) and B (invertible) mapping w to (1, 0, 0 · · · , 0). ThenN = B−1A would map v to w.It follows that (MN)v =M(Nv) =Mw 6= 0.

2. (a) We need to show that T preserves addition and scalar multiplication of matrices. LetA,B ∈M2(C) and c ∈ C. Then we have

T (A+B) = (A+B) + (A+B)T

= (A+B) +(AT +BT

)= (A+AT ) +

(B +BT

)= T (A) + T (B)

and

T (cA) = (cA) + (cA)T= c(A) + c

(AT

)= c

(A+AT

)= cT (A).

Therefore, T is a linear transformation.(b) Let B ∈ M2(C) such that BT = B (i.e. B is a symmetric matrix). Then consider

A =1

2B and observe that

A+AT =1

2B +

(1

2B

)T=

1

2B +

1

2BT =

1

2B +

1

2B = B

and thus T (A) = B.These calculations show that the range of T contains all matrices inM2(C) such that B = BT .Moreover, notice that

T (A)T =(A+AT

)T= AT +

(AT

)T= AT +A = T (A)

and thus the range of T is exactly the set of all matrices in M2(C) such that B = BT .(c) We have that

ker T = {A ∈M2(C) : T (A) = 0}= {A ∈M2(C) : A+AT = 0}

=

{[a bc d

]∈M2(C) :

[a bc d

]+

[a cb d

]=

[0 00 0

]}=

{[a bc d

]∈M2(C) :

[2a b+ cc+ b 2d

]=

[0 00 0

]}.

Thus if A =

[a bc d

]∈ kerT , then 2a = b + c = 2d = 0, or equivalently, a = d = 0 and

c = −b, for some b ∈ R. Therefore, kerT =

{[0 b−b 0

]: b ∈ C

}. �

3. (a) Let A,B ∈ W and c ∈ R. Clearly, W ⊆M2(R) and 0 ∈ W . Now using that the trace isan additive function we get

tr(A−B) = tr(A)− tr(B) = 0− 0 = 0

tr(cA) = c tr(A) = c · 0 = 0

implies that W is a subspace of M2(R)(b) Since M2(R) has dimension four and there are matrices in M2(R) \W (for instance theidentity matrix), then dim(W ) ≤ 3. Note that S contains three elements, all of them in W .Hence, if they were linearly independent then they would be forced to form a basis of W .But this is immediate, as[

0 00 0

]= x

[1 00 −1

]+ y

[0 10 0

]+ z

[0 01 0

]=

[x yz −x

]yields x = y = z = 0.

4. Let q1 = p1, and

〈p2, q1〉 =ˆ 1

0

(2t− 1)(1) dt = (t2 − t)

∣∣∣∣∣1

0

= 0

So, p2 and q1 are already orthogonal. Hence, p2 = q2. Now we compute

〈p3, q1〉 =ˆ 1

0

(12t2)(1) dt = (4t3)

∣∣∣∣∣1

0

= 4

〈q1, q1〉 =ˆ 1

0

(1)(1) dt = t

∣∣∣∣∣1

0

= 1

〈p3, q2〉 =ˆ 1

0

(12t2)(2t− 1) dt = 2

〈q2, q2〉 =ˆ 1

0

(12t2)2 dt =1

6(2t− 1)3

∣∣∣∣∣1

0

=1

3

Then, the projection of p3 onto W2 = span{q1, q2} is given by

〈p3, q1〉〈q1, q1〉

q1 +〈p3, q2〉〈q2, q2〉

q2 =4

1q1 +

2

1/3q2 = 4q1 + 6q2

Hence,q3 = p3 − (4q1 + 6q2)

and thus q3(t) = 12t2 − 4− 6(2t− 1) = 12t2 − 12t+2. It follows that the orthogonal basis is{q1, q2, q3}.

5. (a) Let p,q ∈ P2 and c ∈ R. Then

T (p + q) =

[(p + q)(0)(p + q)(2)

]=

[p(0)p(2)

]+

[q(0)q(2)

]= T (p) + T (q)

T (cp) =[

(cp)(0)(cp)(2)

]=

[c(p(0))c(p(2))

]= c

[p(0)p(2)

]= cT (p).

Hence, T is a linear transformation.(b) Using the definition of the kernel of a linear transformation, we have:

Ker(T ) =

{p ∈ P2 | T (p) =

[p(0)p(2)

]=

[00

]}=

{a+ bt+ ct2 ∈ P2 |

[a

a+ 2b+ 4c

]=

[00

]}=

{a+ bt+ ct2 ∈ P2 | a = 0, a+ 2b+ 4c = 0

}

If a = 0, the equation a+2b+4c =0 is equivalent to b+2c = 0, or to b = −2c, where c is free.Therefore, we arrive at:

Ker(T ) = {(−2c)t+ ct2 | c ∈ R} = Span{−2t+ t2}

which implies that {−2t+ t2} forms a basis for Ker(T ).(c) The standard basis for P2 is {1, t, t2}, thus we need to compute the images under T ofthe polynomials 1, t and t2.

T (1) =

[11

]T (t) =

[02

]T (t2) =

[04

]Therefore the matrix for T relative to the basis {1, t, t2} for P2 and the standard basis for

R2 is[

1 0 01 2 4

]. �

6. (a) True. Matrices A and AT have the same characteristic polynomial, because

det(AT − λI) = det(AT − (λI)T ) = det(A− λI)T = det(A− λI)

and thus the same eigenvalues, counting multiplicities.(b) True. If A is an n × n diagonalizable matrix, then A has n linearly independent eigen-vectors v1, . . . ,vn in Rn. Since dim(Rn) = n, the set {v1, . . . ,vn} forms a basis for Rn, andtherefore each vector in Rn can be written as a linear combination of v1, . . . ,vn.(c) False. If A is a diagonal matrix with (at least one) 0 on the diagonal then the columnsof A are not linearly independent (since a set of vectors containing the zero vector is linearlydependent). �

7. We find first [T ]B , the matrix of T relative to the standard basis B = {1, t} for P1. SinceT (1) = −2 + t and T (t) = 1 + 2t, the B-matrix of T has the following form:

[T ]B =

[−2 11 2

]The eigenvalues of T are exactly the eigenvalues of the matrix A = [T ]B . The characteristicpolynomial of A is χA = λ2 − 5, thus the eigenvalues of A (and hence the eigenvalues of T )are λ = ±

√5.

For λ =√5: A−

√5I =

[−2−

√5 1

1 2−√5

]and the equation (A−

√5I)x = 0 amounts

to(−2−

√5)x1 + x2 = 0 x1 + (2−

√5)x2 = 0

Observe that these two equations are equivalent.So, x2 = (2 +

√5)x1, and x1 free. The general solution to the equation (A −

√5I)x = 0 is

x = x1

[1

2 +√5

], with x1 ∈ R.

This tells us that the eigenspace Vλ=√5 is 1-dimensional, and that if {p(t)} is a basis for

Vλ=√5, then the B-coordinate vector of p(t) is [p(t)]B =

[1

2 +√5

].

Therefore, {p(t) = 1 + (2 +√5)t} is a basis for the eigenspace Vλ=√5 corresponding to the

eigenvalue λ =√5.

The computations for λ = −√5 are similar. �

8. We will denote the change-of-coordinates matrix from the standard basis C to the basis Bby PB←C .

The set B contains 2 vectors, and dim P1 = 2, thus to show that B forms a basis for P1 itsuffices to show that it is a linearly independent set.If a(t− 1) + b(t+ 1) = 0 = 0t+ 1, for some scalars a and b, then a+ b = 0 and −a+ b = 0.Then a = 0 = b, implying that B is linearly independent and, therefore, a basis for P1.Recall that the change-of-coordinates matrix from the basis C = {1, t} to the basis B ={t− 1, t+ 1} is given by PB←C = [ [1]B [t]B ], so we need to find the B-coordinate vectors ofthe polynomials/vectors contained in the basis C.

Since 1 = − 12 (t− 1) + 1

2 (t+ 1) we have [1]B =

[−1/21/2

].

Similarly, since t = 12 (t− 1) + 1

2 (t+ 1), it implies that [t]B =

[1/21/2

].

Then PB←C = [ [1]B [t]B ] =

[−1/2 1/21/2 1/2

]. �

Mathematics Department Qualifying Exam : Algebra Fall 2011


1. Let G and H be groups.

(a) Prove thatS(G,H) = {f : G→ H; f is a function}

is a group with the operation ∗ defined by

(f ∗ g)(x) = f(x) ? g(x)

for all x ∈ G, where ? is the operation in H.

(b) Prove that S(G,H) is abelian if and only if H is abelian.

2. (a) Can there be a group homomorphism from Z6 ⊕ Z6 onto Z12?

(b) Can there be a group homomorphism from Z81 onto Z3 ⊕ Z3?

3. (a) Let a, b and c be elements of a commutative ring with unity, and suppose that a is aunit. Prove that b divides c in this ring if and only if ab divides c in this ring.

(b) Let a belong to a ring R. Let S = {x ∈ R | ax = 0}. Show that S is a subring of R.

4. Let G,H be groups and φ : G→ H be a homomorphism.

(a) Prove that φ(gi) = φ(g)i, for all i ∈ Z.

(b) Show that if φ is an isomorphism and G is cyclic, then H is cyclic.

5. Prove the following claims for permutations in Sn.

(a) (a1a2 · · · ak) = (a1ak) · · · (a1a3)(a1a2) = (a1a2)(a2a3) · · · (ak−1ak), where a1, a2, . . . , akare distinct.Conclude that every permutation is the product of transpositions.

(b) Any permutation can be written as a product of the transpositions (12), (13), · · · , (1n).

6. Let S = R\{−1}. Define ∗ on S by a ∗ b = a+ b+ ab. Show that S is a group.

7. Show, in full detail, that I = {p(x) ∈ R[x]; p(0) = p(1) = 0} is a principal ideal of R[x].

8. Let d be a positive integer. Prove that Q[√d] = {a+ b

√d | a, b ∈ Q} is a field.



1. Let P2 be the vector space of real polynomials of degree 2 or less. Consider the basis B ={1, x, x2} and B′ = {1, x+ 1, x2 + x+ 1} for P2.

(a) Find the transition matrix from B to B′.

(b) Let p(x) = 3− x+ 2x2 ∈ P2. Find the coordinates of p(x) relative to B′.

2. Find an orthonormal basis for the subspace of R4 spanned by the vectors u1 = (0, 0,−1, 0),u2 = (1, 3, 0, 0), and u3 = (1, 0, 1, 1).

3. Find a square root of the matrix

A =

1 3 −30 4 50 0 9

.

4. Let V be an n-dimensional vector space and W a subspace of V . Let B′ = {b1, b2, . . . , bm}be a basis for W . Prove that there exist vectors bm+1, bm+1, . . . , bn in V such that B ={b1, b2, . . . , bm, bm+1, . . . , bn} is a basis for V .

5. Gaussian elimination changes Ax = b to a reduced Rx = d. The complete solution is

x =

400

+ s

210

+ t

501

.(a) What is the reduced row echelon form matrix R?

(b) What is d?

6. Consider the linear transformation with matrix

A =

1 0 −3 00 1 4 00 0 0 1

.Find a basis for the kernel and a basis for the image of the transformation.

7. Let A be a nonzero n× n skew-symmetrix real matrix.

(a) Show that if n is odd, then A is not invertible.

(b) What happens if n is even? Justify your answer.

8. (a) Show that if v is an eigenvector of the matrix product AB and Bv 6= 0, then Bv is aneigenvector of BA.

(b) Suppose that v is an eigenvector of A corresponding to an eigenvalue λ. Show that vis an eigenvector of 5I − 3A+A2, where I is the identity matrix of the same size as A.What is the corresponding eigenvalue?



1. Let G and H be groups.

(a) Prove thatS(G,H) = {f : G→ H; f is a function}

is a group with the operation ∗ defined by

(f ∗ g)(x) = f(x) ? g(x)

for all x ∈ G, where ? is the operation in H.

(b) Prove that S(G,H) is abelian if and only if H is abelian.

Solution.

(a) Since this operation is weird, then we need to check everything about this group (tobe).(i) Closure of ∗ in S(G,H): We take f, g ∈ S(G,H), we want to show that f ∗ g is anelement of S(G,H).First of all, note that since f, g ∈ S(G,H) and x ∈ G then both f(x) and g(x) are inH. Moreover ? being closed in H forces f(x) ? g(x) ∈ H. Hence, (f ∗ g)(x) ∈ H for allx ∈ G. So, f ∗ g takes elements in G to elements in H, but is it a function? We needto check that images are unique! But this is immediate, as both f and g are functions,and thus f(x) and g(x) are uniquely defined... and also is their product. It follows thatS(G,H) is closed under ∗.(ii) Associativity: Let f, g, h ∈ S(G,H), then for any x ∈ G we get

[f ∗ (g ∗ h)](x) = f(x) ? (g ∗ h)(x) = f(x) ? [g(x) ? h(x)]

but now using that ? is associative in H we get

f(x) ? [g(x) ? h(x)] = [f(x) ? g(x)] ? h(x)

which is [(f ∗ g) ∗ h](x). Done.(iii) Identity? Let f ∈ S(G,H) and let e be the constant function e(x) = eH , where eHis the identity of H. Note that

(f ∗ e)(x) = f(x) ? e(x) = f(x) ? eH = f(x)

for al x ∈ G, and thus f ∗e = f . Similarly, e∗f = f . Hence, e is the identity of S(G,H).(iv) Inverses? Let f ∈ S(G,H). We want to find g ∈ S(G,H) such that f ∗ g = e.Consider g(x) = f(x)−1, where the inverse is taken in H. Then,

(f ∗ g)(x) = f(x) ? g(x) = f(x) ? f(x)−1 = eH

and thus f ∗ g = e. Done.

(b) This group is Abelian if H is Abelian, as

(f ∗ g)(x) = f(x) ? g(x) = g(x) ? f(x) = (g ∗ f)(x)

for all x ∈ G.If H is non-Abelian, then there are elements a, b ∈ H such that a ? b 6= b ? a. Then, wedefine the constant functions f(x) = a and g(x) = b, which are clearly in S(G,H) anddo not commute. Hence, S(G,H) is Abelian if and only if H is Abelian.

2. (a) Can there be a group homomorphism from Z6 ⊕ Z6 onto Z12?

(b) Can there be a group homomorphism from Z81 onto Z3 ⊕ Z3?

Solution.

(a) No. The largest possible order of an element in Z6⊕Z6 is 6. But Z12 has an element oforder 12 (any generator would do). Since the order of ψ(x) divides the order of x, forall x ∈ Z6 ⊕ Z6 then the largest order of an image under ψ is 6. Hence, the elements oforder 12 is Z12 are not in the range of ψ.

(b) No. The homomorphic image of a cyclic group is cyclic. Having an onto homomorphismfrom Z81 → Z3 ⊕ Z3 would imply Z3 ⊕ Z3 is cyclic. A contradiction.

3. (a) Let a, b and c be elements of a commutative ring with unity, and suppose that a is aunit. Prove that b divides c in this ring if and only if ab divides c in this ring.

(b) Let a belong to a ring R. Let S = {x ∈ R | ax = 0}. Show that S is a subring of R.

Solution.

(a) Let R be a commutative ring with unity 1, and let a be a unit in R.

Suppose that b|c for some b, c ∈ R; that is, c = br for some r ∈ R. Then c = (a−1a)(br) =(ab)(a−1r) (we used here that R is commutative, that a−1 exists in R and that multi-plication is associative is a ring). But R is closed under multiplication, thus a−1r ∈ R,and then ab|c, as desired.

Conversely, suppose that ab|c for some b, c ∈ R; that is c = (ab)s for some s ∈ R. SinceR is commutative, it implies that c = b(as), thus b|c (since as ∈ R).

(b) First note that 0 ∈ S (since a0 = 0), thus S 6= ∅.Let x, y ∈ S. Then ax = 0 = ay, and we have:

a(x− y) = ax− ay = 0− 0 = 0 and a(xy) = (ax)y = 0y = 0

thus x− y ∈ S and xy ∈ S. Then by the Subring Test, S is a subring of R.

4. Let G,H be groups and φ : G→ H be a homomorphism.

(a) Prove that φ(gi) = φ(g)i, for all i ∈ Z.

(b) Show that if φ is an isomorphism and G is cyclic, then H is cyclic.

Solution.

(a) For i positive we get that φ(gi) = φ(gi−1g) = φ(gi−1)φ(g). So, if we knew that φ(gi−1) =φ(g)i−1 then we would be done, as

φ(gi) = φ(gi−1)φ(g) = φ(g)i−1φ(g) = φ(g)i

Note that we have just set an induction argument. That is, proving the i-case byassuming the (i − 1)-case. Since the case i = 2 holds trivially, then the induction isdone.For i negative, let us first look at φ(g−1) = φ(g)−1. Then putting this case togetherwith the positive i case above will solve the case for all negative i’s.Recall that φ(e) = e. Then, e = φ(e) = φ(gg−1) = φ(g)φ(g−1), which means thatφ(g−1) is the inverse of φ(g). Done.Now let us consider −i for i positive. We need to show that φ(g−i) = φ(g)−i:

φ(g−i) = φ((gi)−1) = φ(gi)−1 = (φ(g)i)−1 = φ(g)−i

The case i = 0 is trivial.

(b) Assume that G =< g >. We know, from part (a) that φ(gi) = φ(g)i. Since everyelement in the domain of φ looks like gi, for some i ∈ Z, then every element in theimage of G must look like φ(g)i. But since φ is onto then every element in H looks likeφ(g)i. It follows that φ(g) generates H.

5. Prove the following claims for permutations in Sn.

(a) (a1a2 · · · ak) = (a1ak) · · · (a1a3)(a1a2) = (a1a2)(a2a3) · · · (ak−1ak).Conclude that every permutation is the product of transpositions.

(b) Any permutation can be written as a product of the transpositions (12), (13), · · · , (1n).

Solution.

(a) Consider the functions σ = (a1a2 · · · ak) and τ = (a1ak) · · · (a1a3)(a1a2). We want toshow σ = τ , so we need to show σ(x) = τ(x), for all x ∈ {1, 2, 3, · · · , n}.If x = ai for some i, then σ(ai) = ai+1 mod k. On the other hand, we have three cases:(i) If x = a1 then τ(a1) = a2, as a2 only appears in the far-most right transpositiononly. This is consistent with having σ(a1) = a2.(ii) If x = ak then τ(ak) = a1, as ak only appears in the far-most left transposition only.This is consistent with having σ(ak) = a1.(iii) If x 6= a1, ak then ai appears exactly once in the transpositions of τ . Hence, aiwill be fixed by whatever is to the right of (a1ak) · · · (a1ai+1)(a1ai) (in the product oftranspositions of τ). Now it is easy to see that ai will be first moved to a1 and then toai+1. Since ai+1 does not appear anymore in this product then τ(ai) = ai+1, which isconsistent wit what σ does.If x 6= ai, for all i then both σ and τ fix this element.Now consider the functions σ = (a1a2 · · · ak) and τ = (a1a2)(a2a3) · · · (ak−1ak). Wewant to show σ = τ , so we need to show σ(x) = τ(x), for all x ∈ {1, 2, 3, · · · , n}.If x = ai for some i, then σ(ai) = ai+1 mod k. On the other hand, we notice that everyai 6= a1, ak appears in exactly two cycles in τ . Hence, we need to take cases again:(I) If x = a1 then τ(a1) = a2, as only the cycle (a1a2) affects a1. We are done becauseσ(a1) = a2.(II) If x = ak, then the first transposition on the right sends ak to ak−1, but then thenext one sends ak−1 to ak−2, and so on, until reaching a1. It follows that τ(ak) = a1,which is exactly what σ does to ak.(III) If x 6= a1, ak then the first cycle (on the right) that moves ai is (aiai+1), whichsends ai to ai+1. At this point there are no ai+1’s on the cycles to the left of (aiai+1),and thus τ(ai) = ai+1. Just like σ.Since every permutation is a product of cycles, and each cycle is a product of transpo-sitions, then it follows that each permutation is a product of transpositions.

(b) If we can get all transpositions out of products of (12), (13), · · · , (1n) then, by part (a),we would be done.We already have all transposition of the form (1b), so let us consider (ab), where a andb are both different from 1 and, of course a 6= b. But that is easy, as

(1a)(1b)(1a) = (ab)

6. Let S = R\{−1}. Define ∗ on S by a ∗ b = a+ b+ ab. Show that S is a group.

Solution. See problem 6 in Spring ’08 exam and problem 1 in spring ’11 exam.

7. Show, in full detail, that I = {p(x) ∈ R[x]; p(0) = p(1) = 0} is a principal ideal of R[x].

Solution. Let p(x) ∈ R[x]. The factor theorem says that if p(α) = 0, for some α ∈ R thenp(x) = (x− α)q(x), for some q(x) ∈ R[x]. Hence, using this theorem twice we can re-write Ias

I = {p(x) ∈ R[x]; p(x) = x(x− 1)r(x), for some r(x) ∈ R[x]}

which is the set of all multiples (in R[x]) of the polynomial t(x) = x(x− 1). It follows that Iis the principal ideal of R[x] generated by t(x) = x(x−1), which is most of the times denotedt(x)R[x].

8. Let d be a positive integer. Prove that Q[√d] = {a+ b

√d | a, b ∈ Q} is a field.

Solution. If√d ∈ Q, then Q[

√d] ⊆ Q (by closure of Q). On the other hand, if considering

the elements of Q[√d] with b = 0 we get Q ⊆ Q[

√d]. It follows that, in this case, Q = Q[

√d].

For√d /∈ Q. It suffices to show that Q[

√d] is a subfield of R. First we observe that

0 = 0 + 0 · d ∈ Q[√d], so Q[

√d] 6= ∅.

Let a1 + b1√d, a2 + b2

√d ∈ Q[

√d], thus a1, a2, b1, b2 ∈ Q. Then

(a1 + b1√d)− (a2 + b2

√d) = (a1 − a2) + (b1 − b2)

√d ∈ Q[

√d].

Assume that a2 + b2√d 6= 0, i.e., a2 6= 0 and b2 6= 0. Then we observe that

(a2 + b2√d)−1 =

1

a2 + b2√d

=a2 − b2

√d

a22 − db22=

a2a22 − db22

+−b2

a22 − db22

√d ∈ Q[

√d]

since a22 − db22 6= 0 (here using that d is not a square) anda2

a22 − db22,−b2

a22 − db22∈ Q.

Then we have

(a1 + b1√d)(a2 + b2

√d)−1 = (a1 + b1

√d)

(a2

a22 − db22− b2a22 − db22

√d

)=

a1a2 − b1b2da22 − db22

+−a1b2 + b1a2a22 − db22

√d ∈ Q[

√d].

Then, by the subfield test, Q[√d] is a subfield of R. Done.



1. Let P2 be the vector space of real polynomials of degree 2 or less. Consider the basis B ={1, x, x2} and B′ = {1, x+ 1, x2 + x+ 1} for P2.

(a) Find the transition matrix from B to B′.

(b) Let p(x) = 3− x+ 2x2 ∈ P2. Find the coordinates of p(x) relative to B′.

Solution.

(a) To find the first column vector of the transition matrix, we must find scalars a1, a2 anda3 such that

a1(1) + a2(x+ 1) + a3(x2 + x+ 1) = 1.

By inspection we see that the solution is a1 = 1, a2 = 0, and a3 = 0. Therefore,

[1]B′ =

100

.The second and third column vectors of the transition matrix can be found by solvingthe equations

b1(1) + b2(x+ 1) + b3(x2 + x+ 1) = x

andc1(1) + c2(x+ 1) + c3(x2 + x+ 1) = x2,

respectively. The solutions are given by b1 = −1, b2 = 1, and b3 = 0, and c1 = 0, c2 =−1, and c3 = 1. Hence, the transition matrix is

PB′←B =

1 −1 00 1 −10 0 1

.Remark: This can also be done by finding the inverse of PB←B′ =

1 1 10 1 10 0 1

,

which is obtained by ‘hanging’ the coordinates of the vectors {1, 1 + x, 1 + x + x2} inthe standard basis.

(b) The coordinate vector of p(x) = 3− x+ 2x2 relative to B is given by

[p(x)]B =

3−1

2

,and therefore

[p(x)]B′ = PB′←B [p(x)]B =

1 −1 00 1 −10 0 1

3−1

2

=

4−3

2

is the coordinate vector of p(x) = 3− x+ 2x2 relative to B′.

Notice also that 3− x+ 2x2 = 4(1)− 3(x+ 1) + 2(x2 + x+ 1).

2. Find an orthonormal basis for the subspace of R4 spanned by the vectors u1 = (0, 0,−1, 0),u2 = (1, 3, 0, 0), and u3 = (1, 0, 1, 1).

Solution. Using the Gram-Schmidt process, we have that v1,v2,v3 is an orthogonal basis,where v1 = u1 = (0, 0,−1, 0), v2 = u2 = (1, 3, 0, 0) (since u1 ·u2 = 0), and v3 = (1, 0, 1, 1)−−11 (0, 0,−1, 0)− 1

10 (1, 3, 0, 0) = ( 910 ,−

310 , 0, 1).

We have |v1| = 1, |v2| =√

10, and |v3| =√19010 . Therefore an orthonormal basis is

{(0, 0,−1, 0), 1√10

(1, 3, 0, 0), 10√190

( 910 ,−

310 , 0, 1)}.

3. Find a square root of the matrix

A =

1 3 −30 4 50 0 9

.Solution. Since A is upper triangular, its eigenvalues are its diagonal entries, and A can thus

be diagonalized (since it has distinct eigenvalues). Thus, we will have S−1AS =

1 0 00 4 00 0 9

,

where S is a matrix whose columns are eigenvectors of A for the respective eigenvalues.

The matrix B = S

1 0 00 2 00 0 3

S−1 will then be the square root of A. Carrying out the

computations, one obtains

S =

1 1 10 1 10 0 1

, S−1 =

1 −1 00 1 −10 0 1

, giving B =

1 1 −10 2 10 0 3

.Remark: Another approach to this problem would be to assume that a square root of A,call it B, is upper triangular (not so crazy to assume, as these matrices form a subring ofM3(C). Since det(A) = 36 then det(B) = 6. It follows that

B =

a b c0 d e0 0 6/(ad)

Easy computations show that a = 1 and d = 2. After that, all other values follow easily.

4. Let V be an n-dimensional vector space and W a subspace of V . Let B′ = {b1, b2, . . . , bm}be a basis for W . Prove that there exist vectors bm+1, bm+1, . . . , bn in V such that B ={b1, b2, . . . , bm, bm+1, . . . , bn} is a basis for V .

Solution. Since V is n-dimensional, there exist a linearly independent set {v1, v2, . . . , vn}that spans V . We wish to add some of the v1, v2, . . . , vn to B′ to form a basis of V . Ifv1 ∈ span(B′), then let S = B′; otherwise, set S = {b1, b2, . . . , bm, v1}. Do this for each ofthe vectors v2, v3, . . . , vn (i.e., if vk ∈ S, then leave S unchanged; otherwise, add vk to S).After each step, the set S is still linearly independent, since vk was only added to the setif vk was not in the span of the previous set of vectors. After n steps, vk ∈ span(S) for allk = 1, 2, . . . , n. Since {v1, v2, . . . , vn} spanned V , S spans V , and thus, S is a basis for V .So, letting bm, bm+1, . . . , bn be equal to the vectors from {v1, v2, . . . , vn} that were added toS and defining B = S, we are done.

5. Gaussian elimination changes Ax = b to a reduced Rx = d. The complete solution is

x =

400

+ s

210

+ t

501

.(a) What is the reduced row echelon form matrix R?

(b) What is d?

Solution. Since R is in reduced row echelon form, we must have

d =

400

.

The other two vectors provide solutions to the homogeneous system, and show that R hasrank 1. Since R is in reduced row echelon form, the bottom two rows must be all 0, and thetop row must be orthogonal to the vectors2

10

501

A few computations show that this vector could be [1− 2− 5]T . Hence,

R =

1 −2 −50 0 00 0 0

.

6. Consider the linear transformation with matrix

A =

1 0 −3 00 1 4 00 0 0 1

.Find a basis for the kernel and a basis for the image of the transformation.

Solution. Set Ax = 0. Then the solution space is

3z−4zz0

∣∣∣∣z ∈ R

. Therefore it is of

dimension 1, and a basis is

3−4

10

.

Since A is 3 × 4, the image is a subspace of R3. Moreover, the dimension of the image is 3since the dimension of the kernel is 1. A 3-dimensional subspace of R3 must be R3 itself, sowe may use the standard basis

{[100]T , [010]T , [001]T

}.

7. Let A be a nonzero n× n skew-symmetrix matrix.

(a) Show that if n is odd, then A is not invertible.

(b) What happens if n is even? Justify your answer.

Solution.

(a) Since the matrix −A is obtained from A by multiplying each of the rows of A by -1,and there are an odd number of rows, det(−A) = (−1)n det(A) = − det(A). Since A isskew-symmetrix, det(A) = det(AT ) = det(−A) = −det(A). Thus, det(A) = 0, and Ais not invertible.

(b) If n is even, then it could go either way. For example, the matrix

A =

[0 1−1 0

]is a 2× 2 skew-symmetric matrix that is invertible. However, the matrix

A =

0 1 0 1−1 0 0 00 0 0 0−1 0 0 0

is skew-symmetrix, but is clearly not invertible.

8. (a) Show that if v is an eigenvector of the matrix product AB and Bv 6= 0, then Bv is aneigenvector of BA.

(b) Suppose that v is an eigenvector of A corresponding to an eigenvalue λ. Show that vis an eigenvector of 5I − 3A+A2, where I is the identity matrix of the same size as A.What is the corresponding eigenvalue?

Solution.

(a) Suppose that Bv 6= 0 and ABv = λv for some λ ∈ R. Then A(Bv) = λv. Left-multiplyeach side by B, and obtain BA(Bv) = B(λv) = λ(Bv). This equation says that Bv isan eigenvector of BA, because Bv 6= 0.

(b) Suppose that Av = λv, with v 6= 0. Then

(5I − 3A+A2)v = 5v− 3Av +A(Av)

= 5v− 3λv + λ2v

= (5− 3λ+ λ2)v.

Hence v is an eigenvector of 5I − 3A+A2 with eigenvalue 5− 3λ+ λ2.



1. Let G = {x ∈ R | x 6= −1}. For x, y ∈ G let x ∗ y = x + y + xy. Prove that ∗ is a binaryoperation on G and that (G, ∗) is a group.

2. Consider the group Sn of permutations on n elements (n ≥ 3), and let An denote the set ofeven permutations in Sn. Prove that An is a normal subgroup of Sn.

3. Prove that there is no ring homomorphism from Z4 ⊕ Z12 onto Z8 ⊕ Z6.

4. (a) Let G be a group, and let a ∈ G. Prove that the function φa : G → G, definedby φa(x) = axa−1 for all x ∈ G is an automorphism of G (it is called the innerautomorphism of G induced by a).

