mat3-alg algebra 2008-2009 | toby bailey lecture 0 preamblechris/alg/course_2013probs.pdf ·...

MAT3-ALG algebra 2006/7 (tnb) - lecture 0 1

MAT3-ALG algebra 2008-2009 — Toby Baileyhttp://student.maths.ed.ac.uk

lecture 0 preamble

problems

These are “easy” revision problems from year 2 Linear Algebra. It is essential

that you are on top of this earlier material and so you are strongly recommended

to do these exercises. Aim to do all of them by the end of week 2 at the latest.

Throughout, Pn denotes the vector space of polynomials of degree ≤ n in a

variable x and M denotes the vector space of 2× 2 real matrices.

done?problem 0.1 Which of the following are subspaces of the given vector space?

1. {x ∈ R3 | 2x1 − x2 + x3 = 1} ⊆ R3

2. {x ∈ R3 | x1 = 2x2} ⊆ R3

3. {P ∈ P3 |P(1) = 0} ⊆ P3

4. {A ∈M |AT = −A} ⊆M (the “T” denotes matrix transpose).

hint/solution 0.1 All except the first.

done?problem 0.2 For each of the examples in the previous question that is a

subspace give its dimension and write down a basis for the subspace.

hint/solution 0.2 The dimensions are 2, 3, 1 in order for the three examples

which are subspaces. Possible bases are

1. 210

,001

2.

x− 1, x2 − 1, x3 − 1

3. (0 1

−1 0

)

http://student.maths.ed.ac.uk


done?problem 0.3 Calculate the coordinate matrix of x3 with respect to the basis

x3 − x2, x2 − x, x− 1, 1 of P3.

hint/solution 0.3 1

1

1

1

done?problem 0.4 Use the change of basis matrix to find the coordinate matrix of

x in the basis v1, v2 of R2 where

x =

(1

1

), v1 =

(1

−1

), v2 =

(2

1

).

hint/solution 0.4 Set P to be the matrix with v1, v2 as columns. Then the

coordinate matrix of x is P−1x. You can easily check your answer: if k, l are the

two entries in the coordinate matrix (which should be a column matrix) then

you should have x = kv1 + lv2.

done?problem 0.5 What is the span of a set S = {v1, . . . , vk} of vectors? What

does it mean for the set to be linearly independent?

done?problem 0.6 Let U = {x | x1 = 0} and V = {x | x2 = 0} be subspaces of R3.What is the sum U+ V of these subspaces? State the Dimension Theorem for

sums of subspaces and verify it in this example. Is this an example of a direct

sum?

hint/solution 0.6 The sum is all of R3 and the intersection is 1-dimensional.

The Dimension Theorem becomes 3 = 2+ 2− 1.

done?problem 0.7 Which of the following are linear maps?

1. T : M→ P2 where T(A) = the characteristic polynomial of the matrix A.

2. T : M→ R where T(A) = TraceA (Here “Trace” denotes the trace of a

matrix — the sum of the elements on the leading diagonal.)

3. T : P3 → P3 where T : p(x) 7→ p ′(x)

hint/solution 0.7 The second two are linear. The first is not because, for

example, if you multiply a matrix by two the charcteristic polynomial does not

get multiplied by two — even the identity matrix is a counterexample.


done?problem 0.8 Define the kernel and image of a linear map. State the Rank

Theorem (a.k.a. “Rank-Nullity theorem”) for linear maps. For each of the

examples in the previous question that is linear, describe the kernel and image

and verify the theorem.

hint/solution 0.8 Example number two above is surjective and so has image

of dimension 1. The kernel is matrices of zero trace — a 3-dim subspace of M.

Example 3 has 2-dim image (polys of degree ≤ 2) and 1-dim kernel (constant

polys).

done?problem 0.9 Let A be an n × n matrix and let T : Rn → Rn be the linear

map T : x 7→ Ax. Which of the following conditions are equivalent to A having

an inverse?

1. detA 6= 0

2. ker T = {0}

3. im T = Rn

4. T is a bijection.

5. A is diagonalisable.

hint/solution 0.9 All except the last.

done?problem 0.10 Find the eigenvalues and eigenvectors of

A =

(1 2

1 1

).

Hence diagonalise A.

lecture 1 sets

problems

done?problem 1.1 Is it true that {} ⊂ B for all sets B?

hint/solution 1.1 It is true for all sets except B = {} for which it is false.


done?problem 1.2 For the four properties of set algebra in §1.4.2 above, write down

(next to the original, perhaps) what they reduce to for a family of just two sets.

In each case, give a proof. For at least two of those not proved in lectures or

the text, write down a proof too of the version for families.

done?problem 1.3 Let S be a finite set. Write down a formula for the size of P(S).

Write down P(S) in the case of S being the empty set. Does the formula work

in this case?

hint/solution 1.3 ]P(S) = 2]S. For S empty, P({}) = {{}} so ]P({}) = 1

and the formula does hold since 20 = 1.

done?problem 1.4 Let L denote the set of (straight) lines through the origin in

R2. Let M denote the set of (straight) lines through (1, 1) in R2. How many

elements does L ∩M have?

hint/solution 1.4 One. (The line y = x.)

done?problem 1.5 + Hand-in for tutorial Suppose S = {x, y, z}. True or

False:

1. S ∈ P(S);

2. x ∈ P(S);

3. {x, y} ∈ S;

4. {x, y} ∈ P(S);

5. {{x, y}, {}} ∈ P(S);

6. {{x, y}, {}} ⊆ P(S);

7. {{x, y}, {}} ∈ P(P(S));

8. {{}} ∈ P(P(S)).

What is the size of P(P(S))?

hint/solution 1.5 T,F,F,T,F,T,T,T. P(P(S)) = 2]P(S) = 223

= 256


done?problem 1.6 + Hand-in part 1 only for tutorial Use set algebra to