(b) Let G be a group and Inn(G) = {φa | a ∈ G} be the set of all inner automorphisms ofG. Prove that Inn(G) is a group under the operation of function composition.

5. (a) Let R be a commutative ring with unity 1 and let I be an ideal of R. Prove that r+I isa unit (invertible) in R/I if and only if there is an element s in R such that rs− 1 ∈ I.

(b) Show that the ring Z⊕ Z has infinitely many zero-divisors.

(c) Find all units in the ring Z⊕ Z.

6. Let R be a ring. Recall that a ∈ R is called a nilpotent element if an = 0 for some n ∈ Z+

and it is called an idempotent element if a2 = a.

(a) Show that if a is an idempotent, then 1− a is also an idempotent.

(b) If f : R → S is a ring homomorphism and a ∈ R is nilpotent, prove that f carries theelement a to a nilpotent element in the ring S.

(c) If R is an integral domain, prove that the only idempotents in R are 0 and 1.

(d) Show that 0 6= a ∈ R is a zero-divisor if and only if aba = 0 for some b 6= 0.

7. Let G be a group and let Z(G) = {g ∈ G | gx = xg, ∀x ∈ G}.

(a) Show that if G/Z(G) is cyclic then G is Abelian.

(b) Show that if G is non-Abelian with |G| = p3, where p is a prime, and Z(G) 6= {e}, then|Z(G)| = p.

8. Suppose G is a cyclic group with at least 3 elements. Without using the Euler φ function,prove that G has an even number of generators.



1. Let

W = span

1321

,

1211

,

1221

.

Find an orthonormal basis for W .

2. Suppose that A is a square matrix of size n and B = {x1,x2,x3, ...,xn} is a basis of Rn.Show that if A is nonsingular, then C = {Ax1, Ax2, Ax3, ..., Axn} is a basis of Rn.

3. Find a basis for both the kernel and the range of the linear transformation represented by

A =

1 3 11 2 00 1 1

.4. Given the matrix

A =

[3 −51 −3

],

find A2011.

5. Find the eigenvalues of the matrix 1 2 3 4 51 2 3 4 51 2 3 4 51 2 3 4 51 2 3 4 5

6. Find all values of λ so that the following matrix is nonsingular: −2 λ 3

1 2 λ1 11 18

.7. Given the 2× 3 matrix

A =

[1 2 32 3 4

],

(a) For which vectors b =

[b1b2

]does Ax = b have at least one solution?

(b) Solve Ax =

[35

].

(c) What is the rank of A?

8. Consider the function T : P2 → R3 defined by T (p(x)) =

p(−1)p(0)p(1)

(a) Prove that T is an invertible linear transformation.

(b) Find T−1

abc



1. Let G = {x ∈ R | x 6= −1}. For x, y ∈ G let x ∗ y = x + y + xy. Prove that ∗ is a binaryoperation on G and that (G, ∗) is a group.Solution. Let x, y ∈ G. Then x ∗ y = x + y + xy ∈ R. Suppose x ∗ y = −1. Thenx+ xy = −1− y, which implies x(y+1) = −(y+1). Since y 6= −1, we have y+1 6= 0; hencex = −1, a contradiction. Therefore x ∗ y ∈ G, and ∗ is a binary operation on G.

It remains to show that (G, ∗) is a group. Let x, y, z ∈ G. Then

(x ∗ y) ∗ z = (x+ y + xy) + z + (x+ y + xy)z

= x+ y + z + xy + yz + xz + xyz

= x+ (y + z + yz) + x(y + z + yz)

= x ∗ (y ∗ z).

Hence ∗ is associative.

We claim that 0 is an identity element for (G, ∗). For x ∈ G we have

x ∗ 0 = x+ 0 + x · 0 = x = 0 + x+ 0 · x = 0 ∗ x.

Finally, for x ∈ G we claim that x−1 = −xx+1 and that such x−1 belongs to G. We have

x ∗ −xx+ 1

= x+−xx+ 1

+−x2

x+ 1=x(x+ 1)− x− x2

x+ 1= 0

and similarly−xx+ 1

∗ x = 0.

Clearly x−1 belongs to R. Moreover x−1 6= −1, else −x = −(x + 1) which is impossible.Thus x−1 ∈ G.

2. Consider the group Sn of permutations on n elements (n ≥ 3), and let An denote the set ofeven permutations in Sn. Prove that An is a normal subgroup of Sn.Solution. First of all e ∈ An, and thus An 6= ∅. Now consider σ = (a1 b1) . . . (ak bk) andτ = (c1 d1) . . . (cl dl) where k and l are even (meaning that σ, τ ∈ An). Then στ−1 =(a1 b1) . . . (ak bk)(cl dl) . . . (c1 d1) is also even and hence belongs to An. Thus An is asubgroup of Sn.

Now let σ ∈ An as above, and suppose α ∈ Sn. Write α as a product of m transpositions.Then ασα−1 can be written as a product of m + k +m = 2m + k transpositions, which iseven since k is even. Thus ασα−1 ∈ An, and An is normal in Sn.

3. Prove that there is no ring homomorphism from Z4 ⊕ Z12 onto Z8 ⊕ Z6.Proof. By contradiction, suppose that there is an onto homomorphism ψ : Z4⊕Z12 → Z8⊕Z6.By the First Isomorphism Theorem, (Z4 ⊕ Z12)/ kerψ ≈ Z8 ⊕ Z6. Since 4 · 12 = 8 · 6 = 48,it follows that | kerψ| = 1. That is, ψ is one-to-one.But then ψ must be an isomorphism. Note, however, that Z4 ⊕Z12 cannot be isomorphic toZ8⊕Z6 since Z8⊕Z6 has an element of order 24 but Z4⊕Z12 does not. Hence there can beno homomorphism from Z4 ⊕ Z12 onto Z8 ⊕ Z6. �

4. (a) Let G be a group, and let a ∈ G. Prove that the function φa : G → G, definedby φa(x) = axa−1 for all x ∈ G is an automorphism of G (it is called the innerautomorphism of G induced by a).

(b) Let G be a group and Inn(G) = {φa | a ∈ G} be the set of all inner automorphisms ofG. Prove that Inn(G) is a group under the operation of function composition.

Solution.

(a) Let x, y ∈ G. Then φa(xy) = a(xy)a−1 = (ax)(a−1a)(ya−1) = (axa−1)(aya−1) =φa(x)φa(y). Thus φa is a group homomorphism. Now assume that φa(x) = φa(y), thatis, axa−1 = aya−1. Then by left and right cancellation, x = y, which implies thatφa is one-to-one. Moreover, if z ∈ G then x = a−1za ∈ G (since G is closed underinverses and group operation) and φa(x) = a(a−1za)a−1 = z, which shows that φa isonto. Therefore φa is an automorphism of G.

(b) Let φa, φb ∈ Inn(G) and let x ∈ G. Then (φa ◦ φb)(x) = φa(bxb−1) = a(bxb−1)a−1 =

(ab)x(ab)−1. Thus φa ◦ φb = φab ∈ Inn(G), and Inn(G) is closed.It is easy to see that φe is the identity element for Inn(G) (where e is the identityelement of G) since φa ◦ φe = φae = φa = φea = φe ◦ φa.To this end, we note that since φa is bijective it has an inverse φ−1

a . Moreover, φ−1a (x) =

a−1xa, for all x ∈ G [since (φa ◦ φ−1a )(x) = x = (φ−1

a ◦ φa)(x)]. Furthermore, φ−1a (x) =

a−1xa = a−1x(a−1)−1. Thus φ−1a = φa−1 ∈ Inn(G), and φa−1 ◦ φa = φe = φa ◦ φa−1 ,

which implies that the inverse of φa in Inn(G) is φa−1 . Hence Inn(G) is a group. �

5. (a) Let R be a commutative ring with unity 1 and let I be an ideal of R. Prove that r+I isa unit (invertible) in R/I if and only if there is an element s in R such that rs− 1 ∈ I.

(b) Show that the ring Z⊕ Z has infinitely many zero-divisors.

(c) Find all units in the ring Z⊕ Z.Solution.

(a) Since R is a ring with unity 1, we know that 1+ I is the unity in the quotient ring R/I.Then an arbitrary element r + I ∈ R/I is a unit if and only if there exists an elements+I ∈ R/I (thus s ∈ R) such that (r+I)(s+I) = rs+I = 1+I. But this is equivalentto rs ∈ (1 + I), which is equivalent to rs− 1 ∈ I.So, we have shown that r+I ∈ R/I is a unit if and only if there exists an element s ∈ Rsuch that rs− 1 ∈ I.

(b) Let a, b ∈ Z such that a 6= 0 and b 6= 0. Then (a, 0), (0, b) ∈ Z ⊕ Z with (a, 0) 6= (0, 0)and (0, b) 6= (0, 0). Observe that (a, 0)(0, b) = (a · 0, 0 · b) = (0, 0). Since (0, 0) is thezero element in the ring Z⊕ Z, we have shown that (a, 0) and (0, b) are zero-divisors inZ⊕ Z, for all nonzero integers a and b.

(c) We know that if R1 and R2 are rings and a ∈ R1 and b ∈ R2, then (a, b) is a unit in thering R1 ⊕R2 if and only if a in a unit in R1 and b is a unit in R2. Since the only unitsin Z are 1 and −1, we have that (1, 1), (1,−1), (−1, 1) and (−1,−1) are all the units inZ⊕ Z. �

6. Let R be a ring. Recall that a ∈ R is called a nilpotent element if an = 0 for some n ∈ Z+

and it is called an idempotent element if a2 = a.

(a) Show that if a is an idempotent, then 1− a is also an idempotent.

(b) If f : R → S is a ring homomorphism and a ∈ R is nilpotent, prove that f carries theelement a to a nilpotent element in the ring S.

(c) If R is an integral domain, prove that the only idempotents in R are 0 and 1.

(d) Show that 0 6= a ∈ R is a zero-divisor if and only if aba = 0 for some b 6= 0.

Solution.

(a) Let a ∈ R be an idempotent. So a2 = a.Then (1− a)2 = (1− a)(1− a) = 1 · 1− 1 · a− a · 1 + a2 = 1− 2a+ a = 1− a. So 1− ais an idempotent element of R.

(b) Let f : R → S be a ring homomorphism and let a ∈ R such that an = 0R for somepositive integer n. Then f(an) = (f(a))n, which can be rewritten as 0S = f(0R) =(f(a))n, thus f(a) is an idempotent element of S.

(c) Now assume that R is an integral domain (that is, R is a commutative ring with unity,call it 1, and no zero-divisors) and let a ∈ R be an idempotent. Then a2 = a, orequivalently, a(a − 1) = 0. Since R has no zero-divisors, the last equality implies thata = 0 or a = 1.

(d) Let 0 6= a ∈ R be a zero-divisor. Then there exists 0 6= b ∈ R such that ab = 0,which implies that aba = 0 · a = 0. Conversely, let b ∈ R, b 6= 0 such that aba = 0.Clearly ba ∈ R by the closure property of R with respect to multiplication, ba 6= 0 and0 = aba = a(ba), thus a is a zero-divisor. �

7. Let G be a group and let Z(G) = {g ∈ G | gx = xg, ∀x ∈ G}.

(a) Show that if G/Z(G) is cyclic then G is Abelian.(b) Show that if G is non-Abelian with |G| = p3, where p is a prime, and Z(G) 6= {e}, then|Z(G)| = p.

Solution.

(a) Assume that G/Z(G) is cyclic and let gZ(G) be a generator of G/Z(G), for some g ∈ G.Let a, b ∈ G. Then aZ(G) = (gZ(G))i = giZ(G) and bZ(G) = (gZ(G))j = gjZ(G) forsome i, j ∈ Z. But then a = gix and b = gjy for some x, y ∈ Z(G).Thus we have

ab = (gix)(gjy)

= gi(xgj)y

= gi(gjx)y

= (gjgi)(yx)

= gj(giy)x

= gj(ygi)x

= (gjy)(gix) = ba,

where we used that x and y commute with everything in G and that gigj = gi+j =gj+i = gjgi. So we showed that ab = ba for all a, b ∈ G, thus G is Abelian.

(b) Now assume that |G| = p3, where p is a prime, that Z(G) 6= {e} and that G is non-Abelian. Since Z(G) is a subgroup of G, |Z(G)| divides p3. Thus |Z(G)| is equalto 1, p, p2 or p3 (since p is a prime number). Since Z(G) is non-trivial, |Z(G)| 6= 1.Moreover, since G is non-Abelian, Z(G) is a proper subgroup of G, thus |Z(G)| 6= p3.Finally, if |Z(G)| = p2, then |G/Z(G)| = p3/p2 = p, and then G/Z(G) must be cyclic(since any group of prime order is cyclic). But then part (a) implies that G is Abelian,which is a contradiction. Therefore the only possible case is |Z(G)| = p. �

8. Suppose G is a cyclic group with at least 3 elements. Without using the Euler φ function,prove that G has an even number of generators.Solution. Let G = 〈a〉 with |a| = n ≥ 3, and let m be an integer with 1 ≤ m < n. Weclaim that am is a generator of G if and only if a−m is also a generator. In the forwarddirection we have (am)n = amn = e, so (a−m)n = (amn)−1 = e−1 = e. If there is an integerk with 1 ≤ k < n such that (a−m)k = e, then (amk)−1 = e−1 = e; hence amk = (am)k = e, acontradiction. The converse is similar. Finally, note that for m as above, if am is a generator,then am 6= a−m, since otherwise a2m = e = an, yielding |am| ≤ 2 < n, a contradiction. Thusthe set of distinct generators of G is {am, a−m | am is a generator}, which clearly has an evennumber of elements.


Part B. Solve five of the following eight problems:

1. Let

W = span

1321

,1211

,1221

.

Find an orthonormal basis for W .

We use the formula

vi = ui −v1 · ui

v1 · v1v1 −

v2 · ui

v2 · v2v2 − . . .−

vi−1 · ui

vi−1 · vi−1vi−1

with v1 = u1 =

1321

. We then obtain v2 =

130− 1

313

and v3 =

15− 2

52515

, which forms an

orthogonal basis for W . Dividing by the magnitude of each vector we obtain an orthonormalbasis for W :

1√15

1321

,√3

3

10−11

,

12−1112

2. Suppose that A is a square matrix of size n and B = {x1,x2,x3, ...,xn} is a basis of Rn.Show that if A is nonsingular, then C = {Ax1, Ax2, Ax3, ..., Axn} is a basis of Rn.

Solution. First, we need to show that C is linearly independent.

0 = c1Ax1 + c2Ax2 + c3Ax3 + ...+ cnAxn

= Ac1x1 +Ac2x2 +Ac3x3 + ...+Acnxn

= A (c1x1 + c2x2 + c3x3 + ...+ cnxn) .

Since A is nonsingular, we obtain

0 = c1x1 + c2x2 + c3x3 + ...+ cnxn.

Since B is a basis, it is linearly independent. Thus, c1 = c2 = ... = cn = 0, and C is linearlyindependent.

Next, we need to show that C spans Rn. Given an arbitrary vector y ∈ Rn. Since A isnonsingular, we can define the vector w to be the unique solution of Aw = y. Since w ∈ Rn,we can write w as a linear combination of vectors in basis B. So,

w = c1x1 + c2x2 + c3x3 + ...+ cnxn,

where c1, c2, c3, ..., cn are scalars. Thus,

y = Aw

= A (c1x1 + c2x2 + c3x3 + ...+ cnxn)

= Ac1x1 +Ac2x2 +Ac3x3 + ...+Acnxn

= c1Ax1 + c2Ax2 + c3Ax3 + ...+ cnAxn.

So, C spans Rn.

3. Find a basis for both the kernel and the range of the linear transformation represented by

A =

1 3 11 2 00 1 1

.

Solution. The kernel is the solution space of the homogeneous system Ax = 0. PerformingGaussian elimination on A gives the matrix1 3 1

0 1 10 0 0

.Thus, a basis for the kernel of the linear transformation is {[2,−1, 1]}. Since the range is thecolumn space of A, a basis for the range of the linear transformation is {[1, 0, 0], [1, 1, 0]}.

4. Given the matrixA =

[3 −51 −3

],

find A2011.

Solution. First, we need to find the eigenvalues and eigenvectors of A. The characteristicequation is found from

0 =

∣∣∣∣ 3− λ −51 −3− λ

∣∣∣∣= λ2 − 4

= (λ+ 2)(λ− 2).

We find an eigenvector for each eigenvalue, obtaining:

v1 =

[11

], v2 =

[51

].

Thus, A = QDQ−1, so

A2011 = QD2011Q−1

=

[1 51 1

] [(−2)2011 0

0 22011

] [1 −5−1 1

](−1

4

)=

[3 · 22010 −5 · 2201022010 −3 · 22010

]since

Q =

[1 51 1

], and D =

[−2 00 2

].

5. Find the eigenvalues of the matrix 1 2 3 4 51 2 3 4 51 2 3 4 51 2 3 4 51 2 3 4 5

.

Solution. The matrix has characteristic polynomial λ4 · (λ− 15)giving the eigenvalues 0 and 15.

Another way to do this is to realize that this matrix has rank 1, and thus λ = 0 is aneigenvalue of multiplicity 4 (as this is a 5× 5 matrix). In order to find the other eigenvalue

we notice that this matrix times the vector

11111

yields

1515151515

and thus λ = 15 is the fifth

eigenvalue.

6. Find all values of λ so that the following matrix is nonsingular: −2 λ 31 2 λ1 11 18

.Solution: Set det

−2 λ 31 2 λ1 11 18

= 0 = λ2 + 4λ − 45 which has roots: −9 and 5; hence

the matrix is non-singular for all values of λ other than those roots.

7. Given the 2× 3 matrix

A =

[1 2 32 3 4

],

(a) For which vectors b =

[b1b2

]does Ax = b have at least one solution?

(b) Solve Ax =

[35

].

(c) What is the rank of A?

Solution.

(a) It is easy to check that rank(A) = 2. Thus, the column space of A spans R2; hence Ax =b has at least one solution for all vectors in R2. Note: the solution will not be uniquebecause the rank of A does not equal 3 (clearly impossible for this underdeterminedsystem which will be impossible to obtain 3 pivots upon row reduction!)

(b) Since solutions to systems represent the scalars for linear combinations of the columns

of the coefficient matrix, by inspection one solution is

110

. Upon reduction of the

augmented system, we obtain the complete parameterized infinite solution (since we

MUST have a free variable):

xyz

=

110

+ t ·

1−21

.(c) 2 of course.

8. Consider the function T : P2 → R3 defined by T (p(x)) =

p(−1)p(0)p(1)

(a) Prove that T is an invertible linear transformation.

(b) Find T−1

abc

Solution.

(a) T is linear, as

T (p+q) =

(p+ q)(−1)(p+ q)(0)(p+ q)(1)

=

p(−1) + q(−1)p(0) + q(0)p(1) + q(1)

=

p(−1)p(0)p(1)

+ q(−1)

q(0)q(1)

= T (p)+T (q)

The matrix for T , A, is given by the image of the columns for the basis for P2, namely{1, x, x2

}; hence A =

1 −1 11 0 01 1 1

. All matrices represent linear transformations,

so linearity is given. det(A) =

∣∣∣∣∣∣1 −1 11 0 01 1 1

∣∣∣∣∣∣ 6= 0 ; hence the linear transformation is

invertible.

(b) A−1 =

1 −1 11 0 01 1 1

−1

=1

2

1 2 0−1 0 11 −2 1

; hence

A−1 =1

2

0 2 0−1 0 11 −2 1

abc

=

1

2

2b−a+ c

a− 2b+ c

=

bc−a2

a−2b+c2

.Thus,

T−1

abc

= b+

(c− a2

)x+

(a− 2b+ c

2

)x2.



1. Let G be a group.

(a) If a, b ∈ G and ab = e, prove that ba = e.

(b) If |G| = 4, prove that G is Abelian.

(c) If a, b ∈ G, prove that o(bab−1) = o(a).

2. Let σ ∈ Sn be a permutation. For x, y ∈ {1, 2, . . . , n} let x ∼ y if and only if σk(x) = y forsome k ∈ Z. Show that ∼ is an equivalence relation on the set {1, 2, . . . , n}.Hint. Note that σ is fixed; it is k that changes.

3. Let R ={[

a 2bb a

]; a, b ∈ Z

}and S = {a + b

√2 | a, b ∈ Z}. Define ϕ : R → S by

ϕ

([a 2bb a

])= a+b

√2. Prove that ϕ is a ring isomorphism. You may assume that R is a

commutative ring with identity using the usual matrix addition and multiplication, that S isa commutative ring with identity using the usual addition and multiplication of real numbers.

4. True or False (prove or give a counterexample)

(a) The elements in Zn are either zero, invertible or zero divisors.

(b) Part (a) is true for all rings.

5. Let G be a group of (finite) order n, and φ : G→ C∗ a homomorphism of groups. Prove thatφ(g) is an nth root of 1 (and thus, lays on the unit circle), for all g ∈ G.

6. How many elements of order 7 does the group Z21 ⊕ Z35 have? Justify your answer.

7. (a) Let R be a commutative ring with a ∈ R. The annihilator of a is defined by

Ann(a) = {x ∈ R |xa = 0}.

Prove that Ann(a) is an ideal of R.

(b) Show that the direct sum of two nonzero rings is never an integral domain.

8. Recall that the center of a group G is the subset Z(G) = {a ∈ G | ax = xa for all x ∈ G}.

(a) Prove that Z(G) is a subgroup of G;

(b) Prove that Z(G) is a normal subgroup of G.



1. Let W be the subspace of R4 with basis

0−100

,

3010

,

1101

. Use the Gram-

Schmidt process to find an orthonormal basis for W .

2. Show that A ∈Mnn is invertible if and only if 0 is not an eigenvalue of A.

3. Consider the linear transform T : P 2 → P 2 defined by

T (a+ bx+ cx2) = (a+ b+ c) + 2(b+ c)x+ 3cx2.

Find the matrix of T with respect to the basis {1, 1 + x, 3 + 4x+ 2x2}.

4. Let U be the subspace of R4 spanned by {(1, 0, 1, 0), (−1, 2, 0, 1)} and let V be the subspaceof R4 spanned by {(0, 2, 1, 1), (0, 0, 1, 1)}.

(a) Find a basis of U⋂V .

(b) Extend (a) to a basis for U .

5. Give the rank and nullity of the matrix

A =

1 0 1 −2 50 1 −1 4 10 0 0 0 30 0 0 0 1

.

6. Suppose that W is a vector space with dimension 5, and U and V are subspaces of W , eachof which has dimension 3. Prove that U

⋂V contains a nonzero vector.

7. Diagonalize 3 −2 20 1 20 2 1

.

8. Let α, β, γ be real numbers. Evaluate the determinant∣∣∣∣∣∣sin2 α sin2 β sin2 γcos2 α cos2 β cos2 γ

1 1 1

∣∣∣∣∣∣


Part A.

1. (a) Note that ab = e implies that a−1(ab) = a−1 (by multiplying by a−1 onthe left). Then b = a−1. But now, by multiplying by a on the right, weget that ba = a−1a. Finally we see that ba = e, as required. �

(b) Let G = {e, a, b, c} be a group of order 4. By contradiction, suppose thatab 6= ba. Then we must have ab = e and ba = c (or vice versa). Note thathere we are using that the only solution in G of an equation of the formax = a is x = e. But then, this contradicts part (a). Thus if x, y ∈ G,then xy = yx. That is, every group of order 4 is abelian. �

(c) First we show that o(bab−1) =∞ iff o(a) =∞:To this end, suppose that o(bab−1) = ∞, but o(a) = m < ∞. Then(bab−1)m = bamb−1 = beb−1 = e. But this contradicts the fact thato(bab−1) =∞. Thus, if o(bab−1) =∞, then o(a) =∞.

Now suppose that o(a) =∞, but o(bab−1) = n <∞. Then (bab−1)n = e,by definition. Hence banb−1 = e. It follows from the last equation thatan = e, a contradiction. Therefore, o(bab−1) = ∞ iff o(a) = ∞. Bycontrapositive, we also have that o(bab−1) <∞ iff o(a) <∞.

To finish up the proof, we show that if bab−1 and a both have finite order,then o(bab−1) = o(a). Let o(bab−1) = m and o(a) = n (m,n ∈ Z). Firstnote that (bab−1)n = banb−1 = beb−1 = e, since o(a) = n. Thus, bydefinition of m, m ≤ n. Now observe that (bab−1)m = e implies thatbamb−1 = e and this implies that am = e. Thus, by definition of n,n ≤ m. Therefore m = n, as required. �

2. Let x, y, z ∈ {1, 2, . . . , n}.

∼ is reflexive: Clearly x ∼ x since σ0(x) = x.

∼ is symmetric: Suppose x ∼ y. Then there exists k ∈ Z such thatσk(x) = y. Thus σ−k(y) = x, and therefore y ∼ x.

∼ is transitive: Suppose x ∼ y and y ∼ z. Then there exist k, l ∈ Z suchthat σk(x) = y and σl(y) = z. Thus σk+l(x) = σlσk(x) = σl(y) = z, andtherefore x ∼ z.

3. First of all we note that every element of both R and S has a unique repre-sentation. This is trivial for R, but for S we need to look at

a+ b√

2 = c+ d√

2

Since this forces a− c = (d− b)√

2, which if d− b 6= 0 implies√

2 =a− cd− b

∈ Z,

which is a contradiction. Hence, d = b, and a = c as well.

Let us assume for a minute that ϕ is a homomorphism.

Now suppose ϕ

([a 2bb a

])= 0 = 0 + 0

√2. Then a = 0 and b = 0 (using

the uniqueness of the representation). Hence ker(ϕ) = 0, and ϕ is one-to-one. Next, ϕ is clearly onto, for all elements of S are of the form a + b

√2 =

ϕ

([a 2bb a

]); hence im(ϕ) = S.

Now, in order to check that ϕ is a homomorphism, let

A =

[a 2bb a

]and B =

[c 2dd c

].

We have

ϕ(A+B) = ϕ

([a+ c 2(b+ d)b+ d a+ c

])= a+ c+ (b+ d)

√2

= ϕ(A) + ϕ(B).

Similarly,

ϕ(AB) = ϕ

([ac+ 2bd 2(ad+ bc)ad+ bc ac+ 2bd

])= ac+ 2bd+ (ad+ bc)

√2

= (a+ b√

2)(c+ d√

2)

= ϕ(A)ϕ(B).

Hence the operations are preserved.

4. (a) True. Let [a] be a nonzero element in Zn. Since a ∈ Z \ {0} it followsthat gcd(a, n) = 1 or gcd(a, n) = d 6= 1.

In the first case, Bezout’s lemma assures the existence of two integers, mand n, such that am + bn = 1. Hence, reducing both sides module n weobtain am ≡ 1 (mod n), which means [a][m] = [1] in Zn, and thus [m] isthe inverse of [a] in Zn.

In the second case, d 6= 1 divides a and n, and thus a = dk and n = dq,where k, q ∈ Z. Note that 0 < k < a and 0 < q < n.

Hence,qa = q(dk) = (qd)k = nk

and thus qa is a multiple of n, i.e.[q][a] = [0] in Zn. It follows that [a] isa zero divisor in Zn because 0 < q < n, which means that [q] 6= [0] in Zn.

(b) False. In Z, the element 2 is neither invertible nor a zero divisor.

5. Let g ∈ G, since |G| = n, then gn = e. using that φ is a homomorphism, andthus that φ(e) = 1 (the identity of the multiplicative group C∗ we get that

1 = φ(e) = φ(gn) = φ(g)n

and thus φ(g) is an nth root of 1. It follows that φ(g) = e2kπi/n, for somek = 1, 2, · · · , n. In particular, the norm of the complex number φ(g) must beone, which means that φ(g) lays on the unit circle, for all g ∈ G.

6. Consider a generic element (a, b) ∈ Z21⊕Z35. Since 7 is prime, it follows thatthe possibilities for the orders of a and b are 1 or 7.

Case 1 |a| = 7 and |b| = 1 or 7. Since Z21 has a unique cyclic group of order7 and any cyclic group of order 7 has six generators, there are six choices fora. Similarly, there are seven choices for b. This gives 42 choices for (a, b).

Case 2 |a| = 1 and |b| = 7. Since Z35 has a unique cyclic group of order 7and any cyclic group of order 7 has six generators, there are six choices for b.There is only one choice for a. So, this case yields six more possibilities for(a, b).

Thus Z21 ⊕ Z35 has 48 elements of order 7.

7. (a) Let A = Ann(a). Since 0 = 0 · a, we have that 0 ∈ A and A 6= ∅. Letx, y ∈ A. Then xa = 0, ya = 0, and so (x ± y)a = xa ± ya = 0. Thusx± y ∈ A. Now let r ∈ R and x ∈ A. Then xa = 0 and (rx)a = r(xa) =r · 0 = 0 and so rx ∈ A. Hence A is an ideal of R. �

(b) Let R = S ⊕ T be the direct sum of the nonzero rings S and T . Thenfor all s ∈ S, s 6= 0 and all t ∈ T, t 6= 0 we have (s, 0)(0, t) = (0, 0). Thus(s, 0) (and (0, t)) is a zero-divisor for R, implying in R is not an integraldomain. �

8. (a) First note that 1 ∈ Z(G) since 1 · x = x · 1 for all x ∈ G. Thus Z(G)is nonempty. Let m and n be elements of Z(G), and let x ∈ G. Sincexn = nx, we have x = nxn−1 and thus n−1x = xn−1. Therefore xmn−1 =mxn−1 = mn−1x, and mn−1 ∈ Z(G). Thus, Z(G) ≤ G.

(b) Let n ∈ Z(G) and g ∈ G, then g and n commute (n is in the center!).Hence, gng−1 = gg−1n = n, and thus gZ(G)g−1 = Z(G), for all g ∈ G,and thus Z(G) �G.

Part B.

1. We use the formula

vi = ui −v1 · uiv1 · v1

v1 −v2 · uiv2 · v2

v2 − . . .−vi−1 · ui

vi−1 · vi−1

vi−1

with v1 = u1 =

0−1

00

. We then obtain v2 =

3010

and v3 =

10−310

, which

forms an orthogonal basis for W . Dividing by the magnitude of each vectorwe obtain an orthonormal basis for W :

0−1

00

, 1√10

3010

, 1√110

10−310

2. Consider the linear transformation Φ : v 7→ Av. We know that A is invertibleif and only if Φ is bijective.

(⇒): Assume A is invertible, i.e. Φ is bijective.

If λ = 0 were an eigenvalue of A, then there would be an eigenvector of Φ,call it v, such that Φ(v) = 0. It follows that v ∈ Ker(Φ). Hence, Φ is notinjective, and thus not bijective. Contradiction.

(⇐): Assume that 0 is not an eigenvalue of A.

It follows that there is no vector such that Φ(v) = 0, and thus Ker(Φ) istrivial. The dimension formula

dim(Domain)− dim(Kernel) = dim(Range)

implies that dim(Range) = n, which means that Φ is also onto. Done.

3. First, since T (1) = 1, T (x) = 1 + 2x, and T (x2) = 1 + 2x+ 3x2, with respectto the standard basis S = {1, x, x2}, the matrix of T ,

[T ]S,S =

1 1 10 2 20 0 3

.

Consider the basis C = {1, 1 + x, 3 + 4x+ 2x2}. The transition matrix from Cto S is the matrix whose columns are the standard coordinates of the elementsof C. So,

[T ]S,C =

1 1 30 1 40 0 2

.

Then, the transition matrix from S to C is given by

[T ]C,S =

1 −1 12

0 1 −20 0 1

2

.

Thus, the matrix P of T with respect to the basis C is given by

P =

1 −1 12

0 1 −20 0 1

2

1 1 10 2 20 0 3

1 1 30 1 40 0 2

=

1 0 00 2 00 0 3

.

4. (a) Any element x ∈ U ∩ V can be written as

x = a(1, 0, 1, 0) + b(−1, 2, 0, 1) = c(0, 2, 1, 1) + d(0, 0, 1, 1)

It follows that

a− b = 0 2b− 2c = 0 a− c− d = 0 b− c− d = 0

Since the first equation implies a = b we get

a = b 2a− 2c = 0 a− c− d = 0

The second equation forces a = c, and thus

a = b a = c d = 0

It follows that the vectors in the intersection look like

x = a(1, 0, 1, 0) + a(−1, 2, 0, 1) = a(0, 2, 1, 1)

which forms the set U ∩ V =< (0, 2, 1, 1) >. So, a basis of U ∩ V wouldbe {(0, 2, 1, 1)}.

(b) Since U has dimension two, then we need to find one more vector in U tocomplete {(0, 2, 1, 1)} to a basis for U . Take (1, 0, 1, 0) (taken from thegiven spanning set of U), which is clearly independent from (0, 2, 1, 1).Hence, B = {(0, 2, 1, 1), (1, 0, 1, 0)} is a basis of U .

5. It is easily verified that A is row equivalent to the matrix1 0 1 −2 50 1 −1 4 10 0 0 0 10 0 0 0 0

which has rank 3; hence rank(A) = 3. Since A is 4×5, we have rank + nullity= 5; thus nullity(A) = 2.

6. The dimension formula says that

dim(U ⊕ V ) = dim(U) + dim(V )− dim(U ∩ V )

and since U ⊕ V < W , then dim(U ⊕ V ) ≤ 5. It follows that

5 ≥ dim(U) + dim(V )− dim(U ∩ V )

Since dim(U) = dim(V ) = 3, then

5 ≥ 3 + 3− dim(U ∩ V )

which implies that dim(U ∩ V ) ≥ 1, and thus U ∩ V is nontrivial (meaning itis not just the zero vector).

7. We want to diagonalize A =

3 −2 20 1 20 2 1

.

The characteristic polynomial of A is

χA(λ) = det(A− λI) = det

3− λ −2 20 1− λ 20 2 1− λ

= (3− λ)det

(1− λ 2

2 1− λ

)= (3− λ)[(1− λ)2 − 22]

= (3− λ)(λ2 − 2λ− 3)

= −(λ+ 1)(λ− 3)2.

It follows that the eigenvalues of A are λ = −1 and λ = 3.

What is the dimension of the eigenspace of λ = 3? We set the equationAv = 3v: 3 −2 2

0 1 20 2 1

xyz

=

3x3y3z

which forces

−2y + 2z = 0−2y + 2z = 0

2y − 2z = 0

It follows that y = z and thus the eigenspace of λ = 3 is spanned by {(1, 0, 0), (0, 1, 1)}.Hence, A is diagonalizable to the matrix3 0 0

0 3 00 0 −1

.

8. Let α, β, γ be real numbers. We will start by subtracting row two from rowthree, we get∣∣∣∣∣∣

sin2 α sin2 β sin2 γcos2 α cos2 β cos2 γ

1 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣sin2 α sin2 β sin2 γcos2 α cos2 β cos2 γ

1− cos2 α 1− cos2 β 1− cos2 γ

∣∣∣∣∣∣but since 1 − cos2 x = sin2 x, for all x ∈ R. Then, the third row of the latterdeterminant is equal to its first row. Hence, the determinant is equal to zero.

Another way to see this is to realize that the sum of the first two rows equalsthe third, as sin2 x + cos2 x = 1, for all x ∈ R. Hence, the three rows beinglinearly dependent forces the determinant to be zero.



1. True or False (prove or give a counterexample):

Let A,N,G be groups such that A�N and N �G, then A�G.

2. Suppose you are given the operation ∗ on the set G = {x ∈ R | x 6= −1}, defined bya ∗ b = ab+ a+ b. Show that under this operation G is a group.

3. Suppose that φ : Z50 → Z15 is a group homomorphism with φ(7) = 6.

(a) Determine φ(x).

(b) Determine the image of φ.

(c) Determine the kernel of φ.

4. True or False (prove or give a counterexample):

If a ring R has more than two idempotent elements (e is idempotent if e2 = e), then R is notan integral domain.

5. Let N = 〈([2], (123))〉� Z4 × S3.

(a) Find the order of the factor group (Z4 × S3)/N .

(b) Find the order of the element ([3], (12))N in (Z4 × S3)/N . Is (Z4 × S3)/N cyclic?

6. Let G be a finite group of odd order. Prove that every element in G has a square root (soyou have to show that for all g ∈ G, there exists x ∈ G such that x2 = g).

Hint: Show that the map θ : G→ G : g → g2 is one-to-one.

7. Suppose that a and b are group elements that commute and have orders m and n. If 〈a〉∩〈b〉 ={e}, prove that the group contains an element whose order is the least common multiple ofm and n. Show that this need not be true if a and b do not commute.

8. Let R be a ring with 1, and let U(R) be the set of all units in R.

(a) Show that U(R) is a group.

(b) If I is an ideal (left, right, or two-sided) in R such that I ∩U(R) 6= ∅, show that I = R.



1. Consider a system of equations Ax = b where A is n × k. Give a proof or counterexamplefor the following:

(a) For given A and b, if n = k then there is always at most one solution.

(b) For given A and b, if n > k then there is always at least one solution.

(c) For given A, if n < k then there exists a vector b for which the system has no solution.

2. Let V be an n-dimensional vector space and T : V → V a linear transformation such thatthe image and kernel of T are identical.

(a) Prove that n is even.

(b) Give an example of such a linear transformation T .

3. Let P2 be the set of all real polynomials of degree at most 2. It is given that the map

f : P2 → P2 : p(x)→ p(x)− p ′(x) + p ′′(x)

is a linear transformation and that β = {1, 1 + x, 1 + x + x2} is a basis for P2. Find thematrix representation of f with respect to the basis β.

4. Let A be a square matrix with the property that the sum of the elements in each of itscolumns is 1. Show that λ = 1 is an eigenvalue of A.

5. Let θ : R2 → R3 be a linear transformation such that θ(2,−1) = (1, 0, 1) and θ(−5, 3) =(0,−1, 1). Find an expression for θ(x, y).

6. Let W1 and W2 be subspaces of a finite-dimensional vector space V . Prove the following:

(a) W1 +W2 = {w1 + w2 : w1 ∈W1, w2 ∈W2} is a subspace of V .

(b) W1 ∩W2 is a subspace of V .

(c) dim (W1) + dim (W2) = dim (W1 +W2) + dim (W1 ∩W2).

7. Find all values of α for which the following matrix is nonsingular −2 α 31 2 α1 11 18

.

8. Suppose that v1 and v2 are arbitrary vectors in Rn. Prove that

span {v1,v2} = span {v1 + v2,v1 − v2} .


Part A.

1. False: Consider G = A4, N to be the Klein group

N = {e, (12)(34), (13)(24), (14)(23)}

and A =< (12)(34) >. It is easy to see that A�N �G, but A is not normalin G because (123)A(123)−1 =< (14)(23) >6= A.

2. We need to check all the axioms of a group. First we realize that the product ofany two elements of R yields an element of R, thus closure is (almost) granted.We just need to check that the product of any two elements in G is not −1.If a ∗ b = −1 then

−1 = a+ b(1 + a) or − (1 + a) = b(1 + a)

which forces either 1 + a = 0 (impossible as a 6= −1) or b = −1 (then againimpossible). It follows that a ∗ b 6= −1, and thus we have closure for G.

In the search for an identity e we set a ∗ e = a, we get a = a + e + ae, whichmeans that e(1 + a) = 0. But, since a 6= −1 then e = 0. So, e = 0 is the onlycandidate to be the identity for this product... but it is easy to check thata ∗ 0 = 0 ∗ a = a for all a ∈ G, so we have found the identity.

Now let us find the inverse of an element a, we want to find b such that a∗b = 0.It is ease to see that a+ b+ ab = 0 forces

b = − a

1 + a

which then again needs a 6= −1 to be well defined. Now note that this fractioncan never be −1 (as a and 1+a are never equal to each other), thus the (right)inverse of any element in G is well defined and an element of G. Now it is easyto check that a∗b = b∗a = 0m and thus every element in G has a double-sidedinverse.

The only thing left to check is associativity. This follows simply from

a ∗ (b ∗ c) = a ∗ (b+ c+ bc)

= a+ (b+ c+ bc) + a(b+ c+ bc)

= a+ b+ c+ bc+ ab+ ac+ abc

= (a+ b+ ab) + c+ (a+ b+ ab)c

= (a+ b+ ab) ∗ c= (a ∗ b) ∗ c

Hence, G is a group. Moreover, it is easy to see that it is an Abelian group.

3. First note that since gcd(50, 7) = 1 then 7 is a generator of (Z50,+). Thusdefining φ(7) will determine uniquely ALL the values φ takes in Z50.

(a) In order to find the images we need to write any element in Z50 as amultiple (modulo 50) of 7. Since,

50 + 7 · (−7) = 1

then, using the homomorphism properties of φ we get

φ(1) = φ(50 + 7 · (−7))

= φ(50) + φ(7 · (−7))

= φ(0)− 7φ(7)

= −7φ(7)

= −42

= 3

where the last step was reduction modulo 15 (the images of φ live in Z15).

It follows that for every integer x,

φ(x) = xφ(1) = 3x (mod 15)

(b) Since 3 divides 15 then the multiples of 3 (modulo 15) yields a propersubgroup of Z15, which is

φ(Z50) = {0, 3, 6, 9, 12}

which is isomorphic to Z5.

(c) Since 15 = 3 · 5, then any x that has a five in its prime factorization(divisible by 5) will be in Ker(φ). Hence, the kernel is the set of multiplesof 5 modulo 50, that is

Ker(φ) = {0, 5, 10, 15, 20, 25, 30, 35, 40, 45}

which is isomorphic to Z10.

Note that the first isomorphism theorem would say, in this case, somethinglike

Z50/Z10∼= Z5

4. TRUE: e2 = e implies e(e − 1) = 0, the two obvious solutions are e = 0, 1,if there were another solution, then e and e− 1 would be both non-zero withproduct equal to zero. Hence, e and e− 1 would be zero divisors.

5. Since [2] has order 2 in Z4 and (123) has order 3 in S3, then σ = ([2], (123))has order 6, and thus |N | = 6 (and cyclic).

(a) First of all notice that N is normal in G = Z4×S3, and that |G| = 4 ·6 =24. It follows that G/N has 24/6 = 4 elements.

(b) First notice that ([3], (12)) /∈ N . Since 2 · [3] = [2] and (12)2 = e, then([3], (12))2 = ([2], e) = ([2], (123))3, which lives in N . It follows that theorder of ([3], (12))N in G/N is 2.

Now, note that ([3], (123)) /∈ N , and that

([3], (123))([3], (12))−1 = ([0], (13)) /∈ N

and thus ([3], (123))N 6= ([3], (12))N . Moreover, since

([3], (123))2 = ([2], (132)) = ([2], (123))5 ∈ N

then G/N contains at least two elements of order 2, thus it cannot becyclic.

6. Consider θ : G → G defined by θ(g) = g2. Let |G| = n < ∞. We want toshow that θ is onto.

Using that gcd(n, 2) = 1 we get α, β ∈ Z such that 1 = 2α + nβ, then

x = x2α+nβ = x2αxnβ = x2α(xn)β = x2α = (xα)2

for all x ∈ G. So, θ(xα) = x.

7. Let G be the group that contains the elements a and b.

Consider x = (ab)gcd(m,n) ∈ G. Since lcm(m,n)gcd(m,n) = mn then

xlcm(m,n) =((ab)gcd(m,n)

)lcm(m,n)= abmn = ambn = e

Hence, the order of x is a divisor of lcm(m,n). If the order d of x were lessthan that, then

e = xd =((ab)gcd(m,n)

)d= adgcd(m,n)bdgcd(m,n)

which forces that the element adgcd(m,n) = b−dgcd(m,n) is in the intersection ofthe groups generated by a and b. This forces adgcd(m,n) = b−dgcd(m,n) = e, butthis implies that n|d gcd(m,n) and m|d gcd(m,n), but since d < lcm(m,n)we get a contradiction. Hence, the order of x is exactly lcm(m,n).

For the counterexample, consider a = (123) and b = (12), both elements inS3. Since S3 is not cyclic then there is no element of order 6 in this group.

8. Let R be a ring with 1, and U(R) be the set of all its units.

(a) We want to show that U(R) is a multiplicative group. Since R has a 1and associativity of the multiplication is inherited from R, then we justneed to check closure under multiplication and inverses.

Let r, s ∈ U(R), then ar = ra = 1 and bs = sb = 1 for some a, b ∈ R.Then,

(ba)(rs) = b(ar)s = b · 1 · s = 1 (rs)(ba) = r(sb)a = r · 1 · a = 1

which means that rs is invertible in R.

Closure for inverses is straight out of the definition of inverses.

(b) Let r ∈ I ∩U(R), then there is an element a ∈ R such that ar = ra = 1.But since I is an ideal (left/right/double-sided), then this forces that1 ∈ I, which immediately implies that all the elements in the ring arecontained in I (because of the “absorption” property of ideals).

Part B.

1. For these solutions consider a system of equations Ax = b where A is n× k.

(a) False. For n = k = 2, the system[1 −11 −1

]x =

[00

]has infinitely many solutions (the subspace spanned by (1, 1)).

(b) False. For n = 3 and k = 2, the system 1 −11 11 1

x =

001

has no solutions (the first two equations force x = y = 0 and the thirdasks for x+ y = 1).

(c) True. Note that in this case no b is given, thus we need to look at thisin a different way. Since A has more columns than rows, then the rowdimension will always be less than the number of columns. But the rowdimension is equal to the column dimension (and this is less than k),which forces the columns of A to be linearly dependent, and thus not abasis of Rk.

Now, the search for an x such that Ax = b can be read as the searchfor coefficients (the components of x) such that b is a linear combinationof the columns of A. But since the column dimension is less than kthen there is always at least one b that will not be written as a linearcombination of the columns of A. This means that for that b the systemhas no solutions.

2. Let V be an n-dimensional vector space and T : V → V a linear transformationsuch that the image and kernel of T are identical.

(a) Since V is finitely dimensional, then we can use the first isomorphismtheorem to get a “dimension formula”, that is

dim(V ) = dim(Ker(T )) + dim(Im(T ))

Since the kernel and image are identical, then they have the same dimen-sion k, and thus n = 2k.

(b) Consider T : R2 → R2 defined by T (x, y) = (0, x). This map hasKer(T ) =< (0, 1) >= Im(T ).

3. We compute the images of the elements of β,

f(1) = 1− 0 + 0 = 1 f(1 + x) = (1 + x)− 1 + 0 = x

f(1 + x+ x2) = (1 + x+ x2)− (1 + 2x) + 2 = 2− x+ x2

It follows that the matrix that represents f from the basis β to the standardbasis of P2 is given by 1 0 2

0 1 −10 0 1

The change of basis matrix from basis β to the standard basis of P2 is 1 1 1

0 1 10 0 1

Hence, the matrix that represents f from basis β to basis β is 1 1 1

0 1 10 0 1

−1 1 0 20 1 −10 0 1

=

1 −1 30 1 −20 0 1

4. Let A be an n × n matrix such that the sum of the elements in each of itscolumns is 1. Note that the transpose of A is such such that the sum of theelements in each of its rows is 1. It follows that if we consider the vectorv with 1’s in all its components. The product ATv is equal to v becausethe components of the product are the sums of the elements in each row ofAT . Thus v is an eigenvector of AT with eigenvalue 1. But we know that aneigenvalue of AT is also an eigenvalue of A, and so we are done proving that 1is an eigenvalue of A.

In fact if λ is an eigenvalue of AT then

det(AT − λId) = 0

but(AT − λId)T = (AT )T − (λId)T = A− λId

and thusdet(A− λId) = det(AT − λId) = 0

which means that λ is also an eigenvalue of A.

5. Consider θ : R2 → R3 be a linear transformation such that θ(2,−1) = (1, 0, 1)and θ(−5, 3) = (0,−1, 1).

Let us write (x, y) as a linear combination of (2,−1) and (−5, 3).

(x, y) = α(2,−1) + β(−5, 3) = (2α− 5β,−α + 3β)

which yieldsx = 2α− 5β y = −α + 3β

and thus β = x+ 2y and α = 3x+ 5y. It follows that

(x, y) = (3x+ 5y)(2,−1) + (x+ 2y)(−5, 3)

and thus

θ(x, y) = (3x+ 5y)θ(2,−1) + (x+ 2y)θ(−5, 3)

= (3x+ 5y)(1, 0, 1) + (x+ 2y)(0,−1, 1)

= (3x+ 5y,−x− 2y, 4x+ 7y)

6. Let W1 and W2 be subspaces of a finite-dimensional vector space V .

(a) We first check that zero is in W1 + W2, this is clear because 0 + 0 = 0and 0 ∈ W1, 0 ∈ W2.

Now we check closure for sum. Let w1 + w2 and v1 + v2 be two elementsof W1 +W2.Then

(w1 + w2) + (v1 + v2) = (w1 + v1) + (w2 + v2)

which is in W1 +W2 because w1 + v1 ∈ W1 and w2 + v2 ∈ W2 (closure ofaddition in these subspaces).

Finally we check closure under scalar multiplication. Let w1 + w2 be anelement of W1 +W2, and α ∈ R. Then,

α(w1 + w2) = αw1 + αw2

which is in W1 +W2 because αw1 ∈ W1 and αw2 ∈ W2 (closure of scalarmultiplication in these subspaces).

(b) We first check that zero is in W1 ∩W2, this is clear because 0 ∈ W1 and0 ∈ W2.

Now we check closure for sum. Let w and v be two elements of W1 ∩W2,then w+v ∈ W1 and w+v ∈ W2 (closure of addition in these subspaces),and thus w + v ∈ W1 ∩W2.

Finally we check closure under scalar multiplication. Let w be an elementof W1 ∩W2, and α ∈ R, then αw ∈ W1 and αw ∈ W2 (closure of scalarmultiplication in these subspaces), and thus αw ∈ W1 ∩W2.

(c) We first take a basis of W1∩W2, then we complete this basis to two otherbases, one for W1 and one for W2. This can always be done by a knowntheorem (which requires the axiom of choice to be proved).

So, let{w1, w2, · · · , wk, wk+1, · · · , wn}

be a basis of W1 (dimension of W1 is n),

{w1, w2, · · · , wk, vk+1, vk+2, · · · , vm}

be a basis for W2 (dimension of W2 is m), and

{w1, w2, · · · , wk}

be a basis for W1 ∩W2 (dimension of W1 ∩W2 is k),

Since the union of the bases of W1 and W2 is a spanning set of W1 +W2,then

span{w1, w2, · · · , wk, wk+1, · · · , wn, vk+1, vk+2, · · · , vm} = W1 +W2

But this set (with n + (m − k) elements) must be linearly independentotherwise we get a contradiction with either the given bases being linearlydependent or we would find vectors not on the basis ofW1∩W2 that shouldbe in the intersection. So, the dimension of W1 +W2 is n+(m−k), done.

7. We need to compute the determinant of this matrix∣∣∣∣∣∣−2 α 31 2 α1 11 18

∣∣∣∣∣∣ =

∣∣∣∣∣∣−2 α 31 2 α0 9 18− α

∣∣∣∣∣∣ by subtracting R2 from R3

=

∣∣∣∣∣∣0 α + 4 3 + 2α1 2 α0 9 18− α

∣∣∣∣∣∣ by adding 2R2 to R1

= −∣∣∣∣ α + 4 3 + 2α

9 18− α

∣∣∣∣ expansion into cofactors, using the first column

= − [(α + 4)(18− α)− (3 + 2α)(9)]

= α2 + 4α− 45

= (α− 5)(α + 9)

So, the values for which the matrix is nonsingular are α 6= 5,−9.

8. Since v1 + v2 and v1 − v2 are linear combinations of v1 and v2, then thesubspace generated by the first two vectors is a subset of the subspace spannedby the latter two.

Now we will show that v1 and v2 are linear combinations of v1+v2 and v1−v2,and thus we will get the other inclusion that is needed to get

span {v1,v2} = span {v1 + v2,v1 − v2} .

It is easy to check that

v1 =1

2(v1 + v2) +

1

2(v1 − v2)

v2 =1

2(v1 + v2)−

1

2(v1 − v2)

We are done.



1. Let G = {x ∈ R |x > 0 and x 6= 1}. Define the operation ∗ on G by a ∗ b = aln b, for alla, b ∈ G.

(a) Prove that G is an abelian group under the operation ∗.(b) Show that G is isomorphic to the multiplicative group R×.

2. Let G1, G2 be groups.

(a) If H1 ≤ G1 and H2 ≤ G2 prove that H1 ×H2 ≤ G1 ×G2.

(b) TRUE/FALSE : If H ≤ G1 × G2 then H = H1 × H2 for some H1 ≤ G1 and someH2 ≤ G2. Prove your answer!

3. If φ : S3 → Z3 is a group homomorphism, show that φ(g) = 0 for all g ∈ S3.

4. Define f : Zmn → Zm ×Zn by f([x]mn) = ([x]m, [x]n). Show that f is well-defined, and thatf is bijective if and only if gcd(m,n) = 1.

5. Let H � G and for any g ∈ G define nH(g) to be the least positive integer such thatgnH(g) ∈ H. Show that nH(g) divides the order of g.

6. Let R be a commutative ring. An element r ∈ R is called nilpotent if rn = 0 for some integern > 0. Prove that a+ b is nilpotent if a and b are nilpotent elements of R.

7. Show that the set of matrices A ∈ Mn(R) such that Av = 0 for some fixed v ∈ Rn is a leftideal of Mn(R).

8. Let

S ={[

a b−b a

]| a, b ∈ R

}⊂M2(R)

(a) Show that S is a subring of M2(R), the ring of 2× 2 matrices with real entries.

(b) Show that S and C are isomorphic rings.



1. Recall that P5 is the set containing the zero polynomial and all polynomials of degree atmost five with real coefficients. Show that the derivative defines a linear transformation fromP5 to itself. Is it onto? Find the matrix for this map in the standard basis.

2. Show that

U ={[

a bc d

]| a, b, c, d ∈ R and a+ b+ c+ d = 0

}⊂M2(R)

is a subspace of M2(R). Find a basis for U .

3. Find a basis for the orthogonal complement of the subspaceW = span{(1, 2,−1, 0), (0, 1, 1, 3)}of R4.

4. Let T be the linear transformation of R3 with standard matrix

1 5 22 1 31 1 4

. Find the

matrix of T with respect to the basis B = {(1, 1, 1), (1, 1, 0), (1, 0, 0)}.

5. Let F : R4 → R3 be any linear transformation such that

KerF =

(x1, x2, x3, x4) ∈ R4

∣∣∣∣∣∣x1 − 2x2 + x3 − x4 = 0x1 − x2 − 2x3 + x4 = 0

x1 − 3x2 + 4x3 − 3x4 = 0

.

(a) Find the dimension of KerF and a basis for it.

(b) Give an example of such a linear transformation F .

(c) For the example you gave in (b), find a basis for the range of F .

6. A square matrix B is skew-symmetric if BT = −B. Suppose that the square matrix A isskew-symmetric and invertible. Prove that A−1 is also skew-symmetric.

7. Diagonalize the following matrix

A =

1 −2 −1−1 1 1

1 0 −1

Then give a basis of R3 for which A ‘becomes’ diagonal.

8. Consider the subspace U = {(x, y, z) ∈ R3| 2x − y − 3z = 0} ⊂ R3 and the set of vectorsS = {(1,−1, 1), (4, 2, 2)} ⊂ R3.

(a) Complete S to a basis in R3.

(b) Show that U = span(S).


Part A.

1. Let G = {x ∈ R |x > 0 and x 6= 1}. Define the operation ∗ on G by a ∗ b = aln b,for all a, b ∈ G.

1. Prove that G is an abelian group under the operation ∗.

2. Show that G is isomorphic to the multiplicative group R×.

Solution.

1. First note that the product of any two elements in G is well defined, as x 7→ lnxis a well-defined function.

Since aln b = 1 only when a = 1 or ln b = 0 (which forces b = 1), and 1 /∈ G,then closure of ∗ holds.

The identity must be an element a (using e will get us into notation problems,as ln x = loge x) such that b = a ∗ b = aln b. If we apply ln both sides we get

ln b = ln aln b = aln b ln a

thus ln a = 1, and thus a = e. I guess that notation problem is solved by now.

It is clear that b = b ∗ e = bln e for all b ∈ G. Thus e is the identity of G.

The inverse of b ∈ G is found by solving e = a ∗ b = aln b, which after applyingln both sides yields 1 = ln a ln b. This equation can always be solved in G,as every non-zero real has a unique inverse in R× and ln : G → R× has aninverse function. We will use this in part 2.

If a and b solve 1 = ln a ln b then it is easy to see that b ∗ a must also be e.

Associativity follows from

(a ∗ b) ∗ c =(aln b)∗ c

=(aln b)ln c

= aln b ln c

= aln c ln b

= aln (bln c)

= aln (b∗c)

= a ∗ (b ∗ c)

So, G is group. Let us check that it is Abelian. This follows from

a ∗ b = aln b

=(eln a)ln b

= eln a ln b

= eln b ln a

=(eln b)ln a

= bln a

= b ∗ a

2. Consider ln : G→ R×. We know this function has an inverse, the restrictionof ex to R×. But, is this a homomorphism? Yes, it is!

ln (a ∗ b) = ln(aln b)

= ln b ln a

2. Let G1, G2 be groups.

1. If H1 ≤ G1 and H2 ≤ G2 prove that H1 ×H2 ≤ G1 ×G2.

2. TRUE/FALSE : If H ≤ G1 × G2 then H = H1 × H2 for some H1 ≤ G1 andsome H2 ≤ G2. Prove your answer!

Solution.

1. AssumeH1 ≤ G1 andH2 ≤ G2, thusH1×H2 is non-empty. Let (h1, h2), (g1, g2) ∈H1 ×H2, then

(h1, h2)(g1, g2)−1 = (h1, h2)(g

−11 , g−1

2 ) = (h1g−11 , h2g

−12 )

which is an element of H1 ×H2 because of closure of H1 and H2.

2. False: Consider G = Z × Z, and H the group generated by (1, 1), which isclearly isomorphic to Z.

Since a subgroup of G of the form H1×H2 with H1, H2 ≤ Z can be isomorphicto Z only if one of H1 or H2 is trivial, then our group H must be contained inZ, which is impossible, as H is generated by (1, 1).

3. If φ : S3 → Z3 is a group homomorphism, show that φ(g) = 0 for all g ∈ S3.

Solution. Since S3 is generated by 2-cycles then we just need to see what φ does tothese elements. Now, since 2-cycles have order two, then the order of their imagesunder φ must have order two or one (divisors of 2). It follows that φ must send2-cycles to the identity of Z3 because this group does not have element of order 2.Since every generator of S3 is mapped to 0 then every element in S3 is mapped tozero under φ.

4. Define f : Zmn → Zm×Zn by f([x]mn) = ([x]m, [x]n). Show that f is well-defined,and that f is bijective if and only if gcd(m,n) = 1.

Solution. Let [x]mn = [y]mn, that is x = y +mnα for some α ∈ Z. Then,

f([x]mn) = ([x]m, [x]n) = ([y +mnα]m, [y +mnα]n) = ([y]m, [y]n) = f([y]mn)

So, f is well-defined.In order to see if f is onto we need to check if for ([a]m, [b]n) ∈ Zm × Zn there isan [x]mnZmn such that f([x]mn) = ([a]m, [b]n). In other words, we are looking for anx ∈ Z that solves the congruences

x ≡ a (mod m) x ≡ b (mod n)

simultaneously.We know that such an x exists if gcd(m,n) = 1 because of the Chinese RemainderTheorem.Since Zmn and Zm×Zn have the same number of elements, then assuming gcd(m,n) =1 implies that f is bijective.If we assume that f is bijective, then we get a unique solution modulo n for thetwo congruences above. But, if gcd(m,n) = d 6= 1 then for x a solution of thecongruences above we get another solution (distinct from x modulo mn), namelyy = x+ mn

d. This is a contradiction, so d = 1.

5. Let H � G and for any g ∈ G define nH(g) to be the least positive integer suchthat gnH(g) ∈ H. Show that nH(g) divides the order of g.

Solution. Note that we will not use the normality of H for this proof. There areother proofs that might use it, though.Let H ≤ G and for g ∈ G define nH(g) as above. Let m be the order of g.Since gm = e, then gm ∈ H. It follows that nH(g) ≤ m, thus we can use the division(Euclidean) algorithm to get

m = nH(g)q + r

where q, r ∈ Z and 0 ≤ r < nH(g).

Note thatgm = gnH(g)q+r = gnH(g)qgr

which implies (gnH(g)q

)−1gm = gr

Since both gnH(g)q and gm live in H, then so does gr. If r were non-zero then therewould be a positive integer that is less than nH(g) such that gr ∈ H. This is acontradiction, so r = 0 and thus nH(g) | m.

6. Let R be a commutative ring. An element r ∈ R is called nilpotent if rn = 0 forsome integer n > 0. Prove that a + b is nilpotent if a and b are nilpotent elementsof R.

Solution. Assume an = bm = 0, then since the ring is commutative we get

(a+ b)m+n =m+n∑k=0

(m+ nk

)akbm+n−k

=n∑

k=0

(m+ nk

)akbm+n−k +

m+n∑k=n+1

(m+ nk

)akbm+n−k

For 0 ≤ k ≤ n, m + n − k > m and thus bm+n−k = 0. Thus the first sum is n zerosummands. Similarly, for n+ 1 ≤ k ≤ m+ n we get ak = 0, which implies that thesecond sum is also zero. Hence, (a+ b)m+n = 0.

7. Show that the set of matrices A ∈Mn(R) such that Av = 0 for some fixed v ∈ Rn

is a left ideal of Mn(R).

Solution. Fix v ∈ Rn. Let I = {A ∈Mn(R); Av = 0}.It is clear that the zero matrix is in I, and thus I is non-empty. Now let A,B ∈ I,then

(A−B)v = Av −Bv = 0− 0 = 0

which means that A−B ∈ I.Now let A ∈ I and B ∈Mn(R), then

(BA)v = B(Av) = B · 0 = 0

which means that BA ∈ I.It follows that I is a left ideal.

8. Let

S =

{[a b−b a

]| a, b ∈ R

}⊂M2(R)

1. Show that S is a subring of M2(R), the ring of 2×2 matrices with real entries.

2. Show that S and C are isomorphic rings.

Solution.

1. S is non-empty as the zero matrix and the identity matrix are elements of S.Now take two elements in S and subtract them[

a b−b a

]−[

α β−β α

]=

[a− α b− β

−(b− β) a− α

]which is an element of S.

Similarly, the product of two elements of S is[a b−b a

] [α β−β α

]=

[aα− bβ aβ + bα

−(bα + aβ) aα− bβ

]which is also an element of S.

2. Consider the map φ : S → C defined by

φ

[a b−b a

]= a+ bi

Since

φ

[a b−b a

]+

[α β−β α

]=

[a+ α b+ β

−(b+ β) a+ α

]= (a+ α) + (b+ β)i

= (a+ bi) + (α + βi)

and

φ

[a b−b a

] [α β−β α

]= φ

[aα− bβ aβ + bα

−(bα + aβ) aα− bβ

]= (aα− bβ) + (aβ + bα)i

= (a+ bi)(α + βi)

then φ is a homomorphism of rings.

The kernel of φ is

Ker(φ) =

{[a b−b a

]; φ

[a b−b a

]= 0

}=

{[a b−b a

]; a+ bi = 0

}= {0}

So, φ is one-to-one. Checking onto is easy, as for any a+ bi ∈ C

φ

[a b−b a

]= a+ bi

Part B.

1. Recall that P5 is the set containing the zero polynomial and all polynomials ofdegree at most five with real coefficients. Show that the derivative defines a lineartransformation from P5 to itself. Is it onto? Find the matrix for this map in thestandard basis.

Solution. Since for any pair of differentiable functions (including polynomials) fand g, and any constant C we have

(f + g)′ = f ′ + g′ and (Cf)′ = Cf ′

then the derivative behaves linearly. Moreover, since the derivative of a polynomialof degree at most five is of degree at most four, and 0′ = 0, then the function isfrom P5 to P5. Also, since there is no way to obtain p(x) = x5 as the derivative of apolynomial of degree at most five (its anti-derivative has degree six), then φ is notonto.The standard basis for P5 is B = {1, x, x2, x3, x4, x5}. Since

φ(1) = 0 φ(x) = 1 φ(x2) = 2x φ(x3) = 3x2 φ(x4) = 4x3 φ(x5) = 5x4

then the matrix of φ with respect to B is0 1 0 0 0 00 0 2 0 0 00 0 0 3 0 00 0 0 0 4 00 0 0 0 0 50 0 0 0 0 0

2. Show that

U =

{[a bc d

]| a, b, c, d ∈ R and a+ b+ c+ d = 0

}⊂M2(R)

is a subspace of M2(R). Find a basis for U .

Solution. Probably the easiest way to do this is to go ahead and re-write U as thespan of a set of vectors (which will turn out being its basis).

We use a+ b+ c+ d = 0 to get d = −a− b− c. Thus an element of U looks like[a bc d

]=

[a bc −a− b− c

]=

[a 00 −a

]+

[0 b0 −b

]+

[0 0c −c

]= a

[1 00 −1

]+ b

[0 10 −1

]+ c

[0 01 −1

]Since the coefficients a, b and c can take any values, then U is the span of

B =

{[1 00 −1

],

[0 10 −1

],

[0 01 −1

]}which makes U a subspace of M2(R).The matrices in B are clearly linearly independent, thus B is a basis of U .

3. Find a basis for the orthogonal complement of the subspaceW = span{(1, 2,−1, 0), (0, 1, 1, 3)}of R4.

Solution. Any vector that is orthogonal to every element of W must be orthogonalto both (1, 2,−1, 0) and (0, 1, 1, 3) (and vice-versa). Thus, a vector (x, y, z, w) ∈ W⊥

must be a solution of the system of equations

x+ 2y − z = 0 y + z + 3w = 0

Solving for x in the first equation and for w in the second we get

x = −2y + z w = −y3− z

3

Thus, the vectors (x, y, z, w) ∈ W⊥ look like

(x, y, z, w) =(−2y + z, y, z,−y

3− z

3

)=

(−2y, y, 0,−y

3

)+(z, 0, z,−z

3

)= y

(−2, 1, 0,−1

3

)+ z

(1, 0, 1,−1

3

)which says, given that y and z are free to roam all over R, that

W⊥ = span

{(−2, 1, 0,−1

3

),

(1, 0, 1,−1

3

)}Since the two vectors that span W⊥ are linearly independent (because one is not amultiple of the other), then they are a basis of W⊥.

4. Let T be the linear transformation of R3 with standard matrix

1 5 22 1 31 1 4

.

Find the matrix of T with respect to the basis B = {(1, 1, 1), (1, 1, 0), (1, 0, 0)}.

Solution. Let us denote the standard basis of R3 by S. The change of basis matrix(or transition matrix) from B to S is given by just ‘hanging’ the vectors of B to get 1 1 1

1 1 01 0 0

It follows that the matrix representing T with respect to B is 1 1 1

1 1 01 0 0

−1 1 5 22 1 31 1 4

1 1 11 1 01 0 0

Since 1 1 1

1 1 01 0 0

−1

=

0 0 10 1 −11 −1 0

then the matrix representing T with respect to B is 0 0 1

0 1 −11 −1 0

1 5 22 1 31 1 4

1 1 11 1 01 0 0

=

6 2 10 1 12 3 −1

5. Let F : R4 → R3 be any linear transformation such that

KerF =

(x1, x2, x3, x4) ∈ R4

∣∣∣∣∣∣x1 − 2x2 + x3 − x4 = 0x1 − x2 − 2x3 + x4 = 0

x1 − 3x2 + 4x3 − 3x4 = 0

.

1. Find the dimension of KerF and a basis for it.

2. Give an example of such a linear transformation F .

3. For the example you gave in (b), find a basis for the range of F .

Solution.

1. We need to solve the system given. The matrix that represents this (homoge-neous) system is

A =

1 −2 1 −11 −1 −2 11 −3 4 −3

Let us do some row operations in A.

A =

1 −2 1 −11 −1 −2 11 −3 4 −3

now we subtract R1 from R3 and R2

→

1 −2 1 −10 1 −3 20 −1 3 −2

now we add R2 to R3, and add 2R2 to R1

→

1 0 −5 30 1 −3 20 0 0 0

It follows that x1 = 5x3 − 3x4 and that x2 = 3x3 − 2x4. Thus the vectors inKer(F ) look like

(x1, x2, x3, x4) = (5x3 − 3x4, 3x3 − 2x4, x3, x4)

= (5x3, 3x3, x3, 0) + (−3x4,−2x4, 0, x4)

= x3(5, 3, 1, 0) + x4(−3,−2, 0, 1)

Since x3 and x4 have no restrictions then

Ker(F ) = span{(5, 3, 1, 0), (−3,−2, 0, 1)}

These vectors are linearly independent, thus they form a basis of Ker(F ).dim(Ker(F )) = 2.

2. The map F : R4 → R3 given by

F (x1, x2, x3, x4) = (x1− 2x2 +x3−x4, x1−x2− 2x3 +x4, x1− 3x2 + 4x3− 3x4)

3. The range of F can be computed by looking at the column space of the matrixA used in part 1. of this problem. So, now we will do some column operations.

on A.

A =

1 −2 1 −11 −1 −2 11 −3 4 −3

now we subtract C1 from C3 and add C1 to C4

→

1 −2 0 01 −1 −3 21 −3 3 −2

now we add2

3C3 to C4, and C3 to C2

→

1 −2 0 01 −4 −3 01 0 3 0

Since (1, 1, 1) = −12(−2,−4, 0) + 1

3(0,−3, 3), then the column space is

span{(−2,−4, 0), (0,−3, 3)} = span{(1, 2, 0), (0,−1, 1)}

These two vectors are linearly independent, thus they form a basis of the rangeof F .

6. A square matrix B is skew-symmetric if BT = −B. Suppose that the squarematrix A is skew-symmetric and invertible. Prove that A−1 is also skew-symmetric.

Solution. We know that AT = −A and that A is invertible. Since (AT )−1 = (A−1)T ,then AT is invertible. It follows that if we inverse both sides of AT = −A we get

(A−1)T = (AT )−1 = (−A)−1 = −(A−1)

which proves that A−1 is also skew-symmetric.

7. Diagonalize the following matrix

A =

1 −2 −1−1 1 1

1 0 −1

Then give a basis of R3 for which A ‘becomes’ diagonal.

Solution. We first find the characteristic polynomial of A.

χA(λ) = det(A− λI)

=

∣∣∣∣∣∣1− λ −2 −1−1 1− λ 11 0 −1− λ

∣∣∣∣∣∣R2 7→ R2 +R3

∣∣∣∣∣∣1− λ −2 −1

0 1− λ −λ1 0 −1− λ

∣∣∣∣∣∣C3 7→ C3 − C2

∣∣∣∣∣∣1− λ −2 1

0 1− λ −11 0 −1− λ

∣∣∣∣∣∣= −(1− λ)2(1 + λ) + 2− (1− λ)

= −(1− λ)2(1 + λ) + (λ+ 1)

= (1 + λ)(1− (1− λ)2)

= (1 + λ)(2λ− λ2) = λ(1 + λ)(2− λ)

It is easy to see that λ = 0,−1, 2 are the eigenvalues of χA(λ). Since the eigenvaluesare distinct the matrix is diagonalizable.We know that the base where A ‘becomes’ diagonal is given by the eigenvectors ofA. Let us find them.For λ = 0

[A|0] =

1 −2 −1 0−1 1 1 01 0 −1 0

now we subtract R3 from R1 and add R3 to R2

→

0 −2 0 00 1 0 01 0 −1 0

It follows that the eigenspace of λ = 0 is span(1, 0, 1).For λ = −1

[A+ I|0] =

2 −2 −1 0−1 2 1 01 0 0 0

now we subtract 2R3 from R1 and add R3 to R2

→

0 −2 −1 00 2 1 01 0 0 0

It follows that the eigenspace of λ = −1 is span(0, 1,−2).For λ = 2

[A− 2I|0] =

−1 −2 −1 0−1 −1 1 01 0 −3 0

now we add R3 to R1 and R2

→

0 −2 −4 00 −1 −2 01 0 −3 0

It follows that the eigenspace of λ = 2 is span(3,−2, 1).Thus, the basis of R3 where A becomes 0 0 0

0 −1 00 0 2

is

B = {(1, 0, 1), (0, 1,−2), (3,−2, 1)}

8. Consider the subspace U = {(x, y, z) ∈ R3| 2x− y − 3z = 0} ⊂ R3 and the set ofvectors S = {(1,−1, 1), (4, 2, 2)} ⊂ R3.

1. Complete S to a basis in R3.

2. Show that U = span(S).

Solution.

1. If we show that U = span(S) (part 2. of this problem), then a basis for thethe orthogonal complement of U would complete S to a basis of R3. Since Uis a plane through the origin in R3, then its orthogonal complement is givenby its normal vector, which can be obtained by just looking at the coefficientsof the equation that defines the plane, namely N = (2,−1,−3).

So, S ∪ {(2,−1,−3)} is a basis of R3.

2. We plug the vectors in S into 2x − y − 3z = 0 (the equation that defines U)to check that these vectors belong to U , we get

2(1)− (−1)− 3(1) = 0 2(4)− (2)− 3(2) = 0

Thus S ⊂ U . Since the two vectors in S are linearly independent and U is aplane (dimension 2), then the vectors in S must span all U .


Part A. Do five of the following eight problems :

1. Show there are no integers n such that

(a) S2 × S5 is isomorphic to Sn.

(b) S3 × Z/Z4 is isomorphic to Sn.

2. Let G be a cyclic group of order 8. Prove that G has exactly one element oforder 2. Does there exist a nonabelian group of order 8 with this property?

3. Prove that the map f : G → G, given by f(a) = a−1, is a group homomor-phism if and only if the map g : G → G, given by g(a) = a2, is a grouphomomorphism.

4. What is the order of σ = (1 10 4)(2 13)(1 12 8)(5 7)(6 9)(5 11)?

5. Let

G =

{[a b0 c

]; a, b, c ∈ R, ac 6= 0

}Show that G is a group under matrix multiplication, and that

H =

{[1 b0 1

]; b ∈ R

}is a subgroup of G.

6. Let G be a group and H a subgroup of G. Let NH be the set of all x ∈ G suchthat xHx−1 = H. Show that NH is a subgroup of G containing H, and thatH is normal in NH . (The group NH is called the normalizer of H.)

7. Define a relation ∼ on the set N of natural numbers by a ∼ b if and only ifa2 + b2 is even. Prove that ∼ is an equivalence relation.

8. Let F be a field and R any ring, and let ϕ : F → R be a nonzero ring homo-morphism. Prove that ker(ϕ) = 0.


1. Let v1, v2, v3 be linearly independent vectors in a vector space V . Show thatthe vectors v1, 2v1 + v2, 3v1 + 2v2 + v3 are also linearly independent.

2. Find explicitly a linear map f : R2 → R3 such that f(2, 3) = (1, 2, 3) andf(1, 2) = (4, 5, 6). Is this linear function unique?

3. Give examples of three of the following:

(a) Three vectors in R3 so that any two are linear independent, but the setof all three is linearly dependent.

(b) Matrices M and N such that MN 6= NM .

(c) Linear mapsf : R2 → R4 g : R4 → R2

such that g ◦ f is invertible.

(d) A matrix M such that M3 = 0 but M2 6= 0.

4. Let P2 = {p(x) = a+ bx+ cx2; a, b, c ∈ R}. Show that the map T : P2 → P2

given by

T (p) =d2

dx2

((x2 + 1)p(x)

)is linear. Then find the matrix of T in the basis {1, x, x2}.

5. Consider the following vector subspaces of R3

A = {(x, y, z) ∈ R3| 2x+y+3z = 0}, B = span{(1,−2, 0)}, C = span{(1, 1,−1)}.

Show that A is the direct sum of B and C, that is A = B ⊕ C.

6. Consider the subspace U = {(x, y, z) ∈ R3|x + y − 2z = 0} ⊂ R3 and thesystem of vectors S = {v1 = (1, 1, 1), v2 = (5,−1, 2)} ⊂ R3.

(a) Show that U is the span of S.

(b) Complete S to a basis in R3.

7. Let T be a linear operator on a vector space V , and let x be an eigenvector ofT corresponding to the eigenvalue λ. For any positive integer m, prove that xis an eigenvector of Tm corresponding to the eigenvalue λm.

8. If an n×n matrix A is diagonalizable, show that det(A) is equal to the productof the eigenvalues of A.


Part A.

1. Show there are no integers n such that

(a) S2 × S5 is isomorphic to Sn.

(b) S3 × Z/Z4 is isomorphic to Sn.

Solution:

(a) The order of S2 × S5 is 2! · 5! = 240, which is not an n!

(b) The order of G = S3 × Z/Z4 is 3! · 4 = 4!, so the only case to check iswhether or notG is isomorphic to S4. However, the element ((123), 1) ∈ Ghas order 12, but there are no elements of order 12 in S4. So, the groupscannot be isomorphic.

2. Let G be a cyclic group of order 8. Prove that G has exactly one element oforder 2. Does there exist a nonabelian group of order 8 with this property?

Solution: A (multiplicative) cyclic group G of order 8 can be written as

G = {e, g, g2, g3, g4, g5, g6, g7}

where g8 = e.

It is easy to see that the orders of the non-trivial elements are

o(g) = 8 o(g2) = 4 o(g3) = 8 o(g4) = 2

o(g5) = 8 o(g6) = 4 o(g7) = 8

Thus, g4 is the only element of order two in G.

There are exactly two (non-isomorphic) non-abelian groups of order 8, namelyD4 and Q8. If one thinks D4 as the group of symmetries of a square, it iseasy to see that the square admits 4 different reflections, these reflections haveorder two, so D4 has at least 4 elements of order 2. On the other hand

Q8 = {1,−1, i,−i, j,−j, k,−k}

where i2 = j2 = k2 = −1 (and some other multiplication rules that we don’tneed right now). It follows that x2 = −1 for all x ∈ Q8 \ {±1}. Hence, −1 isthe only element of order two in Q8.

3. Prove that the map f : G → G, given by f(a) = a−1, is a group homomor-phism if and only if the map g : G → G, given by g(a) = a2, is a grouphomomorphism.

Solution: The easiest way to prove this could be to show that the two state-ments we want to show are equivalent are also equivalent with G being Abelian.

First of all, if G is Abelian, then the two functions are clearly homomorphisms.

Now assume that f is a group homomorphism, then

b−1a−1 = (ab)−1 = f(ab) = f(a)f(b) = a−1b−1

Since every element in a group is the inverse of some element in G, thenb−1a−1 = a−1b−1 forces G to be Abelian.

Finally, if g is a group homomorphism, then

(ab)2 = g(ab) = g(a)g(b) = a2b2

Now we multiply (ab)2 = a2b2 by a−1 by the left and b−1 by the right, then weget ba = ab.

4. What is the order of σ = (1 10 4)(2 13)(1 12 8)(5 7)(6 9)(5 11)?

Solution: As the cycles are not disjoint we must first perform the multiplica-tion of the cycles, we get

σ = (1 10 4)(2 13)(1 12 8)(5 7)(6 9)(5 11) = (1 12 8 10 4)(2 13)(5 11 7)(6 9)

Now that we have disjoint cycles we look at the order of each individual cycle,we get orders 5, 2, and 3. Thus the order of σ is 2 · 3 · 5 = 30 (as the ordersare relatively prime).

5. Let

G =

{[a b0 c

]; a, b, c ∈ R, ac 6= 0

}Show that G is a group under matrix multiplication, and that

H =

{[1 b0 1

]; b ∈ R

}is a subgroup of G.

Solution: First let us check closure of G.[a b0 c

] [d e0 f

]=

[ad ae+ bf0 cf

]The product is in G because its entries are in R and (ad)(cf) 6= 0, as ac 6= 0and df 6= 0.

Note that if a = c = d = f = 1, then the diagonal entries of the product arealso 1’s. This shows closure of H.

Now note that [a b0 c

]−1

=

[a−1 −ba−1c−1

0 c−1

]The inverse is in G because its entries are in R and a−1c−1 6= 0, as ac 6= 0.

Note that if a = c = 1, then the diagonal entries of the inverse are also 1’s.This shows the needed inverse property for H.

Associativity of G follows directly from the associativity of R.

Both G and H have identity equal to the identity 2× 2 matrix. Hence, G is agroup and H (clearly a subset of G) is a subgroup of G.

6. Let G be a group and H a subgroup of G. Let NH be the set of all x ∈ G suchthat xHx−1 = H. Show that NH is a subgroup of G containing H, and thatH is normal in NH . (The group NH is called the normalizer of H.)

Solution: First of all, if x ∈ H, then xHx−1 = H because of the closure ofH. So, H ⊂ NH .

Let x, y ∈ NH , then

(xy)H(xy)−1 = x(yHy−1

)x−1 = xHx−1 = H

So, NH is closed.

Clearly the identity of G is in HN .

Now since xHx−1 = H implies x−1Hx = H (because, for example, of the mapϕ : G → G defined by ϕ(g) = xgx−1 being an isomorphism, and ϕ−1(g) =x−1gx).

Finally, if x ∈ NH then xHx−1 = H, implying that H is normal in NH .

7. Define a relation ∼ on the set N of natural numbers by a ∼ b if and only ifa2 + b2 is even. Prove that ∼ is an equivalence relation.

Solution: We need to show that ∼ is symmetric, reflexive and transitive. Thefirst two are clear as if a2 + b2 is even then b2 + a2 is even, and a2 + a2 = 2a2

being even.

Transitivity follows from the fact that the parity of a2 + c2 equals the parity ofa2 + c2 + 2b2, which is even, as it is the sum of the two even numbers (a2 + b2)and (b2 + c2).

8. Let F be a field and R any ring, and let ϕ : F → R be a nonzero ring homo-morphism. Prove that ker(ϕ) = {0}.

Solution: Since the kernel of ϕ is an ideal of the domain, then ker(ϕ) is anideal of the field F . Since the only ideals in a field are {0} and the fielditself, then ϕ is either 1− 1 or the zero function. In this case, ϕ is a non-zerohomomorphism, thus ker(ϕ) = {0}.


1. Let v1, v2, v3 be linearly independent vectors in a vector space V . Show thatthe vectors v1, 2v1 + v2, 3v1 + 2v2 + v3 are also linearly independent.

Solution: A linear combination of the vectors v1, 2v1 +v2, 3v1 +2v2 +v3 equalto zero yields

0 = αv1 + β(2v1 + v2) + γ(3v1 + 2v2 + v3)

= (α + 2β + 3γ)v1 + (β + 2γ)v2 + γv3

So, using that v1, v2, v3 be linearly independent vectors we get the system

α + 2β + 3γ = 0β + 2γ = 0

γ = 0

which has solution α = β = γ = 0. Hence, the vectors v1, 2v1+v2, 3v1+2v2+v3

are linearly independent.

2. Find explicitly a linear map f : R2 → R3 such that f(2, 3) = (1, 2, 3) andf(1, 2) = (4, 5, 6). Is this linear function unique?

Solution: First we write a generic element of R2 as a linear combination ofthe vectors (2, 3) and (1, 2),

(x, y) = α(2, 3) + β(1, 2)

= (2α + β, 3α + 2β)

So, 2α+β = x and 3α+2β = y. It follows that α = 2x−y and β = −3x+2y.Thus,

(x, y) = (2x− y)(2, 3) + (−3x+ 2y)(1, 2)

So,

f(x, y) = f [(2x− y)(2, 3) + (−3x+ 2y)(1, 2)]

= f [(2x− y)(2, 3)] + f [(−3x+ 2y)(1, 2)]

= (2x− y)f(2, 3) + (−3x+ 2y)f(1, 2)

= (2x− y)(1, 2, 3) + (−3x+ 2y)(4, 5, 6)

= (2x− y + 4(−3x+ 2y), 2(2x− y) + 5(−3x+ 2y), 3(2x− y) + 6(−3x+ 2y))

= (−10x+ 7y,−11x+ 8y,−12x+ 9y)

The construction of f shows that the function is unique. Or, using that{(2, 3), (1, 2)} is a basis of R2 then the function, being defined on a basis,must be unique.

3. Give examples of three of the following:

(a) Three vectors in R3 so that any two are linear independent, but the setof all three is linearly dependent.

(b) Matrices M and N such that MN 6= NM .

(c) Linear mapsf : R2 → R4 g : R4 → R2

such that g ◦ f is invertible.

(d) A matrix M such that M3 = 0 but M2 6= 0.

Solution:

(a){(1, 0, 0), (0, 1, 0), (1, 1, 0)}

(b) Take, for example,

M =

[1 00 0

]N =

[0 10 0

](c) Consider

f(x, y) = (x, y, 0, 0) g(x, y, z, w) = (x, y)

Then g ◦ f is the identity.

(d) Consider

M =

0 1 00 0 10 0 0

4. Let P2 = {p(x) = a+ bx+ cx2; a, b, c ∈ R}. Show that the map T : P2 → P2

given by

T (p) =d2

dx2

((x2 + 1)p(x)

)is linear. Then find the matrix of T in the basis {1, x, x2}.

Solution: Let us check linearity, we will use that the derivative is linear. Letα ∈ R and p(x), q(x) ∈ P2.

T (αp(x) + q(x)) =d2

dx2

[(x2 + 1)(αp(x) + q(x))

]=

d2

dx2

[α(x2 + 1)p(x) + (x2 + 1)q(x)

]=

d2

dx2

[α(x2 + 1)p(x)

]+

d2

dx2

[(x2 + 1)q(x)

]= α

d2

dx2

[(x2 + 1)p(x)

]+

d2

dx2

[(x2 + 1)q(x)

]= αT (p(x)) + T (q(x))

Now we look at the matrix in the standard basis

T (1) =d2

dx2

(x2 + 1

)= 2

T (x) =d2

dx2

(x3 + x

)= 6x

T (x2) =d2

dx2

(x4 + x2

)= 12x2 + 2

So, the matrix is given by 2 0 20 6 00 0 12

5. Consider the following vector subspaces of R3

A = {(x, y, z) ∈ R3| 2x+y+3z = 0}, B = span{(1,−2, 0)}, C = span{(1, 1,−1)}.

Show that A is the direct sum of B and C, that is A = B ⊕ C.

Solution: First of all notice that (1,−2, 0) and (0,−3, 1) are linearly indepen-dent because one is not a multiple of the other. Also, since A is the solutionset f a homogeneous linear equation in R3 then it is a plane, which is two-dimensional. So, if both (1,−2, 0) and (0,−3, 1) belong to A then the resultfollows.

Since both (1,−2, 0) and (0,−3, 1) clearly satisfy the equation 2x+y+ 3z = 0then we are done.

6. Consider the subspace U = {(x, y, z) ∈ R3|x + y − 2z = 0} ⊂ R3 and thesystem of vectors S = {v1 = (1, 1, 1), v2 = (5,−1, 2)} ⊂ R3.

(a) Show that U is the span of S.

(b) Complete S to a basis in R3.

Solution:

(a) Just like the previous problem we first notice that v1 and v2 are linearlyindependent. Secondly, both vectors satisfy the equation that defines U .Done.

(b) Note that (1, 0, 0) is not a solution of the equation that defines U , thus(1, 0, 0) is not a linear combination of v1 and v2. So, {(1, 0, 0), v1, v2} isa linearly independent set in R3, it follows it is a basis (three linearlyindependent vectors in a three-dimensional space).

7. Let T be a linear operator on a vector space V , and let x be an eigenvector ofT corresponding to the eigenvalue λ. For any positive integer m, prove that xis an eigenvector of Tm corresponding to the eigenvalue λm.

Solution: We know T (x) = λx. Note that

Tm(x) = (Tm−1 ◦ T )(x) = Tm−1(λx) = λTm−1(x)

The result follows by just generalizing what is above to Tm(x) = λi Tm−i(x)for all i (or by using induction).

8. If an n×n matrix A is diagonalizable, show that det(A) is equal to the productof the eigenvalues of A.

Solution: If A is diagonalizable, then there is a matrix P (change of basis)such that PAP−1 is a diagonal matrix D = diag(λ1, λ2, · · · , λn−1, λn), wherethe λi’s are the eigenvalues of A.

Since the determinant is a multiplicative function, then

det(D) = det(PAP−1

)= det(P )det(A)det

(P−1

)= det(P )det

(P−1

)det(A)

But, since det (P−1) = det(P )−1, then det(D) = det(A). The result followsfrom the fact that the determinant of a diagonal matrix is given by the productof its diagonal entries.


Part A. Solve five of the following eight problems :

1. Let θ : Z → S5 be a group homomorphism such that θ(1) = (123)(45). Findθ(−4) and Ker θ.

2. Let G be the set of all real-valued functions defined on R. Then (G,+) is a groupwhere + stands for the usual addition of functions. Put N = {f ∈ G | f(2008) = 0}.Prove that N �G and G/N ∼= R.

3. TRUE/FALSE : S6∼= S3 ⊕ S5. Prove your answer!

4. Let G be a finite group and k a natural number that is relatively prime to |G|.Prove that the map θ : G→ G : x→ xk is a bijection.

5. Let G be a finite group, N � G and H ≤ G such that |H| and [G : N ] arerelatively prime. Prove that H ≤ N .

6. Let G = R \ {−1}. We define a binary operation ∗ on G as follows :

a ∗ b = ab+ a+ b for all a, b ∈ GIt is given that G is a group under this operation.

(a) What is the identity element in G?

(b) What is the inverse of a ∈ G?

(c) Solve for x : 2 ∗ x ∗ 3 = 7

7. Let R be a ring such that a2 = a for all a ∈ R.

(a) Prove that a+ a = 0 for all a ∈ R (hint : consider (a+ a)2).

(b) Prove that R is commutative (hint : consider (a+ b)2).

8. Let R be the set of all polynomials with real coefficients. Then (R,+, ·) is aring under the usual addition and multiplication of polynomials. Put S = {p(x) ∈R | p ′(0) = 0} where p ′(x) is the usual derivative of p(x) with respect to x.

(a) Prove that S is a subring of R.

(b) Is S an ideal of R?



1. Let v1,v2, . . . ,vk be linearly independent vectors in Rn and A a non-singularn× n matrix. Prove that Av1, Av2, . . . , Avk are linearly independent.

2. Let

x1 =

122

, x2 =

254

, x3 =

132

,x4 =

274

and x5 =

110

Find a subset of {x1,x2,x3,x4,x5} that is a basis of span{x1,x2,x3,x4,x5}.

3. Let A and B be similar n× n matrices. Prove that there exist n× n matrices Sand T such that S is non-singular, A = ST and B = TS.

4. Let T : R2 → R3 be a linear transformation such that T

(53

)=

123

and

T

(32

)=

101

. Find T

(xy

)for all x, y ∈ R.

5. Let V be the set of all polynomials of degree at most three. For f(x), g(x) ∈ V ,we define the inner product of f(x) and g(x) as

〈f(x), g(x)〉 =

∫ 1

0

f(x)g(x) dx

Find a basis for the subspace W of V of all elements in V that are orthogonal to1− x.

6. Is the matrix

1 0 0−2 −1 −32 2 4

diagonalizable? Justify your answer!

7. Let A be an n × n matrix, λ, µ two different eigenvalues of A, x an eigenvectorfor A corresponding to the eigenvalue λ and y an eigenvector for AT correspondingto the eigenvalue µ. Prove that x and y are orthogonal.

8. Let n be odd and A an n× n matrix whose entries are real numbers. Prove thatA2 + I 6= O where I is the n × n identity matrix and O is the n × n zero matrix(hint : determinants).


Part A.

1. Let θ : Z → S5 be a group homomorphism such that θ(1) = (123)(45). Findθ(−4) and Ker θ.

Solution. θ is a homomorphism going from an additive group into a multiplicativegroup. Thus the homomorphism properties look like

θ(n+m) = θ(n)θ(m) and θ(−n) = θ(n)−1

So, as Z is cyclic with generator 1, and we know what θ(1) is, then

θ(n) = θ(1)n = [(123)(45)]n = (123)n(45)n

where the last step is obtained using that (123) and (45) are disjoint. Then,

θ(−4) = (123)−4(45)−4 = (321)4(45)4 = (321)

Ker(θ) = {n ∈ Z; θ(n) = e}= {n ∈ Z; [(123)(45)]n = e}= {n ∈ Z; n is divisible by the order of (123)(45)}= {n ∈ Z; 6 divides n} → (the order of (123)(45) is 3 · 2 = 6)

= 6Z

2. Let G be the set of all real-valued functions defined on R. Then (G,+) is a groupwhere + stands for the usual addition of functions. Put N = {f ∈ G | f(2008) = 0}.Prove that N �G and G/N ∼= R.

Solution. Consider the map φ : G→ R defined by φ(f) = f(2008).Let f, g ∈ G, then

φ(f + g) = (f + g)(2008) = f(2008) + g(2008) = φ(f) + φ(g)

So, φ is a homomorphism of (additive) groups.For any given y ∈ R consider the map fy : R → R defined by fy(x) = y for allx ∈ R. Since, fy ∈ G and fy(2008) = y, then φ is onto.It is easy to see that Ker(φ) = N . Hence N � G and, by the first isomorphismtheorem, G/N ∼= R.

3. TRUE/FALSE : S6∼= S3 ⊕ S5. Prove your answer!

Solution. False. Note that the element ((123), (12345)) ∈ S3 ⊕ S5 has order 15.On the other hand, S6 does not have elements of order 15.

4. Let G be a finite group and k a natural number that is relatively prime to |G|.Prove that the map θ : G→ G : x→ xk is a bijection.

Solution. Since |G| = n <∞, then we just need to show that θ is onto.We know that there are α, β ∈ Z such that 1 = αk + βn, then

x = xαk+βn = xαkxβn = xαk(xn)β = xαk

for all x ∈ G. So, θ(xα) = x.

5. Let G be a finite group, N � G and H ≤ G such that |H| and [G : N ] arerelatively prime. Prove that H ≤ N .

Solution. Let h ∈ H, then the order of hN ∈ G/N divides the order of h. It followsthat the order of hN is a common divisor of [G : N ] and |H|. Hence, the order ofhN is one, and thus h ∈ N .

6. Let G = R \ {−1}. We define a binary operation ∗ on G as follows :

a ∗ b = ab+ a+ b for all a, b ∈ G

It is given that G is a group under this operation.

(a) What is the identity element in G?

(b) What is the inverse of a ∈ G?

(c) Solve for x : 2 ∗ x ∗ 3 = 7

Solution. First notice that the operation is commutative.

(a) We want to find e such that a = a ∗ e = ae + a + e for all a ∈ G. It followsthat

0 = ae+ e = e(a− 1)

for all a ∈ G. It follows that e = 0.

(b) We look for b such that 0 = e = a ∗ b = ab+ a+ b. We solve for b and we get

b =−aa+ 1

Note that we are using that a+ 1 6= 0, for all a ∈ G

(c) Since 2 ∗ 3 = 2 · 3 + 2 + 3 = 11, then the equation becomes

11 ∗ x = 7

The inverse of 11 is

11−1 =−11

12

So,

x =−11

12∗ 7 =

−11

12· 7 +

−11

12+ 7 =

−22

3+ 7 = −1

3

7. Let R be a ring such that a2 = a for all a ∈ R.

(a) Prove that a+ a = 0 for all a ∈ R (hint : consider (a+ a)2).

(b) Prove that R is commutative (hint : consider (a+ b)2).

Solution.

(a) Using that a2 = a for all a ∈ R we get

a+ a = (a+ a)2 = a2 + a+ a+ a2 = 2(a+ a)

So, a+ a = 0, or a = −a.

(b) We now consider

a+ b = (a+ b)2 = a2 + ab+ ba+ b2 = a+ ab+ ba+ b

So, 0 = ab+ ba. This together with part (a) imply ab = ba for all a, b ∈ R

8. Let R be the set of all polynomials with real coefficients. Then (R,+, ·) is aring under the usual addition and multiplication of polynomials. Put S = {p(x) ∈R | p ′(0) = 0} where p ′(x) is the usual derivative of p(x) with respect to x.

(a) Prove that S is a subring of R.


Solution.

(a) Let p, q ∈ S, then(p− q)′(0) = p′(0)− q′(0) = 0

and(pq)′(0) = p′(0)q(0) + p(0)q′(0) = 0

So, both closure laws and the existence of additive inverses hold in S. It isclear that S is nonempty, as the zero polynomial is in S.

(b) Let p(x) = x2 + 1 ∈ S and q(x) = x ∈ R we get that p(x)q(x) = x3 + x and(pq)′(x) = 3x2 + 1 which is nonzero at x = 0.

So, S is not an ideal of R.

Part B.

1. Let v1,v2, . . . ,vk be linearly independent vectors in Rn and A a non-singularn× n matrix. Prove that Av1, Av2, . . . , Avk are linearly independent.

Solution. Let α1, α2, · · · , αk ∈ R and consider

0 = α1Av1 + α2Av2 + · · ·+ αkAvk

= A(α1v1) + A(α2v2) + · · ·+ A(αkvk)

= A(α1v1 + α2v2 + · · ·+ αkvk)

which forces0 = α1v1 + α2v2 + · · ·+ αkvk

because A is non-singular. But, since the vi’s are linearly independent, then all theαi’s are zero.

2. Let

x1 =

122

, x2 =

254

, x3 =

132

,x4 =

274

and x5 =

110

Find a subset of {x1,x2,x3,x4,x5} that is a basis of span{x1,x2,x3,x4,x5}.

Solution. Call S the space spanned by the xi’s.Note that x4 − x2 = 2e2. Since we are in R, then we have e2 ∈ S. Subtracting thisfrom x5 we obtain e1 ∈ S. Now that we have e1 and e2 we obtain e3 from eitherx4 or x2. It follows that S contains the canonical basis of R3, and thus S = R3.Moreover, we obtained the canonical basis by only playing with x2,x4, and x5, thenthese three vectors define a basis of S (three generators in a space of dimensionthree).

3. Let A and B be similar n× n matrices. Prove that there exist n× n matrices Sand T such that S is non-singular, A = ST and B = TS.

Solution. Since A is similar to B, then there is an invertible matrix P such thatA = PBP−1.Let S = P (invertible) and T = BP−1, then A = ST and B = TS.

4. Let T : R2 → R3 be a linear transformation such that T

(53

)=

123

and

T

(32

)=

101

. Find T

(xy

)for all x, y ∈ R.

Solution. We need to write any vector v =

(xy

)as a linear combination of

v1 =

(53

)and v2 =

(32

).

So we set v = αv1 + βv2, which yields the system of equations (the unknowns are αand β)

x = 5α + 3β y = 3α + 2β

It follows that α = 2x− 3y and β = 5y − 3x. Hence,

T

(xy

)= T

[(2x− 3y)

(53

)+ (5y − 3x)

(32

)]= (2x− 3y)T

(53

)+ (5y − 3x)T

(32

)

= (2x− 3y)

123

+ (5y − 3x)

101

=

−x+ 2y4x− 6y3x− 4y

5. Let V be the set of all polynomials of degree at most three. For f(x), g(x) ∈ V ,we define the inner product of f(x) and g(x) as

〈f(x), g(x)〉 =

∫ 1

0

f(x)g(x) dx

Find a basis for the subspace W of V of all elements in V that are orthogonal to1− x.

Solution. Let p(x) = δx3 + ax2 + bx+ c ∈ W .

〈p(x), 1− x〉 =

∫ 1

0

p(x)(1− x) dx

=

∫ 1

0

−δx4 + (δ − a)x3 + (a− b)x2 + (b− c)x+ c dx

=

(−δ

5x5 +

δ − a4

x4 +a− b

3x3 +

b− c2

x2 + cx+K

)∣∣∣∣10

= −δ5

+δ − a

4+a− b

3+b− c

2+ c

Since we want p(x) to be orthogonal to 1− x, then we are asking for

−δ5

+δ − a

4+a− b

3+b− c

2+ c = 0

or

c = −(δ

10+a

6+b

3

)Thus,

p(x) = δx3 + ax2 + bx−(δ

10+a

6+b

3

)= δ

(x3 − 1

10

)+ a

(x2 − 1

6

)+ b

(x− 1

3

)Hence, a basis for W is {

x3 − 1

10, x2 − 1

6, x− 1

3

}

6. Is the matrix

1 0 0−2 −1 −32 2 4

diagonalizable? Justify your answer!

Solution. The characteristic polynomial of the matrix above (let’s call it A) is

χA(λ) =

∣∣∣∣∣∣1− λ 0 0−2 −1− λ −32 2 4− λ

∣∣∣∣∣∣= (1− λ)

∣∣∣∣ −1− λ −32 4− λ

∣∣∣∣= (1− λ)(λ2 − 3λ− 10)

= (1− λ)(λ− 5)(λ+ 2)

Since all A’s eigenvalues are distinct, then A is diagonalizable.

7. Let A be an n × n matrix, λ, µ two different eigenvalues of A, x an eigenvectorfor A corresponding to the eigenvalue λ and y an eigenvector for AT correspondingto the eigenvalue µ. Prove that x and y are orthogonal.

Solution. We know that Ax = λx (which implies xTAT = λxT ) and that ATy =µy.Note that

λ(x · y) = λ(xTy)

= (λxT )y

= (xTAT )y

= xT (ATy)

= xT (µy)

= µ(xTy)

= µ(x · y)

Since λ 6= µ, then λ(x · y) = µ(x · y) forces x · y = 0.

8. Let n be odd and A an n× n matrix whose entries are real numbers. Prove thatA2 + I 6= O where I is the n × n identity matrix and O is the n × n zero matrix(hint : determinants).

Solution. If A2 + I = O, then A2 = −I, and thus det(A2) = det(−I).Since n is odd det(−I) = −1, then we get det(A)2 = −1, which forces det(A) /∈ R.This contradicts the fact that the entries of A are real.



1. TRUE/FALSE : Let G be a group and a, b ∈ G. If ab has order 3 then ba hasorder 3.

Prove your answer!

2. TRUE/FALSE : Let R be an integral domain and A a proper ideal of R. ThenR/A is an integral domain.

Prove your answer!

3. TRUE/FALSE : D12∼= Z3 ⊕D4

Prove your answer!

4. Let G be a group and H and K finite subgroups of G such that |H| and |K| arerelatively prime. Prove that H ∩K = {1}.

5. Let G be a group. Consider the map f : G → G : a → a−1. Prove that G isabelian if and only if f is a group homomorphism.

6. Let R∗ be the group of nonzero real numbers under multiplication and

H = {g ∈ R∗ |gm ∈ Q for some nonzero integer m}Prove that H is a subgroup of R∗.

7. Find an element of order 10 in A9. Prove that the order is indeed 10.

8. Let R be the set of 2× 2-matrices with real entries :

R =

{[a bc d

]: a, b, c, d ∈ R

}Then R forms a ring under matrix addition and matrix multiplication. Put

S =

{[a 0c d

]∈ R

}• Prove that S is a subring of R.

• Is S an ideal of R?



1. Let A be an n× n-matrix. Prove that det(AAT ) ≥ 0.

2. Let u be a fixed vector in Rn. Show that the set of all vectors in Rn that areorthogonal to u is a subspace of Rn.

3. Let A =

[1 32 2

]. Find a matrix P such that P−1AP is a diagonal matrix.

4. Let A be an n × n-matrix and λ and eigenvalue of A. Prove that λk is aneigenvalue of Ak for all positive integers k.

5. Let L : V → W be a linear transformation. If {v1,v2, . . . ,vk} spans V , showthat {L(v1), L(v2), . . . , L(vk)} spans range(L).

6. Find an orthogonal basis for

S = span{[

1 1 0 1]T,[

0 1 −1 1]T,[

1 0 1 1]T}

7. Let A and B be symmetric n × n-matrices. Prove that AB is symmetric if andonly if AB = BA.

8. Suppose that {v1,v2, . . . ,vk} is a set of linearly independent vectors in Rn. Provethat

{v1,v1 + v2,v1 + v2 + v3, . . . ,v1 + v2 + · · ·+ vk}

is also linearly independent.


Part A.

1. TRUE/FALSE : Let G be a group and a, b ∈ G. If ab has order 3 then ba hasorder 3.

Solution. Note that

(ba)3 = b(ab)2a = b(ab)−1a = b(b−1a−1)a = e

Then, the order of ba is one or three. If it were one, then b = a−1, which contradictsab having order 3.

2. TRUE/FALSE : Let R be an integral domain and A a proper ideal of R. ThenR/A is an integral domain.

Solution. False. For n not a prime number, the ring Z/nZ is not an integraldomain.

3. TRUE/FALSE : D12∼= Z3 ⊕D4

Solution. False. D12 has 13 elements of order 2, but since D4 has exactly 5 elementsof order 2 and Z3 has none, then Z3 ⊕D4 has just 5 elements of order 2.

4. Let G be a group and H and K finite subgroups of G such that |H| and |K| arerelatively prime. Prove that H ∩K = {1}.

Solution. We know that the intersection of two subgroups, H and K, is also asubgroup (that is a subgroup of both H and K). So, the order of H ∩ K dividesboth |H| and |K|, thus |H ∩K| must be one, as (|H|, |K|) = 1.

5. Let G be a group. Consider the map f : G → G : a → a−1. Prove that G isabelian if and only if f is a group homomorphism.

Solution. We know that (ab)−1 = b−1a−1.Assuming f is a homomorphism

b−1a−1 = (ab)−1 = f(ab) = f(a)f(b) = a−1b−1

Since every element in a group has an inverse, then cd = dc for all c, d ∈ G

Assuming that G is Abelian.

f(ab) = (ab)−1 = b−1a−1 = a−1b−1 = f(a)f(b)

6. Let R∗ be the group of nonzero real numbers under multiplication and

H = {g ∈ R∗ |gm ∈ Q for some nonzero integer m}

Prove that H is a subgroup of R∗.

Solution. Let g, h ∈ H, then there are integers n,m such that gn ∈ Q and hm ∈ Q.Now consider

(gh−1)mn = gmn(h−1)mn = (gn)m(hm)−n

which is in Q because Q∗ is a multiplicative group.Since 1 ∈ H, then H is a subgroup of R∗.

7. Find an element of order 10 in A9. Prove that the order is indeed 10.

Solution. Consider σ = (12)(34)(56789).Since

σ = (12)(34567) = (12)(34)(59)(58)(57)(56)

then σ ∈ A9.The order of σ is 10 because the order of a product of disjoint cycles is the lcm ofthe orders of the cycles. In our case, the order of σ is lcm(2, 5) = 10.


R =

{[a bc d

]: a, b, c, d ∈ R

}Then R forms a ring under matrix addition and matrix multiplication. Put

S =

{[a 0c d

]∈ R

}• Prove that S is a subring of R.

• Is S an ideal of R?

Solution. This is problem number 4 in part A in the exam of Spring 2007.

Part B.

1. Let A be an n× n-matrix with entries in R. Prove that det(AAT ) ≥ 0.

Solution. We know that det(A) = det(AT ), and that det(AB) = det(A) det(B).So,

det(AAT ) = det(A) det(AT ) = det(A)2

Since det(A) ∈ R, then det(AAT ) = det(A)2 ≥ 0.

2. Let u be a fixed vector in Rn. Show that the set of all vectors in Rn that areorthogonal to u is a subspace of Rn.

Solution. Let v and w be two vectors that are orthogonal with u, and let α ∈ R.Then,

u · (v −w) = u · v − u · u = 0− 0 = 0

andu · (αv) = α(u · v) = α(0) = 0

Since the zero vector is orthogonal to u, then the set of orthogonal vectors to u isnon-empty.

3. Let A =

[1 32 2

]. Find a matrix P such that P−1AP is a diagonal matrix.

Solution. We first look at the characteristic polynomial of A

χA(λ) =

∣∣∣∣ 1− λ 32 2− λ

∣∣∣∣= (1− λ)(2− λ)− 6

= λ2 − 3λ− 4

= (λ− 4)(λ+ 1)

We know that there is a matrix P such that

P−1AP =

[−1 00 4

]In order to find P we need to find eigenvectors for the two eigenvalues of A.For λ = −1 we have to solve the equation Av = −v, which yields the system ofequations

x+ 3y = −x 2x+ 2y = −y

which has solution space spanned by (3,−2).For λ = 4 we have to solve the equation Av = 4v, which yields the system ofequations

x+ 3y = 4x 2x+ 2y = 4y

which has solution space spanned by (1, 1).It follows that

P =

[3 1−2 1

]4. Let A be an n × n-matrix and λ and eigenvalue of A. Prove that λk is aneigenvalue of Ak for all positive integers k.

Solution. Let v be an eigenvector of A associated to the eigenvalue λ, that isAv = λv.Now note that

Akv = Ak−1(Av) = Ak−1(λv) = λ(Ak−1v)

Repeating the process above we see that Akv = λkv.

5. Let L : V → W be a linear transformation. If {v1,v2, . . . ,vk} spans V , showthat {L(v1), L(v2), . . . , L(vk)} spans range(L).

Solution. Let v = α1v1 + α2v2 + · · ·+ αkvk ∈ V . Then

L(v) = L(α1v1 + α2v2 + · · ·+ αkvk) = α1L(v1) + α2L(v2) + · · ·+ αkL(vk)

So, every element in the range of L is a linear combination of the elements in theset {L(v1), L(v2), . . . , L(vk)}.

6. Find an orthogonal basis for

S = span{[

1 1 0 1]T,[

0 1 −1 1]T,[

1 0 1 1]T}

Solution. Note that v2 =[

0 1 −1 1]T

and v3 =[

1 0 1 1]T

are already

orthogonal. So, what we want to do is to replace v1 =[

1 1 0 1]T

with a vectorin S that is orthogonal to the last two.

Any element in S looks like v =[x+ z, x+ y, z − y, x+ y + z

]Tfor some

x, y, z ∈ R.Note that

v · v2 = x+ y + z − y + x+ y + z = 2x+ y + 2z

andv · v3 = x+ z + z − y + x+ y + z = 2x+ 3z

Since we want v to be orthogonal to both v2 and v3, then

2x+ y + 2z = 0 = 2x+ 3z

which forces y = z, and thus we get 0 = 2x+ 3z. We now may take x = 3 and theny = z = −2. Finally,

v =[

3− 2, 3− 2, 0, 3− 2− 2]T

=[

1 1 0 −1]T

So, {[1 1 0 −1

]T,[

0 1 −1 1]T,[

1 0 1 1]T}

is an orthogonal basis for S.Of course you may solve this using projections, but I thought this way was muchmore fun.

7. Let A and B be symmetric n × n-matrices. Prove that AB is symmetric if andonly if AB = BA.

Solution. Assuming AB = BA, AT = A and BT = B.

(AB)T = BTAT = BA = AB

So, AB is symmetric.Now assuming that AB is symmetric, AT = A and BT = B.

AB = (AB)T = BTAT = BA

8. Suppose that {v1,v2, . . . ,vk} is a set of linearly independent vectors in Rn. Provethat

{v1,v1 + v2,v1 + v2 + v3, . . . ,v1 + v2 + · · ·+ vk}

is also linearly independent.

Solution. This is problem 1 in part B in the exam of Spring 2007

Mathematics Department Qualifying Exam Spring 2007Subject: Algebra

Part A. Do five of the following eight problems.

1. Show that every subgroup of a cyclic group is cyclic.

2. Let G be a group of symmetries of the square with vertices A, B, C, D (that is, the groupof rigid motions of the plane which transform the square into itself). Let H1 ⊂ G be thesubgroup of G consisting of those symmetries which do not move the point A and H2 ⊂ Gthe subgroup of G consisting of rotations. Is H1 a normal subgroup of G? Is H2 a normalsubgroup of G? Justify your answer.

3. Prove that all groups of order 4 are abelian.


R =

{[a bc d

]: a, b, c, d ∈ R

}Then R forms a ring under matrix addition and multiplication. Let

S =

{[a 0c d

]∈ R

}(a) Prove that S is a subring of R.


5. Consider the map θ : R[x]→ R given by f(x) 7→ f ′(1), where f ′ is the derivative of f .

(a) Is θ a group homomorphism from (R[x],+) to (R,+)?

(b) Is θ a ring homomorphism from (R[x],+, ·) to (R,+, ·)?

6. Give an example of a finite group G and an integer n such that n divides the order of Gbut G has no subgroup of order n. Explain why G has no such subgroup.

7. (a) Show that every field is an integral domain.

(b) Give an example of an integral domain that is not a field. Explain.

8. Let Z5[x] be the ring of polynomials over the finite field Z5.

(a) Show that f(x) = x3 + 3x+ 2 is irreducible over Z5.

(b) Express g(x) = x4 + 4 as a product of irreducible polynomials in Z5[x].


Part B. Do five of the following eight problems.

1. Suppose that x1,x2, . . . ,xk are vectors in Rn.

(a) If y = a1x1 + a2x2 + · · · + akxk where a1 6= 0, show that span{x1,x2, . . . ,xk} =span{y,x2, . . . ,xk}.

(b) If {x1,x2, . . . ,xk} is independent, show that {x1,x1 + x2,x1 + x2 + x3, . . . ,x1 + x2 +· · ·+ xk} is also independent.

2. Let V be a vector space.

(a) Show that if S = {v1,v2, . . . ,vr} is a linearly independent set of vectors in V , thenso is every non-empty subset of S.

(b) Show that if {v1,v2, . . . ,vr} is a linearly dependent set of vectors in V and vr+1, . . . ,vn

are any vectors in V , then {v1,v2, . . . ,vr,vr+1, . . . ,vn} is also linearly dependent.

3. Show that A =

[1 3−3 −5

]is not diagonalizable.

4. Let T : R2 → R2 be the linear transformation given by T (x, y) = (x+ ky,−y). Show thatT is one-to-one for every real value of k and that T−1 = T .

5. Let v1 = 〈0, 1, 0〉, v2 = 〈−45, 0, 3

5〉, and v3 = 〈3

5, 0, 4

5〉.

(a) Check that S = {v1,v2,v3} is an orthonormal basis for R3 with the Euclidean innerproduct.

(b) Express the vector u = 〈1, 1, 1〉 as a linear combination of the vectors in S and findthe coordinates of u with respect to the basis S.

6. Find the characteristic polynomial, eigenvalues, and eigenvectors for A =

1 1 10 2 −10 −3 0

.

7. Suppose that L : R3 → R2 is given by L(x, y, z) = (x+ 1, y− z). Is L a linear transforma-tion? Explain.

8. Compute the rank and nullity of A =

1 2 3−1 2 1

3 1 2

.


Part A.

1. Show that every subgroup of a cyclic group is cyclic.

Solution. Let G =< g >, and H a subgroup of G. Let i be the least positive integer suchthat gi ∈ H.

Let h ∈ H, then h = gj for some i ≤ j. Using the division algorithm we get that

j = iq + r

for some integers q, r such that 0 ≤ r < i.

Note that gj = giq+r = giqgr, thus (giq)−1gj = gr, which forces gr ∈ H (as both (giq)−1

and gj live in H). This yields a contradiction unless r = 0. It follows that i divides j, andthus H =< gi >.

2. Let G be a group of symmetries of the square with vertices A, B, C, D (that is, the groupof rigid motions of the plane which transform the square into itself). Let H1 ⊂ G be thesubgroup of G consisting of those symmetries which do not move the point A and H2 ⊂ Gthe subgroup of G consisting of rotations. Is H1 a normal subgroup of G? Is H2 a normalsubgroup of G? Justify your answer.

Solution. The group of symmetries of the square is called D4, its order is 8.

The group of all rotations, H2, is cyclic of order 4. Since its index is two in D4, then it isnormal in D4.

Note that H1 can be considered as the group generated by a reflection (that fixes A), thusH1 ∩H2 = {e} (otherwise a rotation would fix a vertex). It follows that D4 = H1H2. So,if H1 were normal, then D4 would be Abelian. However, D4 is not Abelian, as a clockwiserotation in 90 degrees followed by a reflection with diagonal axis ` is not the same as areflection with axis ` followed by a clockwise rotation in 90 degrees.

3. Prove that all groups of order 4 are abelian.

Solution. Let G = {e, a, b, c} be a group of order 4, and let g ∈ G.

If the order of some nonidentity element in G is 4, then G is cyclic of order four and thusAbelian.

If all nonidentity elements in G have order two. Note that ab cannot be a or b (that wouldforce the existence of a second identity), thus ab = c, similarly ba = c. The same argumentshows that ac = ca = b and that bc = cb = a. Hence, G is abelian.


R =

{[a bc d

]: a, b, c, d ∈ R

}Then R forms a ring under matrix addition and multiplication. Let

S =

{[a 0c d

]∈ R

}(a) Prove that S is a subring of R.


Solution.

(a) Let M,N ∈ S, then the difference M −N is clearly in S. Also, since the zero matrixis in S, then the only thing left to prove is that MN ∈ S. But this follows from[

a 0c d

] [x 0y z

]=

[ax 0

cx+ dy dz

]∈ S

(b) S is not an ideal of R because, for example, using that the identity is in S we get[1 00 1

] [1 23 4

]=

[1 23 4

]/∈ S

5. Consider the map θ : R[x]→ R given by f(x) 7→ f ′(1), where f ′ is the derivative of f .

(a) Is θ a group homomorphism from (R[x],+) to (R,+)?

(b) Is θ a ring homomorphism from (R[x],+, ·) to (R,+, ·)?

Solution.

(a) We need to check if θ preserves the additive group operation of R[x].

θ(p(x) + q(x)) = p′(1) + q′(1) = θ(p(x)) + θ(q(x))

So, θ is a group homomorphism from (R[x],+) to (R,+)

(b) θ is not a ring homomorphism, as for p(x) = x− 1 and q(x) = x2 we get

θ(p(x)q(x)) = p′(1)q(1) + p(1)q′(1) = 1

andθ(p(x))θ(q(x)) = p′(1)q′(1) = 2

6. Give an example of a finite group G and an integer n such that n divides the order of Gbut G has no subgroup of order n. Explain why G has no such subgroup.

Solution. We know that A5 is simple, and that it has order 60. If A5 had a subgroupof order 30, then that subgroup would have index two and, thus, it would be normal. Itfollows that A5 has no subgroup of order 30.

7. (a) Show that every field is an integral domain.

(b) Give an example of an integral domain that is not a field. Explain.

Solution.

(a) A field is clearly a commutative ring.

Let ab = 0, with a 6= 0. Since in a field every nonzero element has an inverse, thenab = 0 implies b = a−10 = 0, so b = 0. It follows that every field has no zero divisors.

(b) Z has no zero divisors and, for example, 2 has no inverse in Z

8. Let Z5[x] be the ring of polynomials over the finite field Z5.

(a) Show that f(x) = x3 + 3x+ 2 is irreducible over Z5.

(b) Express g(x) = x4 + 4 as a product of irreducible polynomials in Z5[x].

Solution.

(a) Since the degree of f(x) is three, then for it to be irreducible over Z5 it is enough tocheck that it has no zeros (roots) in Z5. This is easy to check as (all computationsin Z5!)

f(0) = 2 f(1) = 1 f(2) = 1 f(3) = 3 f(4) = 3

(b) Since 4 ≡ −1 (mod5) then g(x) = x4 − 1 ∈ Z5[x]. Factoring as usual we get

g(x) = (x2 − 1)(x2 + 1) = (x2 − 1)(x2 − 4) = (x− 1)(x+ 1)(x− 2)(x+ 2)

Since all the factors are linear we are done.

Part B.

1. Suppose that x1,x2, . . . ,xk are vectors in Rn.

(a) If y = a1x1 + a2x2 + · · · + akxk where a1 6= 0, show that span{x1,x2, . . . ,xk} =span{y,x2, . . . ,xk}.

(b) If {x1,x2, . . . ,xk} is independent, show that {x1,x1 + x2,x1 + x2 + x3, . . . ,x1 + x2 +· · ·+ xk} is also independent.

Solution.

(a) Note that y ∈ span{x1,x2, . . . ,xk}, then we clearly have

span{y,x2, . . . ,xk} ⊂ span{x1,x2, . . . ,xk}

Now letv = α1x1 + α2x2 + · · ·+ αkxk ∈ span{x1,x2, . . . ,xk}

then

v =α1

a1

(a1x1) + α2x2 + · · ·+ αkxk

=α1

a1

(a1x1 + a2x2 + · · ·+ akxk) + α2x2 + · · ·+ αkxk −α1

a1

(a2x2 + · · ·+ akxk)

=α1

a1

y +

(α2 −

α1

a1

)x2 + · · ·+

(αk −

α1

a1

)xk

which is an element of span{y,x2, . . . ,xk}. So,

span{x1,x2, . . . ,xk} ⊂ span{y,x2, . . . ,xk}

(b) Let

0 = α1x1 + α2(x1 + x2) + α3(x1 + x2 + x3) + · · ·+ αk(x1 + x2 + · · ·+ xk)

be a linear combination of the elements of the set {x1,x1 + x2,x1 + x2 + x3, . . . ,x1 +x2 + · · ·+ xk}.We re-write the previous equation as

0 = (α1 +α2 +α3 + · · ·+αk)x1 +(α2 +α3 + · · ·+αk)x2 + · · ·+(αk−1 +αk)xk−1 +αkxk

which is a linear combination of the elements of the set {x1,x2, . . . ,xk}. Since thisset is linearly independent, then

0 = α1 +α2 +α3 + · · ·+αk 0 = α2 +α3 + · · ·+αk 0 = αk−1 +αk 0 = αk

But, 0 = αk plugged into 0 = αk−1 + αk yields 0 = αk−1, and then we can repeatthis process as many times as necessary to get that all the αi’s have to be zero! Thisimplies that the set {x1,x1 + x2,x1 + x2 + x3, . . . ,x1 + x2 + · · · + xk} is linearlyindependent.

2. Let V be a vector space.

(a) Show that if S = {v1,v2, . . . ,vr} is a linearly independent set of vectors in V , thenso is every non-empty subset of S.

(b) Show that if {v1,v2, . . . ,vr} is a linearly dependent set of vectors in V and vr+1, . . . ,vn

are any vectors in V , then {v1,v2, . . . ,vr,vr+1, . . . ,vn} is also linearly dependent.

Solution.

(a) If a non-empty subset T of S were linearly dependent, then there would be scalars(not all zero) that could be used to write a linear combination of the elements in Tthat is equal to zero. So, now we could extend this linear combination to a linearcombination of the elements in S by just multiplying by zero the elements in S \ T .This would yield a non-trivial linear combination of the elements of S that is equalto zero. A contradiction.

(b) Using the same idea used in part (a). If there is a non-trivial linear combination ofthe elements in {v1,v2, . . . ,vr} that is equal to zero, then we extend it to the set{v1,v2, . . . ,vr,vr+1, . . . ,vn} by just multiplying the elements vr+1, . . . ,vn by zero.

3. Show that A =

[1 3−3 −5

]is not diagonalizable.

Solution. The characteristic polynomial of A is

χA(λ) =

∣∣∣∣ 1− λ 3−3 −5− λ

∣∣∣∣= (1− λ)(−5− λ) + 9

= λ2 + 4λ+ 4

= (λ+ 2)2

We now look at the eigenspace associated to λ = −2. We need to solve the equationAv = −2v, which yields the system of equations

x+ 3y = −2x − 3x− 5y = −2y

which has solution space spanned by (1,−1).

It follows that λ = −2 has geometric multiplicity one and algebraic multiplicity two.Hence, A is not diagonalizable.

4. Let T : R2 → R2 be the linear transformation given by T (x, y) = (x+ ky,−y). Show thatT is one-to-one for every real value of k and that T−1 = T .

Solution. Let us look at the kernel of T .

Ker(T ) = {(x, y) ∈ R2; T (x, y) = (0, 0)}= {(x, y) ∈ R2; (x+ ky,−y) = (0, 0)}= {(x, y) ∈ R2; x+ ky = 0 and − y = 0}= {(0, 0)}

So, T is one-to-one.

It is easy to see that T ◦ T (x, y) = (x, y) for all (x, y) ∈ R2

5. Let v1 = 〈0, 1, 0〉, v2 = 〈−45, 0, 3

5〉, and v3 = 〈3

5, 0, 4

5〉.

(a) Check that S = {v1,v2,v3} is an orthonormal basis for R3 with the Euclidean innerproduct.

(b) Express the vector u = 〈1, 1, 1〉 as a linear combination of the vectors in S and findthe coordinates of u with respect to the basis S.

Solution.

(a) It is easy to check that the norm of v1,v2 and v3 is one. Also,

< v1,v2, >= 0 < v1,v3 >= 0 < v2,v3 >= −12

25+

12

25= 0

Since a set of pairwise orthogonal vectors must be linearly independent, then S isa set having three linearly independent vectors in a three-dimensional vector space,thus S is a basis.

(b) We want to find α, β and γ such that

u = αv1 + βv2 + γv3

Since v2 and v3 have a zero in their second component, then α = 1. So, we just needto solve the system

−4

5β +

3

5γ = 1

3

5β +

4

5γ = 1

It follows that β = −15

and γ == 75. So,

u =

(1,−1

5,7

5

)S

Express the vector u = 〈1, 1, 1〉 as a linear combination of the vectors in S and findthe coordinates of u with respect to the basis S.

6. Find the characteristic polynomial, eigenvalues, and eigenvectors for A =

1 1 10 2 −10 −3 0

.

Solution. We first find the characteristic polynomial of A

χA(λ) =

∣∣∣∣∣∣1− λ 1 1

0 2− λ −10 −3 −λ

∣∣∣∣∣∣= (1− λ)

∣∣∣∣ 2− λ −1−3 −λ

∣∣∣∣= (1− λ)(λ2 − 2λ− 3)

= (1− λ)(λ− 3)(λ+ 1)

Since all the eigenvalues of A are distinct (they are λ = −1, 1, and 3). Then A is diago-nalizable to −1 0 0

0 1 00 0 3

Now we look for eigenvectors. First of all, just by looking at A we realize that Ae1 = e1,so that would be an eigenvector associated to the eigenvalue λ = 1.

For λ = −1 we need to solve the equation Av = −v, which yields the system of equations

x+ y + z = −x 2y − z = −y − 3y = −z

which has solution subspace spanned by (−2, 1, 3). Thus we can consider (−2, 1, 3) as aneigenvector for λ = −1.

For λ = 3 we need to solve the equation Av = 3v, which yields the system of equations

x+ y + z = 3x 2y − z = 3y − 3y = 3z

which has solution subspace spanned by (0, 1,−1). Thus we can consider (0, 1,−1) as aneigenvector for λ = 3.

7. Suppose that L : R3 → R2 is given by L(x, y, z) = (x+ 1, y− z). Is L a linear transforma-tion? Explain.

Solution. It is not. If this function were linear, then its kernel would be a subspace.However, since L(0, 0, 0) = (1, 0), then the zero vector is not in the ‘kernel’ of L. This isa contradiction.

8. Compute the rank and nullity of A =

1 2 3−1 2 1

3 1 2

.

Solution. Since the first two columns are linearly independent, then the rank is at leasttwo. Now note that the sum of the first and second column yield the third column plus

the vector

002

. This implies that the third column is not in the span of the first two

(the two zeros in the ‘extra vector’ do the work). It follows that the rank is three, andthus the nullity is zero.

Mathematics Department Qualifying Exam Fall 2006Subject: Algebra

Part A. Do five of the following eight problems.

1. Let R be the ring of all continuous real valued functions on the closed interval [0, 1]. Prove

that the map φ : R → R defined by φ(f) =∫ 1

0f(t) dt is a homomorphism of additive

groups but not a ring homomorphism.

2. Show that the symmetric group Sn (n ≥ 2) is generated by the 2-cycles (1 2), (2 3), . . . , (n−1 n).

3. Let R× denote the multiplicative group of nonzero real numbers and R denote the additivegroup of real numbers. Show that R× ∼= R× Z2.

4. An element x in a ring R is called nilpotent if xm = 0 for some m ∈ Z+. Let R be acommutative ring with 1 6= 0. Prove that if a is a nilpotent element of R, then 1− ab is aunit for all b ∈ R.

5. (a) Let H = {(1), (2 3)}. Is H normal in S3?

(b) What is the order of the element 14 + 〈8〉 in the quotient group Z24/〈8〉?

6. Prove that any subfield of R must contain Q.

7. Prove that if H and K are finite subgroups of G whose orders are relatively prime thenH ∩K = 1.

8. Consider the additive quotient group Q/Z.

(a) Show that every coset of Z in Q contains exactly one representative q ∈ Q in therange 0 ≤ q < 1.

(b) Show that every element of Q/Z has finite order but that there are elements ofarbitrarily large order.


Part B. Do five of the following eight problems.

1. Determine the dimension of the solution space (over the real numbers) to the system ofequations

x1 + x2 = 0x2 + x3 = 0

x3 + x4 = 0x4 + x5 = 0

−x1 + x5 = 0

2. Find a unit vector in R3 that is mutually perpendicular to v = (1, 2, 3) and w = (3, 2, 1).

3. Suppose

A =

1 −1 0 2−2 0 0 1−1 −1 0 3

.

Determine a basis over the reals for the image of the linear transformation L(v) = Av.

4. Determine the inverse of the matrix 0 0 0 10 0 2 00 3 0 04 0 0 0

5. Suppose A is an invertible matrix. Prove (At)

−1= (A−1)

twhere M t and M−1 represent

the transpose and inverse, respectively, of the matrix M .

6. Determine an orthonormal basis for span{(−1, 0, 0, 1), (0, 0, 1,−1), (1,−1, 0, 0)}.

7. State what it means (the definition) for a finite set of vectors to be linearly independent,and use this definition to prove the set of vectors {(1, 1, 0), (0, 1, 1), (1, 0, 1)} is linearlyindependent.

8. Determine the eigenvalues of the matrix 2 2 −53 7 −151 2 −4


Part A.

1. Let R be the ring of all continuous real valued functions on the closed interval [0, 1]. Prove

that the map φ : R → R defined by φ(f) =∫ 1

0f(t) dt is a homomorphism of additive

groups but not a ring homomorphism.

Solution. We know that ∫(f + g) dt =

∫f dt+

∫g dt

using this it is easy to show that φ is a homomorphism of additive groups. In fact

φ(f + g) =

∫ 1

0

[f(t) + g(t)] dt =

∫ 1

0

f(t) dt+

∫ 1

0

g(t) dt = φ(f) + φ(g)

However, for the functions f(t) = t and g(t) = t we get

φ(fg) =

∫ 1

0

t2 dt =1

36= 1

4=

(∫ 1

0

t dt

)2

= φ(f)φ(g)

2. Show that the symmetric group Sn (n ≥ 2) is generated by the 2-cycles (1 2), (2 3), . . . , (n−1 n).

Solution. First let us show that using these 2-cycles we can construct any 2-cycle. Callthe set of elements above T .

Let (a b) be a 2-cycle in Sn. WLOG assume b = a+ k + 1 for some positive integer k.

Note that (a a+ 1) is one of the cycles we can use, and that

(a+ 1 b) = (a a+ 1)(a b)(a a+ 1)

We now repeat this with (a a+ 2) to get

(a+2 b) = (a+1 a+2)(a+1 b)(a+1 a+2) = (a+1 a+2)(a a+1)(a b)(a a+1)(a+1 a+2)

... etcetera, until we get

(a+k b) = (a+k−1 a+k) · · · (a a+2)(a a+1)(a b)(a a+1)(a+1 a+2) · · · (a+k−1 a+k)

Since b = a+ k+ 1, then the left hand side is in T , and the right hand side is a product ofelements in T and (ab). If we move most of the things to the left hand side (multiplyingby inverses) we get

(a a+1)(a+1 a+2) · · · (a+k−1 a+k)(a+k b)(a+k−1 a+k) · · · (a+1 a+2)(a a+1) = (a b)

which means that (a b) is generated by elements in T .

Since every element in Sn can be written as a product of 2-cycles, then T generates all Sn.

3. Let R× denote the multiplicative group of nonzero real numbers and R denote the additivegroup of real numbers. Show that R× ∼= R× Z2.

Solution. Think Z2 as the multiplicative group {1,−1}. Define a function sign(x) :R× → Z2 by sign(x) = 1 if x is positive, and sign(x) = −1 if x is negative.

Note that sign(xy) = sign(x)sign(y) for all non-zero real numbers x and y.

Consider the map φ : R× → R× Z2 defined by φ(x) = (ln |x|, sign(x)) (I know it is ugly-defined as I am using additive notation for the first component and multiplicative for thesecond).

Now note that

φ(xy) = (ln |xy|, sign(xy)) = (ln |x|+ln |y|, sign(x)sign(y)) = (ln |x|, sign(x))(ln |y|, sign(y))

So, φ is a homomorphism. It is easy to see that φ is onto using that ln |x| is onto R.Moreover,

Ker(φ) = {x ∈ R×; ln |x| = 0 and sign(x) = 1} = {1}

Hence, φ defines an isomorphism.

4. An element x in a ring R is called nilpotent if xm = 0 for some m ∈ Z+. Let R be acommutative ring with 1 6= 0. Prove that if a is a nilpotent element of R, then 1− ab is aunit for all b ∈ R.

Solution. First notice that if am = 0 for some m, then (ab)m = 0 for the same m (hereusing the ring is commutative). So, let’s call ab = x, we know x is nilpotent and that themth power kills it.

Now consider the product

(1− x)(1 + x+ x2 + · · ·+ xm−1) = 1− xm = 1

It follows that the inverse of 1− x is 1 + x+ x2 + · · ·+ xm−1 (whatever that is, anyway itwill be an element of the ring).

5. (a) Let H = {(1), (2 3)}. Is H normal in S3?

(b) What is the order of the element 14 + 〈8〉 in the quotient group Z24/〈8〉?

Solution.

(a) No, as (123)(23)(123)−1 = (13) /∈ H(b) The group H =< 8 > has the elements H = {0, 8, 16}.

Now I will compute the ‘powers’ of 14 all modulo 24

14 /∈ H14 + 14 = 28 ≡ 4 /∈ H14 + 14 + 14 ≡ 4 + 14 ≡ 18 /∈ H14 + 14 + 14 + 14 ≡ 4 + 4 ≡ 8 ∈ HHence, the order of 14+ < 8 > is 4

6. Prove that any subfield of R must contain Q.

Solution. A field always F has a one. By closure of the addition of F we can see thatall the integers must be in F (here we are using that Z ⊂ R). Since we need inverses forall non-zero elements in F , then all the elements of the form 1

x, for x ∈ Z must be in F

as well. Finally, using closure of the multiplication we can construct any rational as theproduct of two elements in F ,

a

b= a · 1

b

and thus Q ⊂ F .

7. Prove that if H and K are finite subgroups of G whose orders are relatively prime thenH ∩K = 1.

Solution. This is problem 4 part A in the exam of Fall 2007

8. Consider the additive quotient group Q/Z.

(a) Show that every coset of Z in Q contains exactly one representative q ∈ Q in therange 0 ≤ q < 1.

(b) Show that every element of Q/Z has finite order but that there are elements ofarbitrarily large order.

Solution.

(a) Let x ∈ Q, we denote the greatest integer that is less or equal to x as [x], thenx − [x] ∈ [0, 1). Since [x] is an integer, then x + Z = x − [x] + Z. So, everycoset of Z in Q contains at least one representative in [0, 1). Moreover, having tworepresentatives in [0, 1) will force a number x to have two decimal parts, that is notpossible.

(b) Let x = ab∈ Q. WLOG we take b > 0. Then, the order of x+ Z is at most b, as

a

b+ · · ·+ a

b(b times)

is equal to a ∈ ZSo, every element in Q/Z has finite order.

Now consider a number N (as large as you wish), we know there are infinitely manyprimes, thus there must be a prime number p that is larger than N . It is easy to seethat x = 1

phas order p > N .

Part B.

1. Determine the dimension of the solution space (over the real numbers) to the system ofequations

x1 + x2 = 0x2 + x3 = 0

x3 + x4 = 0x4 + x5 = 0

−x1 + x5 = 0

Solution. The last equation in the system says x5 = x1, the previous to the last saysx4 = −x5, which together with the last one imply x1 = −x4. Since x3 = −x4 we get thatx1 = x3.

The solution space of this system is spanned by (1,−1, 1,−1, 1). So, the dimension is one.

2. Find a unit vector in R3 that is mutually perpendicular to v = (1, 2, 3) and w = (3, 2, 1).

Solution. We will first find a vector that is perpendicular to v and w, then we willnormalize it.

The vector we are looking for is u = (x, y, z), it follows that

0 = (x, y, z) · (1, 2, 3) = x+ 2y+ 3z and 0 = (x, y, z) · (3, 2, 1) = 3x+ 2y+ z

So, subtracting these two equations we get 2x − 2z = 0, which implies x = z. Pluggingthis into one of the equations we get 4x + 2y = 0, which implies y = −2x. Hence, thesolution space for the system of equations above is spanned by (1,−2, 1).

So, any vector that is a multiple of (1,−2, 1) is orthogonal to both v and w. In particular,we take

1

|(1,−2, 1)|(1,−2, 1) =

1√6

(1,−2, 1)

which has norm one.

3. Suppose

A =

1 −1 0 2−2 0 0 1−1 −1 0 3

.

Determine a basis over the reals for the image of the linear transformation L(v) = Av.

Solution. We know the image of L is spanned by the column vectors of the matrix A.Since one of the column vectors is the zero vector, then we just have to play with

B =

1 −1 2−2 0 1−1 −1 3

.

(which has determinant zero, thus we need to get rid of columns)

We will do some column operations on this matrix to find a basis for Im(L)

B =

1 −1 2−2 0 1−1 −1 3

now we add column 2 to column 1

=

0 −1 2−2 0 1−2 −1 3

now we add column 2 twice to column 3

=

0 −1 0−2 0 1−2 −1 1

Now we can see clearly that column 1 and column 3 are linearly dependent. So, a basisfor Im(L) is given by

{(−1, 0,−1), (0, 1, 1)}

4. Determine the inverse of the matrix0 0 0 10 0 2 00 3 0 04 0 0 0

Solution. Since this is an anti-diagonal matrix then its inverse is also anti-diagonal andgiven by

0 0 0 14

0 0 13

00 1

20 0

1 0 0 0

5. Suppose A is an invertible matrix. Prove (At)

−1= (A−1)

twhere M t and M−1 represent

the transpose and inverse, respectively, of the matrix M .

Solution. We know that (AB)t = BtAt, then

I = (A−1A)t = At(A−1

)tand

I = (AA−1)t =(A−1

)tAt

6. Determine an orthonormal basis for span{(−1, 0, 0, 1), (0, 0, 1,−1), (1,−1, 0, 0)}.

Solution. We will use the Gram-Schmidt process.

Note that the vectors v1 = (0, 0, 1,−1) and v2 = (1,−1, 0, 0) are already orthogonal. So,we just need to find the third vector

Then

v3 = (−1, 0, 0, 1)− (0, 0, 1,−1) · (−1, 0, 0, 1)

|(0, 0, 1,−1)|2(0, 0, 1,−1)− (1,−1, 0, 0) · (−1, 0, 0, 1)

|(1,−1, 0, 0)|2(1,−1, 0, 0)

= (−1, 0, 0, 1) +1

2(0, 0, 1,−1) +

1

2(1,−1, 0, 0)

=

(−1

2,−1

2,1

2,1

2

)

So, v1, v2 and v3 are orthogonal and span the same subspace the original three vectorsspanned. Now we normalize these vectors by dividing them by their norm. The finalanswer is {

1√2

(0, 0, 1,−1),1√2

(1,−1, 0, 0),

(−1

2,−1

2,1

2,1

2

)}

7. State what it means (the definition) for a finite set of vectors to be linearly independent,and use this definition to prove the set of vectors {(1, 1, 0), (0, 1, 1), (1, 0, 1)} is linearlyindependent.

Solution. A set of vectors is linearly independent if the only linear combination of thesevectors that yields the zero vector is the trivial one (i.e. all scalars must be zero).

Assume that

(0, 0, 0) = α(1, 1, 0) + β(0, 1, 1) + γ(1, 0, 1) = (α + γ, α + β, β + γ)

then we get the system

α + γ = 0 α + β = 0 β + γ = 0

which has only one solution α = β = γ = 0.

8. Determine the eigenvalues of the matrix 2 2 −53 7 −151 2 −4

Solution. Let’s play with this determinant

χA(λ) =

∣∣∣∣∣∣2− λ 2 −5

3 7− λ −151 2 −4− λ

∣∣∣∣∣∣ now we multiply row 1 by 3

=1

3

∣∣∣∣∣∣6− 3λ 6 −15

3 7− λ −151 2 −4− λ

∣∣∣∣∣∣ now we subtract row 2 from row 1

=1

3

∣∣∣∣∣∣3− 3λ −1 + λ 0

3 7− λ −151 2 −4− λ

∣∣∣∣∣∣ now we multiply column 2 by 3

=1

32

∣∣∣∣∣∣3− 3λ −3 + 3λ 0

3 21− 3λ −151 6 −4− λ

∣∣∣∣∣∣ now we add column 1 to column 2

=1

32

∣∣∣∣∣∣3− 3λ 0 0

3 24− 3λ −151 7 −4− λ

∣∣∣∣∣∣=

1

32(3− 3λ)

∣∣∣∣ 24− 3λ −157 −4− λ

∣∣∣∣= (1− λ)

∣∣∣∣ 8− λ −57 −4− λ

∣∣∣∣= (1− λ)[(8− λ)(−4− λ) + 35]

= (1− λ)(λ2 − 4λ) + 3)

= (1− λ)(λ− 3)(λ− 1)

So, the eigenvalues are λ = 1 (with multiplicity 2) and λ = 3 (with multiplicity one).


Part A. Do five of the following 8 problems.

1. Show that a nonzero finite commutative ring R with no zero-divisors has a multiplicativeidentity.

2. Let G = {x ∈ R| x > 0 and x 6= 1}. Define the operation ∗ on G by a ∗ b = aln b, for alla, b ∈ G. Prove that G is an Abelian group under the operation ∗.

3. Let R[x] denote the ring of all polynomials with real coefficients. Also, let a ∈ R, andlet f(x) ∈ R[x], with derivative f ′(x). Show that the remainder when f(x) is divided by(x− a)2 is f ′(a)(x− a) + f(a).

4. Let φ : G1 → G2 and θ : G2 → G3 be group homomorphisms. Prove that θφ : G1 → G3 isa homomorphism. prove that ker(φ) ⊂ ker(θφ).

5. Let N be a subgroup of the center of G. Show that if G/N is a cyclic group, then G mustbe Abelian.

6. Show that a relation on R+ defined by x ∼ y iff xy = yx is an equivalence relation.

7. Let F be a field and let φ : F → R be a ring homomorphism. Show that φ is either zeroor one-to-one.

8. Let Sn denote the symmetric group of degree n and let An denote the alternating groupof degree n. For any elements σ, τ ∈ Sn show that στσ−1τ−1 ∈ An.


Part B. Do five of the following 8 problems.

1. Consider the linear transformation with matrix A =

1 0 −2 00 1 3 00 0 0 1

. Find a basis for the

kernel and a basis for the image of the transformation.

2. Let A be an m× n matrix. Consider the set

W ={v ∈ Rn

∣∣ Av = 0}

Prove that W a subspace of Rn.

3. Let A =

[cos θ − sin θsin θ cos θ

]. Prove that the linear transformation T (x) = Ax represents a

rotation of the vector x by an angle of θ.

4. Find an orthonormal basis for the subspace of R4 spanned by the vectors u1 = 〈0,−1, 0, 0〉,u2 = 〈3, 0, 1, 0〉, and u3 = 〈1, 1, 0, 1〉.

5. Let A be an invertible n×n matrix, and let c be a nonzero real number. Prove or disprove:(cA)−1 = 1

cA−1.

6. Let Am×n and Bn×p be matrices. Prove that (AB)T = BTAT , where AT denotes thetranspose of the matrix A.

7. Let A =

0 0 −20 −1 2−1 0 1

. Find the eigenvalues and corresponding eigenvector(s) of A.

8. Let A be an invertible n× n matrix. Prove that det(A−1) =1

det(A).


Part A.

1. Show that a nonzero finite commutative ring R with no zero-divisors has a multiplicativeidentity.

Solution. Fix an element r ∈ R∗. Consider the map φ : R→ R defined by φ(x) = xr.

Note thatφ(x) = φ(y) implies r(x− y) = 0

Since R has no zero divisors, then φ is one-to-one. But, as R is finite then φ is bijective.It follows there is an element xr ∈ R such that rxr = r.

using that φ is onto we represent an element y ∈ R as y = rx for some x ∈ R, then

xry = xr(rx) = (xrr)x = rx = y

So, xr works as the identity for ALL elements in R.

2. Let G = {x ∈ R| x > 0 and x 6= 1}. Define the operation ∗ on G by a ∗ b = aln b, for alla, b ∈ G. Prove that G is an Abelian group under the operation ∗.

Solution. Let a, b ∈ G, then aln b is also a positive real number (a is positive). Moreover,since ln b 6= 0 (as b 6= 1) then aln b 6= 1. So, closure for the product works out.

What is the identity? It is e!! (meaning e ∼ 2.71)... confusing as the standard notationfor the identity in a group is also e. To check this just notice that

eln b = b and aln e = a1 = a

What is the inverse of a ∈ G? We look for an element b such that a ∗ b = e, or in otherwords

aln b = e

Applying natural log both sides we get

ln(aln b)

= ln e

which isln a ln b = 1

So, b is given by the unique positive number such that ln b is the multiplicative inverse ofln a.

Finally, we check that the group is Abelian

a ∗ b = aln b = (eln a)ln b = (eln b)ln a = bln a

3. Let R[x] denote the ring of all polynomials with real coefficients. Also, let a ∈ R, andlet f(x) ∈ R[x], with derivative f ′(x). Show that the remainder when f(x) is divided by(x− a)2 is f ′(a)(x− a) + f(a).

Solution. Since the degree of f ′(a)(x− a) + f(a) is one, then we just need to show that

f(x) = q(x)(x− a)2 + f ′(a)(x− a) + f(a)

for some q(x) ∈ R[x].

The remainder theorem says that

f(x) = p(x)(x− a) + f(a)

for some p(x) ∈ R[x].

we now derive the previous equation both sides to get

f ′(x) = p′(x)(x− a) + p(x)

So, f ′(a) = p(a)

We use the remainder theorem again with p(x) divided by (x− a) to get

p(x) = q(x)(x− a) + p(a)

for some q(x) ∈ R[x].

It follows that

f(x) = p(x)(x− a) + f(a)

= [q(x)(x− a) + p(a)](x− a) + f(a)

= q(x)(x− a)2 + p(a)(x− a) + f(a)

= q(x)(x− a)2 + f ′(a)(x− a) + f(a)

4. Let φ : G1 → G2 and θ : G2 → G3 be group homomorphisms. Prove that θφ : G1 → G3 isa homomorphism. prove that ker(φ) ⊂ ker(θφ).

Solution. Let g, h ∈ G1

θφ(gh) = θ(φ(gh)) = θ(φ(g)φ(h)) = θ(φ(g))θ(φ(h)) = θφ(g)θφ(h)

Hence, θφ is a group homomorphism

Now let g ∈ ker(φ), then

θφ(g) = θ(φ(g)) = θ(eG2) = eG3

So, g ∈ ker(θφ).

5. Let N be a subgroup of the center of G. Show that if G/N is a cyclic group, then G mustbe Abelian.

Solution. Let G/N =< gN >. Take two elements in G, x = gin and y = gjm, wherei, j ∈ Z and n,m ∈ N , then (don’t forget that both n and m live in the center of G)

xy = (gin)(gjm)

= gi(ngj)m

= gi(gjn)m

= (gigj)(nm)

= (gjgi)(mn)

= gj(gim)n

= gj(mgi)n

= (gjm)(gin) = yx

6. Show that a relation on R+ defined by x ∼ y iff xy = yx is an equivalence relation.

Solution. Since xy = yx implies yx = xy and xx = xx then ∼ is both reflexive andsymmetric.

Now assume that xy = yx and yz = zy (note that none of these elements is zero), then

xz = (xy)z/y = (yx)z/y = (yz)x/y = (zy)x/y = zx

Hence, transitivity also holds.

7. Let F be a field and let φ : F → R be a ring homomorphism. Show that φ is either zeroor one-to-one.

Solution. Since F is a field, then it has no proper ideals. However, the kernel of φ is anideal of F , thus ker(φ) = {0} (in which case φ is one-to-one) or ker(φ) = F , in the lattercase φ is the zero function.

8. Let Sn denote the symmetric group of degree n and let An denote the alternating groupof degree n. For any elements σ, τ ∈ Sn show that στσ−1τ−1 ∈ An.

Solution. Note that στ is even if and only if σ−1τ−1 is even. The result follows by notingthat the product of two permutations having the same parity is always even.

Part B.


1 0 −2 00 1 3 00 0 0 1

. Find a basis for the

kernel and a basis for the image of the transformation.

Solution. Set Ax = 0. Then the solution space is

2z−3zz0

∣∣∣∣z ∈ R

. Therefore it is of

dimension 1, and a basis is

2−3

10

.

Since A is 3× 4, the image is a subspace of R3. Moreover, the dimension of the image is 3since the dimension of the kernel is 1. A 3-dimensional subspace of R3 must be R3 itself,

so we may use the standard basis

1

00

,0

10

,0

01

.

2. Let A be an m× n matrix. Consider the set

W ={v ∈ Rn

∣∣ Av = 0}

Prove that W a subspace of Rn.

Solution. Let u, v ∈ W and let c ∈ R. Then

(i) A(u + v) = Au + Av = 0 + 0 = 0. Therefore u + v ∈ W .

(ii) A(cu) = cAu = c · 0 = 0. Therefore cu ∈ W .

Since W is closed under addition and scalar multiplication, W is a subspace of Rn.

3. Let A =

[cos θ − sin θsin θ cos θ

]. Prove that the linear transformation T (x) = Ax represents a

rotation of the vector x by an angle of θ.

Solution. Let a =

[xy

]and let b = A

[xy

]=

[cos θx− sin θysin θx+ cos θy

]. We must show that

a · b = |a||b| cos θ.

We have |a| =√x2 + y2 and

|b| =√

(x cos θ − y sin θ)2 + (x sin θ + y cos θ)2

=

√x2 cos2 θ − 2xy cos θ sin θ + y2 sin2 θ + x2 sin2 θ + 2xy sin θ cos θ + y2 cos2 θ

=√x2 + y2.

Therefore

a · b = x2 cos θ − xy sin θ + xy sin θ + y2 cos θ

= (x2 + y2) cos θ

= |a||b| cos θ.

4. Find an orthonormal basis for the subspace of R4 spanned by the vectors u1 = 〈0,−1, 0, 0〉,u2 = 〈3, 0, 1, 0〉, and u3 = 〈1, 1, 0, 1〉.

Solution. Using the Gram-Schmidt process, we have that v1, v2, v3 is an orthogonalbasis, where v1 = u1 = 〈0,−1, 0, 0〉, v2 = u2 = 〈3, 0, 1, 0〉 (since u1 · u2 = 0), and v3 =〈1, 1, 0, 1〉 − −1

1〈0,−1, 0, 0〉 − 3

10〈3, 0, 1, 0〉 = 〈 1

10, 0,− 3

10, 1〉.

We have |v1| = 1, |v2| =√

10, and |v3| =√

110. Therefore an orthonormal basis is{〈0,−1, 0, 0〉, 1√

10〈3, 0, 1, 0〉, 1√

110〈 1

10, 0,− 3

10, 1〉}.

5. Let A be an invertible n×n matrix, and let c be a nonzero real number. Prove or disprove:(cA)−1 = 1

cA−1.

Solution. (cA)(

1cA−1

)= c · 1

cAA−1 = In. Therefore (cA)−1 = 1

cA−1.

6. Let Am×n and Bn×p be matrices. Prove that (AB)T = BTAT , where AT denotes thetranspose of the matrix A.

Solution. Let A = [aij] and B = [bij]. The (i, j)-entry of (AB)T is the (j, i)-entry of AB,

which is [jth row of A] · [ith column of B] = aj1b1i + . . .+ ajnbni =n∑

k=1

ajkbki.

On the other hand, the (i, j)-entry of BTAT is [ith row of BT ] · [jth column of AT ] =

b1iaj1 + . . .+ bniajn =n∑

k=1

ajkbki.

Since the corresponding entries are equal, we have that (AB)T = BTAT .

7. Let A =

0 0 −20 −1 2−1 0 1

. Find the eigenvalues and corresponding eigenvector(s) of A.

Solution. The characteristic polynomial is

p(λ) = λ ((λ+ 1)(λ− 1))− 2(λ+ 1)

= (λ+ 1) (λ(λ− 1)− 2)

= (λ+ 1)(λ2 − λ− 2)

= (λ+ 1)2(λ− 2);

therefore the eigenvalues are λ1 = −1, λ2 = 2.

For λ1 we have the system (−I − A)x = 0, which gives −1 0 2 00 0 −2 01 0 −2 0

which reduces to 1 0 −2 0

0 0 1 00 0 0 0

.A basis for the solution space is

0

10

. So an eigenvector for λ1 is v1 =

010

.

Similarly for λ2 we have the system (2I − A)x = 0, which gives 2 0 2 00 3 −2 01 0 1 0

which reduces to 1 0 1 0

0 3 −2 00 0 0 0

.A basis for the solution space is

−1

23

1

. So an eigenvector for λ2 is v2 =

−123

1

.

8. Let A be an invertible n× n matrix. Prove that det(A−1) =1

det(A).

Solution. We have 1 = det(In) = det(AA−1) = det(A) det(A−1). Therefore det(A−1) =1

det(A).



1. Prove that if G is any cyclic group, then G is abelian. (Your proof must include the casewhere G is infinite.)

2. Let Sn denote the group of permutations on the set {1, 2, . . . , n}, and let σ ∈ Sn be anodd permutation. Prove that σ−1 is an odd permutation.

3. Let N be a normal subgroup of an abelian group G. Prove that the factor group G/N isabelian.

4. Let G = Q − {−1} be the set of rational numbers except −1. For a, b ∈ G, let a ∗ b bedefined by a ∗ b = a+ b+ ab. Prove that (G, ∗) is a group.

5. Let H be a subgroup of a group G. For a, b ∈ G, define a relation on G by letting a ∼ bif and only if ab−1 ∈ H. Prove that ∼ is an equivalence relation.

6. Prove that the polynomial f(x) = x7−10x4+15x−5 is irreducible over the field of rationalnumbers.

7. An element a of a ring R is called a zero divisor if there exists a nonzero element b ∈ Rsuch that ab = 0. If a is a zero divisor of a commutative ring R and r ∈ R, prove that aris a zero divisor.

8. Let R and S be commutative rings, and let ϕ : R→ S be a non-trivial ring homomorphism.

(a) Prove that the kernel of ϕ is an ideal of R.

(b) If R is a field, prove that ϕ is one-to-one.



1. Let LA : Rn → Rn be the linear transformation given by LA(v) = Av where A is a realn× n matrix. Show that if n is odd then L has a real eigenvalue.

2. If v1,v2, . . . ,vn are distinct eigenvectors corresponding to distinct eigenvalues λ1, λ2, . . . , λn

of a matrix A, prove that {v1,v2, . . . ,vn} is a linearly independent set.

3. Let V be a real vector space of dimension n, and suppose that S = {v1,v2, . . . ,vt} is alinearly independent subset of V . Prove that there is a basis B of V such that S ⊆ B.

4. Let S be the set of all functions f : R→ R that satisfy the differential equation

y′′ − y′ + 2y = 0.

Is S a real vector space? (Assume the usual operations (f + g)(t) = f(t) + g(t) and(c · f)(t) = cf(t) where c ∈ R.) Explain why or why not.

5. Let A and B be n × n matrices. Show that AB is invertible if and only if A and B areinvertible.

6. Prove that a linear transformation of vector spaces L : V → W is one to one if and only ifL maps linearly independent subsets of V to linearly independent subsets of W .

7. Let V be a finite dimensional vector space and L : V → W be a linear transformation tovector space W . Prove that dim(kernel L)+ dim(image L) = dim V .

8. Let A be an n× n matrix. Prove that A is invertible if and only if the determinant of Ais non-zero.


Part A.

1. Prove that if G is any cyclic group, then G is abelian. (Your proof must include the casewhere G is infinite.)

Solution. Let G = 〈a〉, and let x, y ∈ G. Then x = ak and y = al for some k, l ∈ Z.Thus xy = akal = ak+l = alak = yx.

2. Let Sn denote the group of permutations on the set {1, 2, . . . , n}, and let σ ∈ Sn be anodd permutation. Prove that σ−1 is an odd permutation.

Solution. σ can be written as a product of an odd number of transpositions. Suppose

σ = (a1 b1)(a2 b2) · · · (a2k+1 b2k+1).

Then σ−1 = (a2k+1 b2k+1) · · · (a1 b1), again a product of an odd number of transpositions.Therefore σ−1 is odd.

3. Let N be a normal subgroup of an abelian group G. Prove that the factor group G/N isabelian.

Solution. Let aN , bN ∈ G/N . Then aNbN = abN = baN = bNaN .

4. Let G = Q − {−1} be the set of rational numbers except −1. For a, b ∈ G, let a ∗ b bedefined by a ∗ b = a+ b+ ab. Prove that (G, ∗) is a group.

Solution.

Closure. If a, b ∈ G, then clearly a+b+ab ∈ Q. If a+b+ab = −1 then a(1+b) = −(b+1),and hence a = − b+1

b+1= −1, a contradiction. Therefore G is closed under ∗.

Associativity. We have

(a ∗ b) ∗ c = (a+ b+ ab) ∗ c = (a+ b+ ab) + c+ (a+ b+ ab)c

= a+ b+ c+ ab+ ac+ bc+ abc

= a+ (b+ c+ bc) + a(b+ c+ bc) = a ∗ (b+ c+ bc) = a ∗ (b ∗ c).

Identity. We claim that e = 0 is the identity element. For all a ∈ G we have

a+ 0 + a · 0 = a = 0 + a+ 0 · a.

Inverse. Let a ∈ G. We claim that a−1 = − a1+a

, which is an element of G since a 6= 1.We have

a+

(− a

1 + a

)+ a ·

(− a

1 + a

)=a(1 + a)− a− a2

1 + a= 0.

Thus (G, ∗) is a group.

5. Let H be a subgroup of a group G. For a, b ∈ G, define a relation on G by letting a ∼ bif and only if ab−1 ∈ H. Prove that ∼ is an equivalence relation.

Solution.

Reflexivity. Let a ∈ G. Then aa−1 = e ∈ H. Thus a ∼ a.

Symmetry. If ab−1 ∈ H, then (ab−1)−1 = ba−1 ∈ H. Thus a ∼ b ⇒ b ∼ a.

Transitivity. If ab−1, bc−1 ∈ H, then ab−1bc−1 = ac−1 ∈ H. Thus a ∼ b, b ∼ c⇒ a ∼ c.

Thus ∼ is an equivalence relation.

6. Prove that the polynomial f(x) = x7−10x4+15x−5 is irreducible over the field of rationalnumbers.

Solution. Using Eisenstein’s Criterion with p = 5, we see that 5 | ai for 0 ≤ i ≤ 6 whereai is the coefficient of xi; also 5 does not divide the leading coefficient, and 25 does notdivide the constant term. Therefore f(x) is irreducible over Q.

7. An element a of a ring R is called a zero divisor if there exists a nonzero element b ∈ Rsuch that ab = 0. If a is a zero divisor of a commutative ring R and r ∈ R, prove that aris a zero divisor.

Solution. If a is a zero divisor, then there is a nonzero element b ∈ R such that ab = 0.Therefore arb = rab = 0. Thus ar is a zero divisor.

8. Let R and S be commutative rings, and let ϕ : R→ S be a non-trivial ring homomorphism.

(a) Prove that the kernel of ϕ is an ideal of R.

(b) If R is a field, prove that ϕ is one-to-one.

Solution.

(a) Let k1, k2 ∈ ker(ϕ) and r ∈ R. Then ϕ(k1 + k2) = ϕ(k1) + ϕ(k2) = 0 + 0 = 0. Thusk1 + k2 ∈ ker(ϕ). Also ϕ(rk1) = ϕ(r)ϕ(k1) = ϕ(0) · 0 = 0; therefore rk1 ∈ ker(ϕ).Thus ker(ϕ) / R.

(b) The only ideals of a field are {0} and the field itself. Thus ker(ϕ) = {0} or ker(ϕ) = F .But ϕ is given to be nontrivial; therefore ker(ϕ) 6= F . Hence ker(ϕ) = {0}, and ϕ isone-to-one.

Part B.

1. Let LA : Rn → Rn be the linear transformation given by LA(v) = Av where A is a realn× n matrix. Show that if n is odd then L has a real eigenvalue.

Solution. Since the dimension of Rn is odd the characteristic polynomial of A will beof odd degree. Since every polynomial in R[x] of odd degree has a real root, we concludethat LA has a real eigenvalue.

2. If v1,v2, . . . ,vn are distinct eigenvectors corresponding to distinct eigenvalues λ1, λ2, . . . , λn

of a matrix A, prove that {v1,v2, . . . ,vn} is a linearly independent set.

Solution. Assume that v1, v2, . . . , vn form a linearly dependent set. Let k be the small-est positive integer such that vk ∈ span{v1, v2, . . . , vk−1}. Thus there are real numbersc1, . . . , ck−1 such that

vk = c1v1 + c2v2 + · · ·+ ck−1vk−1 (1)

Hence,

Avk = c1Av1 + c2Av2 + · · ·+ ck−1Avk−1 thus (2)

λkvk = c1λ1v1 + c2λ2v2 + · · ·+ ck−1λk−1vk−1 (3)

Now subtract equation (3) from λk times equation (1) to get

0 = c1(λk − λ1)v1 + c2(λk − λ2)v2 + ck−1(λk − λk−1)vk−1

Since the λi are all distinct and the set {v1, v2, . . . , vk−1} is independent, we must havethat all of the ci are zero. It follows that vk is zero, contradicting the hypothesis that vk

is an eigenvector of A.

3. Let V be a real vector space of dimension n, and suppose that S = {v1,v2, . . . ,vt} is alinearly independent subset of V . Prove that there is a basis B of V such that S ⊆ B.

Solution. If t = n then S is a basis. So assume that t < n. Let B = {w1, . . . , wn} be abasis of V . Consider the set B∗{v1, v2, . . . , vt, w1, w2, . . . , wn}. Clearly this spans V . Thuswe can form a basis from B∗ by deleting every vector x that is a linear combination of thevectors preceding x. Since the vi’s are independent of one another, none of the vi’s will bedeleted during this procedure. Thus S is a subset of the resulting basis.

4. Let S be the set of all functions f : R→ R that satisfy the differential equation

y′′ − y′ + 2y = 0.

Is S a real vector space? (Assume the usual operations (f + g)(t) = f(t) + g(t) and(c · f)(t) = cf(t) where c ∈ R.) Explain why or why not.

Solution. S is a vector space. Since the derivative is a linear transformation it is easilyverified that S is closed under addition and scalar multiplication. The other vector spaceproperties are easily seen as well.

5. Let A and B be n × n matrices. Show that AB is invertible if and only if A and B areinvertible.

Solution. (=⇒) Assume that AB is invertible. Suppose that B is not invertible. Thusthere is a non-trivial solution to the homogenous system Bx = 0. This non-trivial solutionwill also solve the system ABx = 0 contradicting the hypothesis that AB is invertible. SoB must be invertible.

Now given that AB and B are invertible, we have that A = (AB)B−1 is the product ofinvertible matrices, whence A is invertible.

(⇐=) If A and B are invertible, then AB is the product of invertible matrices and hasinverse B−1A−1.

6. Prove that a linear transformation of vector spaces L : V → W is one to one if and only ifL maps linearly independent subsets of V to linearly independent subsets of W .

Solution. If V is the zero vector space the result is vacuously true. So assume thatV 6= {0}.(=⇒) Assume that L is one to one. Let {v1, . . . , vn} be a linearly independent set in V .Suppose that there are constants {c1, . . . , cn} such that

c1Lv1 + c2Lv2 + · · ·+ cnLvn = 0

Since L is linear we would then have

L(c1v1 + c2v2 + · · ·+ cnvn) = 0

Since L is one to one, this implies that c1v1 + c2v2 + · · ·+ cnvn = 0 whence all the ci’s mustbe zero because of the linear independence of {v1, . . . , vn}.(⇐=) Assume that L preserves linear independence. Let v ∈ V be a non-zero vector. Thenthe set {v} is linearly independent, whence {Lv} is linearly independent. Thus Lv 6= 0,so L is one to one.

7. Let V be a finite dimensional vector space and L : V → W be a linear transformation tovector space W . Prove that dim(kernel L)+ dim(image L) = dim V .

Solution. Let n =dim V and k =dim kerL. Note that k ≤ n. If k = n then kerL = Vand image L = {0} so the result is true.

So suppose that 0 ≤ k < n. Let {v1, v2, . . . , vk} ⊆ V be a basis for kerL. Since{v1, v2, . . . , vk} is linearly independent, it can be extended to a basis

BV = {v1, v2, . . . , vk, vk+1, . . . , vn}

of V . We will show that T = {Lvk+1, . . . , Lvn} is a basis of the image of L.

To see that T spans let w ∈ image L. Thus w = Lv for some v in V . Write so there areconstants ci(1 ≤ i ≤ n) such that v = c1v1 + c2v2 + · · · ckvk + ck+1vk+1 + · · ·+ cnvn. Thus

w = Lv = 0 + ck+1Lvk+1 + · · ·+ cnLvn

whence T spans image L.

To see that T is linearly independent, suppose that there are constants ci(k + 1 ≤ i ≤ n)such that ck+1Lvk+1 + · · · + cnLvn = 0. Thus ck+1vk+1 + · · · + cnvn is in the kernel of Lwhence there are constants ti(1 ≤ i ≤ k) such that ck+1vk+1 + · · ·+cnvn = t1v1 + · · ·+ tkvk.The fact that all the vi’s are independent then implies that all the ci’s (and all the ti’s)are zero.

Thus we’ve shown that dim(image L) = n− k and the result follows.

8. Let A be an n× n matrix. Prove that A is invertible if and only if the determinant of Ais non-zero.

Solution. (=⇒) If A is invertible then A is the product of elementary matrices. Since thedeterminant respect matrix multiplication and no elementary matrix has zero determinant,it follows that A has non-zero determinant.

(⇐=) If A is not invertible then A is row equivalent to a matrix A that contains a row ofzeros. Hence A = E1E2 · · ·EnA where E1, E2, . . . , En is a sequence of elementary matrices.Thus

detA = det [E1E2 · · ·EnA] = det (E1) det (E2) · · · det (En) det (A) = 0

since det A = 0.



1. Let a and p be integers. If p is prime and a is not divisible by p, prove that the additiveorder of a modulo p is equal to p.

2. Let G and H be groups, and let ϕ : G→ H be a group homomorphism with kernel ker(ϕ).Prove that ker(ϕ) is a normal subgroup of G.

3. Let N be a normal subgroup of a group G. Prove that the factor group G/N is abelian ifand only if aba−1b−1 ∈ N for all elements a, b ∈ G.

4. Let G be any group with no proper nontrivial subgroups, and assume the order of G isgreater than 1. Prove that G is finite cyclic of order p for some prime p.

5. Let G be a group and let D = {(a, a, a) | a ∈ G}.

(a) Prove that D is a subgroup of the direct product G×G×G.

(b) Prove that D is normal in G×G×G if and only if G is abelian.

Hint. If D is normal, then (a, a, b)(b, b, b)(a, a, b)−1 ∈ D for all a, b ∈ G.

6. Let Q[x] be the set of all polynomials in x with rational coefficients. Define a relation ∼on Q[x] by f(x) ∼ g(x) if and only if f(x) − g(x) is divisible by x2 + 1. Prove that ∼ isan equivalence relation.

7. Let R be the ring {m+n√

2 | m, n ∈ Z}, and let I be the subset {m+n√

2 ∈ R | m is even}.Prove that I is an ideal of R.

8. Assume that the set S = {a + b√

3 | a, b are rational numbers} is a commutative ring.Prove that S is a field.



1. For which values of the parameters a, b, and c is the matrix A =

a 1 b0 1 c0 0 1

invertible?

Find the inverse when it exists.


1 2 12 5 31 3 2

. Find a basis for the kernel

and a basis for the image of the transformation.

3. Let A =

[1 −12 4

]. Find all eigenvalues of A and all their corresponding eigenvectors.

4. Find an orthonormal basis for the subspace of R3 spanned by the vectors v1 = 〈1, 0,−1〉and v2 = 〈0, 3, 4〉

5. (a) Show that two non-zero vectors are linearly dependent if and only if one is a scalarmultiple of the other.

(b) Let v1, v2, and v3 be linearly independent vectors in Rn. Are the vectors v1, v2, andv1 + v2 + v3 necessarily linearly independent?

6. Let C denote the field of complex numbers and R the field of real numbers. With theusual operations, C is a vector space over R. Prove that the map ϕ : C → R2 given byϕ(x + iy) = (x, y) is an isomorphism of vector spaces.

7. Prove or disprove: The matrix

A =

1 x x2

1 y y2

1 z z2

over R has determinant equal to (y − x)(z − x)(z − y).

8. Let A =

[1 11 1

]. Prove that An = 2n−1A for all positive integers n.


Part A.

1. Let a and p be integers. If p is prime and a is not divisible by p, prove that the additiveorder of a modulo p is equal to p.

Solution. The additive order of [a]p is at most p since ap ≡ 0 (mod p).

If ax ≡ 0 (mod p), then p | ax. Since p is prime this implies p | a or p | x. But p - a; thusp | x. Since x 6= 0 we have that p ≤ x. Thus x cannot be less than p.

2. Let G and H be groups, and let ϕ : G→ H be a group homomorphism with kernel ker(ϕ).Prove that ker(ϕ) is a normal subgroup of G.

Solution. Let n ∈ ker(ϕ) and g ∈ G. Then ϕ(gng−1) = ϕ(g)ϕ(n)ϕ(g−1) = ϕ(g)ϕ(n)ϕ(g)−1 =ϕ(g)ϕ(g)−1 = 1. Therefore gng−1 ∈ ker(ϕ).

3. Let N be a normal subgroup of a group G. Prove that the factor group G/N is abelian ifand only if aba−1b−1 ∈ N for all elements a, b ∈ G.

Solution. aba−1b−1 ∈ N if and only if Naba−1b−1 = N if and only if Nab = Nba if andonly if NaNb = NbNa.

4. Let G be any group with no proper nontrivial subgroups, and assume the order of G isgreater than 1. Prove that G is finite cyclic of order p for some prime p.

Solution. Let a ∈ G, a 6= 1. Then 〈a〉 is a nontrivial subgroup of G and thus is all of G.So G is cyclic. If G is infinite then G ∼= Z, a contradiction to the hypothesis that G hasno proper nontrivial subgroups since 2Z ≤ Z. Thus G is finite, and is isomorphic to Zn

for some n. Let q be a prime factor of n. If n is not prime then q < n and we have 〈aq〉 aproper, nontrivial subgroup of G of order n/q, a contradiction. Thus n = p is prime.

5. Let G be a group and let D = {(a, a, a) | a ∈ G}.

(a) Prove that D is a subgroup of the direct product G×G×G.

(b) Prove that D is normal in G×G×G if and only if G is abelian.

Hint. If D is normal, then (a, a, b)(b, b, b)(a, a, b)−1 ∈ D for all a, b ∈ G.

Solution.

(a) First, D is nonempty since (e, e, e) ∈ D. Now let (a, a, a), (b, b, b) ∈ D. Then(a, a, a)(b, b, b)−1 = (ab−1, ab−1ab−1) ∈ D.

(b) (⇒) Let a, b ∈ G. Using the hint, if D is normal, then (a, a, b)(b, b, b)(a, a, b)−1 ∈ Dfor all a, b ∈ G by definition of normal subgroup. Now (a, a, b)(b, b, b)(a, a, b)−1 =(aba−1, aba−1, bbb−1) = (aba−1, aba−1, b) ∈ D implies aba−1 = b, or ab = ba. ThusG is abelian.

(⇐) Any subgroup of an abelian group is normal.

6. Let Q[x] be the set of all polynomials in x with rational coefficients. Define a relation ∼on Q[x] by f(x) ∼ g(x) if and only if f(x) − g(x) is divisible by x2 + 1. Prove that ∼ isan equivalence relation.

Solution.

Reflexive. f(x)− f(x) = 0 = 0 · (x2 + 1), so f(x) ∼ f(x).

Symmetric. Suppose f(x)− g(x) = k(x)(x2 + 1). Then g(x)− f(x) = −k(x)(x2 + 1).

Transitive. Suppose f(x) − g(x) = k(x)(x2 + 1) and g(x) − h(x) = m(x)(x2 + 1).Then f(x) − h(x) = f(x) − g(x) + g(x) − h(x) = k(x)(x2 + 1) + m(x)(x2 + 1) =(k(x) +m(x))(x2 + 1).

7. LetR be the ring {m+n√

2 | m,n ∈ Z}, and let I be the subset {m+n√

2 ∈ R | m is even}.Prove that I is an ideal of R.

Solution. Let x = 2m+ n√

2, y = 2p+ q√

2 ∈ I. Then

x± y = (2m± 2p) + (n± q)√

2 = 2(m+ p) + (n+ q)√

2 ∈ I.

Now let x be as above and let r = a+ b√

2 ∈ R. Then

rx = (2am+ 2bn) + (an+ bm)√

2 = 2(am+ bn) + (an+ bm)√

2 ∈ I.

Therefore I is an ideal of R.

8. Assume that the set S = {a + b√

3 | a, b are rational numbers} is a commutative ring.Prove that S is a field.

Solution. Let x = a + b√

3 6= 0. We must show that there are rational numbers c, d forwhich (a+ b

√3)(c+ d

√3) = 1. In other words, ac+ 3bd = 1 and ad+ bc = 0. At least one

of a or b must be nonzero. If a 6= 0 then d = ba· c, and ac− 3b2c

a= 1. Note that a2 − 3b2

cannot be 0 since if so then a = ±b√

3 and a and b are not both rational, a contradiction.Therefore c = a

a2−3b2, d = − b

a· aa2−3b2

= − ba2−3b2

, and x−1 = aa2−3b2

− ba2−3b2

√3. Thus every

nonzero element of S has a multiplicative inverse. If a = 0 then c = −adb

= 0, and 3bd = 1,

or d = 13b

. Thus x−1 = 13b

√3.

Part B.

1. For which values of the parameters a, b, and c is the matrix A =

a 1 b0 1 c0 0 1

invertible?

Find the inverse when it exists.

Solution. Since A is upper triangular then its determinant is just the product of theelements in the diagonal. Thus, in this case det(A) = a. So, A is invertible whenevera 6= 0.

When a 6= 0 the inverse of A is

A−1 =

a−1 −a−1 a−1(c− b)0 1 −c0 0 1


1 2 12 5 31 3 2

. Find a basis for the kernel

and a basis for the image of the transformation.

Solution. Performing row operations on A we get

A =

1 2 12 5 31 3 2

now we subtract R1 from R3

→

1 2 12 5 30 1 1

now we subtract 3R3 from R2, and R3 from R1

→

1 1 02 2 00 1 1

now we subtract 3R3 from R2, and R3 from R1

It follows that the kernel of A is spanned by (1,−1, 1).

Since A is symmetric, we use the matrix obtained by row operations to see that the rangeis spanned by (1, 1, 0) and (0, 1, 1).

3. Let A =

[1 −12 4

]. Find all eigenvalues of A and all their corresponding eigenvectors.

Solution. The characteristic polynomial of A is

χA(λ) =

∣∣∣∣ 1− λ −12 4− λ

∣∣∣∣ = λ2 − 5λ+ 6 = (λ− 3)(λ− 2)

So, the eigenvalues of A are λ = 3 and λ = 2.

For λ = 3 we need to solve the equation Av = 3v, which yields the system

x− y = 3x 2x+ 4y = 3y


For λ = 2 we need to solve the equation Av = 2v, which yields the system

x− y = 2x 2x+ 4y = 2y


4. Find an orthonormal basis for the subspace of R3 spanned by the vectors v1 = 〈1, 0,−1〉and v2 = 〈0, 3, 4〉

Solution. The first step on the Gram-Schmidt process says that we fix v1 and define

u2 = 〈0, 3, 4〉 − 〈1, 0,−1〉 · 〈0, 3, 4〉|〈1, 0,−1〉|2

〈1, 0,−1〉

= 〈0, 3, 4〉+ 2〈1, 0,−1〉= 〈2, 3, 2〉

So, right now we have an orthogonal basis. We obtain an orthonormal basis by dividingv1 and u2 by their norm. The final answer is{

1√2〈1, 0,−1〉, 1√

17〈2, 3, 2〉

}5. (a) Show that two non-zero vectors are linearly dependent if and only if one is a scalar

multiple of the other.

(b) Let v1, v2, and v3 be linearly independent vectors in Rn. Are the vectors v1, v2, andv1 + v2 + v3 necessarily linearly independent?

Solution.

(a) If one vector is a multiple of the other, let us say v = αw, then

v − αw = 0

Hence, the vectors are dependent.

If the vectors are linearly dependent, then

αv + βw = 0

with WLOG α 6= 0. Then we can ‘solve’ for v to get v = −βαw.

(b) Yes, because ifαv1 + βv2 + γ(v1 + v2 + v3) = 0

then(α + γ)v1 + (β + γ)v2 + γv3 = 0

which forces a system

γ = 0 β + γ = 0 α + γ = 0

with a unique (trivial) solution.

6. Let C denote the field of complex numbers and R the field of real numbers. With theusual operations, C is a vector space over R. Prove that the map ϕ : C → R2 given byϕ(x+ iy) = (x, y) is an isomorphism of vector spaces.

Solution. We first check that ϕ is a homomorphism.

ϕ((x+ iy) + (a+ bi)) = ϕ(x+ a+ i(y + b))

= (x+ a, y + b)

= (x, y) + (a, b)

= ϕ(x+ iy) + ϕ(a+ bi)

and for α ∈ R

ϕ(α(x+ iy)) = ϕ(αx+ αiy)

= (αx, αy)

= α(x, y)

= αϕ(x+ iy)

Now, since ϕ is clearly bijective, we are done

7. Prove or disprove: The matrix

A =

1 x x2

1 y y2

1 z z2

over R has determinant equal to (y − x)(z − x)(z − y).

Solution. It is true.

det(A) =

∣∣∣∣∣∣1 x x2

1 y y2

1 z z2

∣∣∣∣∣∣ now we subtract row 3 from row 1 and row 2

=

∣∣∣∣∣∣0 x− z x2 − z2

0 y − z y2 − z2

1 z z2

∣∣∣∣∣∣ now we factor x− z and y − z in rows 1 and 2

= (x− z)(y − z)

∣∣∣∣∣∣0 1 x+ z0 1 y + z1 z z2

∣∣∣∣∣∣ now we subtract row 2 from row 1

= (x− z)(y − z)

∣∣∣∣∣∣0 0 x− y0 1 y + z1 z z2

∣∣∣∣∣∣= (x− z)(y − z)(x− y)

∣∣∣∣ 0 11 z

∣∣∣∣= −(x− z)(y − z)(x− y) = (y − x)(z − x)(z − y)

8. Let A =

[1 11 1

]. Prove that An = 2n−1A for all positive integers n.

Solution. By induction. For n = 1 the result is trivially true.

Assume Ak = 2k−1A, for all k ≤ n. Now we want to check that An+1 = 2nA

An+1 = AnA

= (2n−1A)A

= 2n−1A2

= 2n−1(2A)

= 2nA



1. Let a, b, m, and n be integers, and suppose am+ bn = 1. Prove that a and b are relativelyprime.

2. Let T = R3 − {(0, 0, 0)}. Define a relation ∼ on T by (x1, y1, z1) ∼ (x2, y2, z2) if and onlyif there exists a nonzero real number λ such that x1 = λx2, y1 = λy2, and z1 = λz2. Provethat ∼ is an equivalence relation.

3. Let G and H be groups, and let ϕ : G → H be an onto group homomorphism. SupposeG is abelian. Prove that H is abelian.

4. Let G be a group, and let N be the subset {g ∈ G | gx = xg for all x ∈ G} (N is calledthe center of G). Prove that N is a normal subgroup of G.

5. Let Sn denote the group of permutations on the set {1, 2, . . . , n}, and let An denote thesubset consisting of even permutations.

(a) Prove that An is a normal subgroup of Sn. You may assume An is a subgroup of Sn.

(b) Prove that Sn/An is isomorphic to the group Z2 = {0, 1}.

6. Let R be a ring with identity element 1R, and let I be an ideal of R. Prove that if 1R isin I, then I = R.

7. Let ϕ : C→ C be defined by ϕ(a + bi) = a− bi for all a + bi ∈ C. Prove that ϕ is a ringisomorphism.

8. Prove that the only ideals of a field F are {0F} and F , where 0F denotes the additiveidentity element of F .



1. Let A =

0 1 0 10 0 2 40 0 0 3

, and let R be the reduced row echelon form for A.

(a) Find R, determine the (row) rank of A, and find a basis for the row space of A.

(b) Find a matrix P such that PA = R.

2. Find an orthonormal basis for the subspace of the Euclidean space R3 spanned by thevectors v1 = (1, 0, 1) and v2 = (0, 3, 4).

3. Prove: If S is a finite linearly independent subset of the vector space V and w ∈ V is notin the subspace spanned by S, then the set S ∪ {w} is linearly independent.

4. Let V and W be vector spaces over the field F and let T be a linear transformation fromV into W . Suppose V is finite dimensional. Prove: rank (T ) + nullity (T ) = dim (V ).

5. For each natural number n, determine the value of the determinant of the following matrix:

A =

1 2 3 4 · · · n1 0 3 4 · · · n1 2 0 4 · · · n...

. . . · · · ...1 2 3 4 · · · 0

6. Let A be the symmetric matrix

[1 −1−1 1

]. Find an orthogonal matrix T such that T−1AT

is a diagonal matrix.

7. Let A = (aij) be an n×n matrix over the reals. Show that A can be expressed in a uniqueway as A = S + K where S is symmetric and K is skew-symmetric. (Hint: Consider thematrices 1

2(A+ At) and 1

2(A− At)).

8. Let A and B be n× n matrices and suppose A and B are similar. Show:

(a) det(A) = det(B).

(b) If A is nonsingular, so is B, and A−1 is similar to B−1.


Part A.

1. Let a, b, m, and n be integers, and suppose am+ bn = 1. Prove that a and b are relativelyprime.

Solution. Suppose d = (a, b). Let a = ds, b = dt. Then 1 = am + bn = dsm + dtn =d(sm+ tn). Therefore d = 1.

2. Let T = R3 − {(0, 0, 0)}. Define a relation ∼ on T by (x1, y1, z1) ∼ (x2, y2, z2) if and onlyif there exists a nonzero real number λ such that x1 = λx2, y1 = λy2, and z1 = λz2. Provethat ∼ is an equivalence relation.

Solution. An equivalence relation must be reflexive, symmetric and transitive.

Reflexive. Let λ = 1. Then (x, y, z) = 1 · (x, y, z), so (x, y, z) ∼ (x, y, z).

Symmetric. Suppose (x1, y1, z1) = λ(x2, y2, z2). Then (x2, y2, z2) = 1λ(x1, y1, z1).

Transitive. Suppose (x1, y1, z1) = λ(x2, y2, z2) and (x2, y2, z2) = µ(x3, y3, z3). Then(x1, y1, z1) = λ(µ(x3, y3, z3)) = λµ(x3, y3, z3).

3. Let G and H be groups, and let ϕ : G → H be an onto group homomorphism. SupposeG is abelian. Prove that H is abelian.

Solution. Let h, k ∈ H. Since ϕ is onto there exist a, b ∈ G such that ϕ(a) = h andϕ(b) = k. Thus hk = ϕ(a)ϕ(b) = ϕ(ab) = ϕ(ba) = ϕ(b)ϕ(a) = kh.

4. Let G be a group, and let N be the subset {g ∈ G | gx = xg for all x ∈ G} (N is calledthe center of G). Prove that N is a normal subgroup of G.

Solution 1. First note that 1 ∈ N since 1 · x = x · 1 for all x ∈ G. Thus N is nonempty.Let m and n be elements of N , and let x ∈ G. Since xn = nx, we have x = nxn−1

and thus n−1x = xn−1. Therefore xmn−1 = mxn−1 = mn−1x, and mn−1 ∈ N . ThusN ≤ G.

To show that N is normal, let m and x be as above. Then xm = mx and thusxmx−1 = m ∈ N .

Solution 2. First note that 1 ∈ N since 1 · x = x · 1 for all x ∈ G. Now let m and n beelements of N , and let x ∈ G. Then xmn = mxn = mnx; therefore mn ∈ N . Sincexn = nx, we have x = nxn−1 and thus n−1x = xn−1. Therefore n−1 ∈ N . ThusN ≤ G.

To show that N is normal, let m and x be as above. Then xm = mx and thusxmx−1 = m ∈ N .

5. Let Sn denote the group of permutations on the set {1, 2, . . . , n}, and let An denote thesubset consisting of even permutations.

(a) Prove that An is a normal subgroup of Sn. You may assume An is a subgroup of Sn.

(b) Prove that Sn/An is isomorphic to the group Z2 = {0, 1}.

Solution.

(a) Let τ ∈ An and let σ ∈ Sn. If σ is even, then so is σ−1. Similarly, if σ is odd, thenso is σ−1. Therefore στσ−1 can be written as an even number of transpositions.

(b) Solution 1. Let ϕ : Sn → Z2 be defined by ϕ(σ) =

{0 if σ is even

1 if σ is odd.

Clearly ϕ is an onto homomorphism since even ◦ even = even, odd ◦ odd = even,etc. Moreover, ker(ϕ) = An. Therefore by the First Isomorphism Theorem,Sn/An ∼= Z2.

Solution 2. Let ϕ : Sn/An → Z2 be defined by ϕ(σAn) =

{0 if σ is even

1 if σ is odd.

Clearly ϕ is an onto homomorphism since even ◦ even = even, odd ◦ odd = even,etc. Moreover, |Sn/An| = 2 = |Z2|. Therefore ϕ is an isomorphism.

6. Let R be a ring with identity element 1R, and let I be an ideal of R. Prove that if 1R isin I, then I = R.

Solution. Let r ∈ R. Then since 1R ∈ I we have r · 1R = r ∈ I since I is an ideal of R.

7. Let ϕ : C→ C be defined by ϕ(a + bi) = a− bi for all a + bi ∈ C. Prove that ϕ is a ringisomorphism.

Solution. Clearly ϕ is onto. ϕ is also one-to-one since if ϕ(a + bi) = a − bi = 0, thena = 0 and b = 0, and thus a+ bi = 0. ϕ is a ring homomorphism since

• ϕ((a + bi) + (c + di) = ϕ(a + c + (b + d)i) = a + c − (b + d)i = a − bi + c − di =ϕ(a+ bi) + ϕ(c+ di)

• ϕ((a+ bi)(c+ di) = ϕ(ac− bd+ (ad+ bc)i) = ac− bd− (ad+ bc)i = (a− bi)(c− di) =ϕ(a+ bi)ϕ(c+ di).

Therefore ϕ is a ring isomorphism.

8. Prove that the only ideals of a field F are {0F} and F , where 0F denotes the additiveidentity element of F .

Solution. Suppose I is an ideal of F . If I = {0F} then we are done, so suppose a ∈ I,a 6= 0. Then there is an element a−1 ∈ F since F is a field. Therefore a−1a = 1 ∈ I sinceI is an ideal. But by #6, this implies I = F .

Part B.

1. Let A =

0 1 0 10 0 2 40 0 0 3

, and let R be the reduced row echelon form for A.

(a) Find R, determine the (row) rank of A, and find a basis for the row space of A.

(b) Find a matrix P such that PA = R.

Solution. We need to perform row reduction on A to get to R. The matrices of theelementary operations we will use will yield P .

A =

0 1 0 10 0 2 40 0 0 3

now we do R1 7→ R1 −1

3R3

→

0 1 0 00 0 2 40 0 0 3

now we do R2 7→ R2 −4

3R3

→

0 1 0 00 0 2 00 0 0 3

= R

It follows that

P =

1 0 00 1 −4

3

0 0 1

1 0 −13

0 1 00 0 1

=

1 0 −13

0 1 −43

0 0 1

2. Find an orthonormal basis for the subspace of the Euclidean space R3 spanned by the

vectors v1 = (1, 0, 1) and v2 = (0, 3, 4).

Solution. The first step on the Gram-Schmidt process says that we fix v1 and define

u2 = (0, 3, 4)− (1, 0, 1) · (0, 3, 4)

|(1, 0, 1)|2(1, 0, 1)

= (0, 3, 4)− 2(1, 0, 1)

= (−2, 3, 2)

So, right now we have an orthogonal basis. We obtain an orthonormal basis by dividingv1 and u2 by their norm. The final answer is{

1√2

(1, 0, 1),1√17

(−2, 3, 2)

}3. Prove: If S is a finite linearly independent subset of the vector space V and w ∈ V is not

in the subspace spanned by S, then the set S ∪ {w} is linearly independent.

Solution. Consider a linear combination of the elements of S and w that is equal to zero.If the scalar with w is zero, then we have a linear combination of the elements of S that isequal to zero, thus all the scalars are equal to zero. If the scalar with w is different fromzero, then we can ‘solve’ for w and leave w as a linear combination of the elements of S,which would imply that w ∈ Span(S). A contradiction.

4. Let V and W be vector spaces over the field F and let T be a linear transformation fromV into W . Suppose V is finite dimensional. Prove: rank (T ) + nullity (T ) = dim (V ).

Solution. We know that V/ker(T ) ∼= Im(T ). So, dim(V/ker(T ) = dim(Im(T )) =rank(T ).

Sincedim(V/ker(T )) = dim(V )− dim(ker(T )) = dim(V )− nullity(T )

thendim(V )− nullity(T ) = rank(T )

5. For each natural number n, determine the value of the determinant of the following matrix:

A =

1 2 3 4 · · · n1 0 3 4 · · · n1 2 0 4 · · · n...

. . . · · · ...1 2 3 4 · · · 0

Solution.

det(A) =

∣∣∣∣∣∣∣∣∣∣∣

1 2 3 4 · · · n1 0 3 4 · · · n1 2 0 4 · · · n...

. . . · · · ...1 2 3 4 · · · 0

∣∣∣∣∣∣∣∣∣∣∣now we subtract row 1 to all rows

=

∣∣∣∣∣∣∣∣∣∣∣

1 2 3 4 · · · n0 −2 0 0 · · · 00 0 −3 0 · · · 0...

. . . · · · ...0 0 0 0 · · · −n

∣∣∣∣∣∣∣∣∣∣∣= (−1)n−1n! because the matrix above is upper triangular

6. Let A be the symmetric matrix

[1 −1−1 1

]. Find an orthogonal matrix T such that T−1AT

is a diagonal matrix.

Solution. Since A is symmetric, then its eigenvectors associated to distinct eigenvaluesmust be orthogonal. So, the only thing we have to do is to get the standard diagonalizationof A.

The characteristic polynomial of A is

χA(λ) = (1− λ)2 − 1 = λ2 − 2λ = λ(λ− 2)

For λ = 0 we get the system

x− y = 0 − x+ y = 0

which has solution space spanned by (1, 1).

For λ = 2 we get the system

x− y = 2x − x+ y = 2y


It follows that T =

[1 11 −1

]

7. Let A = (aij) be an n×n matrix over the reals. Show that A can be expressed in a uniqueway as A = S + K where S is symmetric and K is skew-symmetric. (Hint: Consider thematrices 1

2(A+ At) and 1

2(A− At)).

Solution. Note that the matrix 12(A+At) is symmetric and that the matrix 1

2(A−At) is

skew-symmetric. Also, note that

1

2(A+ At) +

1

2(A− At) = A

Now we just have to check the uniqueness of the representation. So, we assume there arematrices M (symmetric) and N (skew-symmetric) such that A = M +N . So,

M +N =1

2(A+ At) +

1

2(A− At)

which implies

M − 1

2(A+ At) =

1

2(A− At)−N

In the previous equation the right hand side is skew-symmetric and the left hand side issymmetric. It follows that

M − 1

2(A+ At) =

1

2(A− At)−N = 0

Hence, the representation is unique.

8. Let A and B be n× n matrices and suppose A and B are similar. Show:

(a) det(A) = det(B).

(b) If A is nonsingular, so is B, and A−1 is similar to B−1.

Solution. Assume B = PAP−1

(a) Since the determinant is multiplicative, then

det(B) = det(PAP−1) = det(P ) det(A) det(P−1) = det(P ) det(A) det(P )−1 = det(A)

(b) If A is nonsingular, then det(A) 6= 0, which implies det(B) 6= 0, and thus B isnon-singular. In this case

B−1 = (PAP−1)−1 = (P−1)−1A−1P−1 = PA−1P−1

So, A−1 is similar to B−1.



1. Prove that if N �G and H is any subgroup of G, then N ∩H �H.

2. Prove that if H and K are finite subgroups of G whose orders are relativelyprime, then H ∩K = {e}

3. Show that the relation on Z defined by a ∼ b iff a2 ≡ b2 mod 6 is an equiva-lence relation.

4. Suppose that φ is a homomorphism from Z30 to Z30 and ker(φ) = {0, 10, 20}.If φ(23) = 9, determine all elements that map to 9. That is, find all k ∈ Z30

such that φ(k) = 9.

5. (a) Show that a =

(0 1−1 −1

)has order 3 in GL(2,R) and b =

(0 −11 0

)has order 4.

(b) Show that ab has infinite order.

6. Prove that σ2 is an even permutation for every permutation σ.

7. Define a new addition and multiplication on Z by

a⊕ b = a+ b− 1 and a⊗ b = ab− (a+ b) + 2

Prove that with these operations Z is an integral domain.

8. Show that a finite commutative ring with no zero-divisors has a multiplicativeidentity.


1. Suppose that A is an n × n matrix such that A3 = 0 but A2 6= 0. Show that{I, A,A2} is independent in the space of all n× n matrices with real entries.

2. Let A be diagonalizable 2 × 2 matrix. If λ4 = 5λ for each eigenvalue λ of A,show that A4 = 5A.

3. If T : V → V is linear, show that T 2 = IV iff T is an isomorphism andT−1 = T .

4. (a) Find A if (A−1 − 3I)T = 2

(−1 25 4

)(b) If det(A) = 2 and det(B) = −3, compute det(A3B−1ATB2).

5. Invertible 2× 2 matrices with determinant one have the form(a bc 1+bc

a

), where a 6= 0 and bc 6= 1, or(

0 b−1

bd

), where b 6= 0 and d 6= 0, or(

a b−1

b0

), where b 6= 0

Determine the form(s) of all 2× 2 matrices that are their own inverses.

6. Let V = {v ∈ R| v > 0}. Show that V is a vector space over R if thevector addition is ordinary multiplication and scalar multiplication is definedby a · v = va.

7. If Rn = span{v1, v2, · · · , vn} and if x and y in Rn satisfy x · vi = y · vi for alli, show that x = y.

8. Let Pn denote the space of all polynomials with real coefficients that havedegree at most n (union the zero polynomial). Find a linear transformationT : P2 → P4 such that

T (1) = x4, T (1 + x) = 1 + x3 T (1 + x2) = 1− x2


Part A.

1. Prove that if N �G and H is any subgroup of G, then N ∩H �H.

Solution. We know the intersection of two subgroups of G is a subgroup ofG. It follows that N ∩H < H.

Let x ∈ N ∩ H, and h ∈ H. Since N is normal in G and H ⊂ G, thenhxh−1 ∈ N . Moreover, by closure in H hxh−1 ∈ H. So, hxh−1 ∈ N ∩H.

2. Prove that if H and K are finite subgroups of G whose orders are relativelyprime, then H ∩K = {e}

Solution. This is problem 7 in part A in the exam of Fall 2006.

3. Show that the relation on Z defined by a ∼ b iff a2 ≡ b2 mod 6 is an equiva-lence relation.

Solution. Since a2 ≡ a2 mod 6, and 6 dividing a2− b2 implies that 6 dividesb2 − a2 then the only thing left to check is transitivity. So, assume a2 ≡b2 mod 6 and b2 ≡ c2 mod 6, that is a2 − b2 = 6x and b2 − c2 = 6y for someintegers x, y. Then,

a2 − c2 = (a2 − b2) + (b2 − c2) = 6x+ 6y = 6(x+ y)

Since x+ y ∈ Z then we are done.

4. Suppose that φ is a homomorphism from Z30 to Z30 and ker(φ) = {0, 10, 20}.If φ(23) = 9, determine all elements that map to 9. That is, find all k ∈ Z30

such that φ(k) = 9.

Solution. We know that the pre-image of any element in the range of ahomomorphism is a coset of the kernel of the homomorphism. So, in this case,

φ−1(9) = 23 + ker(φ) = {23, 23 + 10, 23 + 20} = {23, 33, 43}

5. (a) Show that a =

(0 1−1 −1

)has order 3 in GL(2,R) and b =

(0 −11 0

)has order 4.

(b) Show that ab has infinite order.

Solution.

(a) Easy computations show

a2 =

(−1 −11 0

)a3 =

(1 00 1

)and

b2 =

(−1 00 −1

)b4 =

(1 00 1

)(b) This follows from

ab =

(1 0−1 1

)and thus (ab)k =

(1 0−k 1

)for all k ∈ Z+

6. Prove that σ2 is an even permutation for every permutation σ.

Solution. This follows from the fact that in the integers

even + even = odd + odd = even

In this case the number of 2-cycles in the representation of σ2 is the sum ofthe number of 2-cycles in σ, so it is the sum of two even or two odd numbers.

7. Define a new addition and multiplication on Z by

a⊕ b = a+ b− 1 and a⊗ b = ab− (a+ b) + 2

Prove that with these operations Z is an integral domain.

Solution. We first check that (Z,⊕,⊗) is a ring.

Closure is clear, it is also clear that the additive identity is the number 1, andthe additive inverse of an element a is 2 − a. Note that this ring (to be) hasno multiplicative identity.

Just checking at the formulas for ⊕ and ⊗ we see that both operations are‘commutative’. So, we have ourselves a commutative ring.

Now assume that for a, b ∈ Z we have

0 = a⊗ b

which means1 = ab− (a+ b) + 2

and this implies(a− 1)(b− 1) = 0

Hence, either a = 1 or b = 1. Since 1 is the additive identity in our ting, thenwe are done.

8. Show that a finite commutative ring with no zero-divisors has a multiplicativeidentity.

Solution. This is problem 1 in part A of the exam of Spring 2006.

Part B.

1. Suppose that A is an n × n matrix such that A3 = 0 but A2 6= 0. Show that{I, A,A2} is independent in the space of all n× n matrices with real entries.

Solution. If {I, A,A2} is linearly dependent, then there is a non-trivial linearcombination of then that equals zero, that is

aA2 + bA+ cI = 0

for some a, b, c ∈ R.

But this implies that the minimal polynomial of A, µA(x), divides p(x) =ax2 + bx+ c. However, A3 = 0 implies that µA(x) = x, x2 or x3 and, since thedegree of p(x) is two, then µA(x) = x, x2, which contradicts the assumptionA2 6= 0.

2. Let A be diagonalizable 2 × 2 matrix. If λ4 = 5λ for each eigenvalue λ of A,show that A4 = 5A.

Solution. Since A is diagonalizable, then there is a matrix P such thatPAP−1 = D, where D is a diagonal matrix with diagonal entries the eigen-values of A.

Since λ4 = 5λ for each eigenvalue λ of A, it is easy to see that D4 = 5D. Butthis implies that so does A = P−1DP , in fact

A4 = (P−1DP )4 = P−1D4P = P−1(5D)P = 5(P−1DP ) = 5A

3. If T : V → V is linear, show that T 2 = IV iff T is an isomorphism andT−1 = T .

Solution. T 2 = I means that the composition of T with itself is the identity.Hence T is its own inverse, and thus it is an isomorphism.

The other direction is immediate.

4. (a) Find A if (A−1 − 3I)T = 2

(−1 25 4

)(b) If det(A) = 2 and det(B) = −3, compute det(A3B−1ATB2).

Solution.

(a) Transposing we get

(A−1 − 3I) =

(−2 104 8

)then, adding 3I both sides we get

A−1 =

(1 104 11

)It follows that

A−1 =1

11− 40

(11 −10−4 1

)=

1

29

(−11 10

4 −1

)(b)

det(A3B−1ATB2) = det(A3) det(B−1) det(AT ) det(B2)

= det(A)3 det(B)−1 det(A) det(B)2

= det(A)4 det(B)

= 24(−3) = −48

5. Invertible 2× 2 matrices with determinant one have the form(a bc 1+bc

a

), where a 6= 0 and bc 6= 1, or(

0 b−1

bd

), where b 6= 0 and d 6= 0, or(

a b−1

b0

), where b 6= 0

Determine the form(s) of all 2× 2 matrices that are their own inverses.

Solution. The short way to solve this would be to realize that A2 = I meansthat the minimal polynomial of A divides x2 − 1. It follows that

µA(x) = x+1 or µA(x) = x−1 or µA(x) = (x−1)(x+1)

In any case, the minimal polynomial of A has distinct roots, thus A is diago-nalizable. It follows that A diagonalizes to

B =

(±1 00 ±1

)So, the answer would be any conjugation of the four matrices B in the previousequation. Of course ±Id do not yield anything interesting, the others, though,give a big family of matrices.

6. Let V = {v ∈ R| v > 0}. Show that V is a vector space over R if thevector addition is ordinary multiplication and scalar multiplication is definedby a · v = va.

Solution. Fix v, w ∈ R>0 and a, b ∈ R.

Since both vw and va are positive reals, then closure works out. Also,

a(v + w) = (vw)a = vawa = av + aw

and(a+ b)v = va+b = vavb = av + aw

Finally, 0 in this vector space is the number 1, as v + 1 = v · 1 = v. It followsthat the additive inverse of v in V is v−1, which is also a positive real number.

7. If Rn = span{v1, v2, · · · , vn} and if x and y in Rn satisfy x · vi = y · vi for alli, show that x = y.

Solution. First note that since the spanning set has the same number as thedimension of the space, then the set is a basis. Now let

x = x1v1 + x2v2 + · · ·+ xnvn and x = y1v1 + y2v2 + · · ·+ ynvn

Thenx · vi = x1(v1 · vi) + x2(v2 · vi) + · · ·+ xn(vn · vi)

andy · vi = y1(v1 · vi) + y2(v2 · vi) + · · ·+ yn(vn · vi)

So,

y1(v1 ·vi)+y2(v2 ·vi)+ · · ·+ym(vm ·vi) = x1(v1 ·vi)+x2(v2 ·vi)+ · · ·+xn(vn ·vi)

or

(x1 − y1)(v1 · vi) + (x2 − y2)(v2 · vi) + · · ·+ (xn − yn)(vn · vi) = 0

for all i.

But this implies

[(x1 − y1)v1 + (x2 − y2)v2 + · · ·+ (xn − yn)vn] · vi = 0

for all i.

So, the vector w = (x1 − y1)v1 + (x2 − y2)v2 + · · ·+ (xn − yn)vn is orthogonalto all vectors in the spanning set of Rn. It follows that w = 0. Since w is alinear combination of linearly independent vectors, then xi = yi for all i, andthus x = y

8. Let Pn denote the space of all polynomials with real coefficients that havedegree at most n (union the zero polynomial). Find a linear transformationT : P2 → P4 such that

T (1) = x4, T (1 + x) = 1 + x3 T (1 + x2) = 1− x2

Solution. Since T is linear, then

T (x) = T (1 + x− 1) = T (1 + x)− T (1) = 1 + x3 − x4

andT (x2) = T (1 + x2 − 1) = T (1 + x2)− T (1) = 1− x2 − x4

It follows that

T (a+ bx+ cx2) = T (a) + T (bx) + T (cx2)

= aT (1) + bT (x) + cT (x2)

= a(x4) + b(1 + x3 − x4) + c(1− x2 − x4)

= (b+ c)− cx2 + bx3 + (a− b− c)x4


Part A. Do five of the following eight problems :

1. Let S be a set, and let P(S) = {A| A ⊂ S} be the collection of all subsetsof S. Define a relation ∼ on P(S) by letting A ∼ B if and only if there isa one-to-one correspondence from A to B. Prove that ∼ is an equivalencerelation.

2. Let G be a group and let f : G → G be defined by f(g) = g−1 for all g ∈ G.Under what conditions is f a group homomorphism? Justify your answer.

3. Let f : G → H be a group homomorphism. Prove that ker(f) is a normalsubgroup of G.

4. Show that in a finite cyclic group of order n with identity element e, theequation xm = e has exactly m solutions, for each positive integer m that is adivisor of n.

5. Let G be any group with no proper nontrivial subgroups, and assume the orderof G is greater than 1. Prove that G is cyclic of order p for some prime p.

6. (a) Let R be a commutative ring such that a2 = a for each a ∈ R. Provethat a+ a = 0 for each a ∈ R.

(b) Prove that (a + b)(a − b) = a2 − b2 for all a, b in a ring R if and only ifR is commutative.

7. Let R be the set of all continuous functions from the set of real numbers toitself. Define addition and multiplication of f, g ∈ R by

(f + g)(x) = f(x) + g(x) and (f · g)(x) = f(x)g(x)

for all numbers x.

Prove that R is a commutative ring under these operations.

8. Elements a and b of a ring R are called zero divisors if a and b are nonzeroand ab = 0. Prove that every finite commutative ring with no zero divisors isa field.


1. (a) Show that A =

[a 0b 0

]is not invertible for any choice of a and b.

(b) If A is any matrix, show that AAT and ATA are both symmetric matrices.

2. Find all 3× 3 diagonal matrices A that satisfy A2 − 3A− 4I = 0

3. If A is an n × n diagonalizable matrix with eigenvalues λ1, λ2, · · · , λn, showthat det(A) = λ1λ2 · · ·λn

4. Show that

< (u1, u2), (v1, v2) >=1

4u1v1 +

1

16u2v2

defines an inner product on R2.

5. Let Pn be the vector space of polynomials in x of degree at most n withreal coefficients union the zero polynomial. Determine the dimension of thesubspace of Pn consisting of all polynomials

a0 + a1x+ a2x2 + · · ·+ anx

n

for which a0 = 0.

6. Show that

[2 11 −5

]and

[a bc d

]commute if a− d = 7b.

7. Suppose A is a square matrix. Suppose x is an eigenvector of A with cor-responding eigenvalue λ, and y is an eigenvector of AT with correspondingeigenvalue µ. Show that if λ 6= µ, then x · y = 0.

8. Let B = {va, v2, v3, v4} be a basis for a vector space V . Find the matrix withrespect to B of the linear operator T : V → V defined by T (v1) = v2, T (v2) =v3, T (v3) = v4, and T (v4) = v1.


Part A.

1. Let S be a set, and let P(S) = {A| A ⊂ S} be the collection of all subsetsof S. Define a relation ∼ on P(S) by letting A ∼ B if and only if there isa one-to-one correspondence from A to B. Prove that ∼ is an equivalencerelation.

Solution. Since the identity defines a one-to-one correspondence from a set toitself and if f : S → T defines a one-to-one correspondence then f−1 : T → Sis also a one-to-one correspondence, then we just need to check transitivity.

So, assume f : S → T is a one-to-one correspondence and g : T → U is also aone-to-one correspondence. Consider the map

g ◦ f : S → U

Since both f and g are bijective, then so is fg.

2. Let G be a group and let f : G → G be defined by f(g) = g−1 for all g ∈ G.Under what conditions is f a group homomorphism? Justify your answer.

Solution. If f is a group homomorphism, then for any g, h ∈ G we get

h−1g−1 = (gh)−1 = f(gh) = f(g)f(h) = g−1h−1

So, h−1g−1 = g−1h−1 for all g, h ∈ G. Since every element in G is the inverseof an element in G, then G is Abelian.

Clearly, if G is Abelian, then f is a homomorphism. Hence,

f is a homomorphism ⇔ G is an Abelian group

3. Let f : G → H be a group homomorphism. Prove that ker(f) is a normalsubgroup of G.

Solution. Let us first check it is a subgroup. We know the identity of thegroup is there, so it is not empty. Also, for g, h ∈ ker(f)

f(gh) = f(g)f(h) = e · e = e

ande = f(gg−1) = f(g)f(g−1) = e · f(g−1) = f(g−1)

So, ker(f) is a subgroup of G

Finally, let x ∈ G, then

f(xgx−1) = f(x)f(g)f(x−1) = f(x) · e · f(x)−1 = f(x)f(x)−1 = e

4. Show that in a finite cyclic group of order n with identity element e, theequation xm = e has exactly m solutions, for each positive integer m that is adivisor of n.

Solution. Let G =< g > and let m be a divisor of n. Then the elementh = gn/m ∈ G has order m (otherwise the order of g wouldn’t be n). Thesubgroup < h > has m elements, and all of them (by Lagrange’s theorem)satisfy xm = e. So, we have at least m solutions for the equation xm = e.

Now consider x = gi ∈ G be such that xm = e. Then

e = (gi)m = gim

So, n divides im, and thus i is a multiple of nm

. Hence gi ∈< h >.

5. Let G be any group with no proper nontrivial subgroups, and assume the orderof G is greater than 1. Prove that G is cyclic of order p for some prime p.

Solution. If G contains an element g of infinite order, then Z ∼=< g >< G,and thus G contains infinitely many subgroups. So, all elements of G havefinite order.

Let g ∈ G, if there is an element h ∈ G that does not belong to < g >,then < g > is a proper subgroup of G, a contradiction. It follows that G is a(finite) cyclic group. Hence G ∼= Zm for some m. We know that Zm containssubgroups of order any divisor of m. It follows that G must be isomorphic toZp for some prime number p

6. (a) Let R be a commutative ring such that a2 = a for each a ∈ R. Provethat a+ a = 0 for each a ∈ R.

(b) Prove that (a + b)(a − b) = a2 − b2 for all a, b in a ring R if and only ifR is commutative.

Solution.

(a)a+ a = (a+ a)2 = a2 + 2a2 + a2 = a+ 2a+ a = 2(a+ a)

So, a+ a = 0.

(b) Assume the ring is commutative, then

(a+ b)(a− b) = a2 − ab+ ba− b2 = a2 − ab+ ab− b2 = a2 − b2

Assume (a+ b)(a− b) = a2 − b2 for all a, b in R. Since

(a+ b)(a− b) = a2−ab+ ba− b2 and a2− b2 = a2−ab+ab− b2

then a2 − ab+ ba− b2 = a2 − ab+ ab− b2, which implies ba = ab

7. Let R be the set of all continuous functions from the set of real numbers toitself. Define addition and multiplication of f, g ∈ R by

(f + g)(x) = f(x) + g(x) and (f · g)(x) = f(x)g(x)

for all numbers x.

Prove that R is a commutative ring under these operations.

Solution. Clearly closure holds for both operations (Calculus!!). Also, thezero function is the additive identity and the identity map is the multiplicativeidentity of R.

All the other things needed to check for R to be a commutative ring followtrivially from the fact that R is a commutative ring.

8. Elements a and b of a ring R are called zero divisors if a and b are nonzeroand ab = 0. Prove that every finite commutative ring with no zero divisors isa field.

Solution. In problem 8 of part A in the exam of Spring 2003 it is shownthat a ring R as the one we are considering must have a one (multiplicativeidentity). That proof uses that, for a fixed r ∈ R∗, the function φr : R → Rdefined by φr(x) = rx is bijective.

Since we know that R has a one, then there must be an s ∈ R such thatφr(s) = 1 (φ is onto!). It follows that rs = 1, and thus r has an inverse.

Part B.

1. (a) Show that A =

[a 0b 0

]is not invertible for any choice of a and b.

(b) If A is any matrix, show that AAT and ATA are both symmetric matrices.

Solution.

(a) The determinant of A is zero no matter what a and b are.

(b) (AAT )T = (AT )TAT = AAT .

To show ATA is symmetric we proceed in the same way.

2. Find all 3× 3 diagonal matrices A that satisfy A2 − 3A− 4I = 0

Solution. Since A satisfies A2 − 3A − 4I = 0, then the minimal polynomialof A divides x2 − 3x− 4 = (x− 4)(x+ 1). It follows that the non-zero entriesof A can only be −1 and/or 4 (using that A is diagonal). So, there are 8possibilities for A (too much to type, so I will write that down in an odd way)

A =

1.5± 2.5 0 00 1.5± 2.5 00 0 1.5± 2.5

3. If A is an n × n diagonalizable matrix with eigenvalues λ1, λ2, · · · , λn, show

that det(A) = λ1λ2 · · ·λn

Solution. We know det(AB) = det(A) det(B).

Since A is diagonalizable, then there is a matrix P such that A = PDP−1,where D is diagonal with and diagonal entries given by the n eigenvalues ofA. Hence,

det(A) = det(PDP−1) = det(P ) det(D) det(P−1) = det(D) = λ1λ2 · · ·λn

4. Show that

< (u1, u2), (v1, v2) >=1

4u1v1 +

1

16u2v2

defines an inner product on R2.

Solution. Let u = (u1, u2), v = (v1, v2) ∈ R2.

< u, u >=1

4(u1)

2 +1

16(u2)

2

which is greater or equal to zero, and zero only when both u1 and u2 are zero.Also,

< u, v >=1

4u1v1 +

1

16u2v2 =

1

4v1u1 +

1

16v2u2 =< v, u >

And finally, for α, β ∈ R and w = (w1, w2) ∈ R2

< αu+ βv, w > = < (αu1 + v1, αu2 + v2), (w1, w2) >

=1

4(αu1 + βv1)w1 +

1

16(αu2 + βv2)w2

=1

4αu1w1 +

1

4βv1w1 +

1

16αu2w2 +

1

16βv2w2

=

(1

4αu1w1 +

1

16αu2w2

)+

(1

4βv1w1 +

1

16βv2w2

)= α

(1

4u1w1 +

1

16u2w2

)+ β

(1

4v1w1 +

1

16v2w2

)= α < u,w > +β < v,w >

5. Let Pn be the vector space of polynomials in x of degree at most n withreal coefficients union the zero polynomial. Determine the dimension of thesubspace of Pn consisting of all polynomials

a0 + a1x+ a2x2 + · · ·+ anx

n

for which a0 = 0.

Solution. We are assuming that

S = {a0 + a1x+ a2x2 + · · ·+ anx

n ∈ Pn; a0 = 0}

is a subspace (if you want to show it, consider the homomorphism φ : Pn → Rdefined by φ(p(x)) = p(0))

It is clear that S is spanned by

B = {x, x2, · · · , xn}

Since B is a subset of the standard basis of R[x], then it must be linearlyindependent. It follows that the dimension of S is n− 1.

6. Show that

[2 11 −5

]and

[a bc d

]commute if a− d = 7b.

Solution. We want the following two products to be the same[2 11 −5

] [a bc d

]=

[2a+ c 2b+ da− 5c b− 5d

][a bc d

] [2 11 −5

]=

[2a+ b a− 5b2c+ d c− 5d

]Then, the two 1, 1 entries being the same implies c = b (also implies by thetwo 2, 2 entries being the same). So, now we have just one equation

2b+ d = a− 5b or a− d = 7b

7. Suppose A is a square matrix. Suppose x is an eigenvector of A with cor-responding eigenvalue λ, and y is an eigenvector of AT with correspondingeigenvalue µ. Show that if λ 6= µ, then x · y = 0.

Solution. This is problem 7 in part B in the exam of Spring 2008.

8. Let B = {va, v2, v3, v4} be a basis for a vector space V . Find the matrix withrespect to B of the linear operator T : V → V defined by T (v1) = v2, T (v2) =v3, T (v3) = v4, and T (v4) = v1.

Solution. Since we are looking for the matrix of T with respect to the baseB, then we have to consider the v1’s as if they were the canonical basis of, inthis case, R4. So,

[T ]B =

0 0 0 11 0 0 00 1 0 00 0 1 0

part a. solve ve of the following eight problems... find the characteristic of ... part b. solve ve...

Documents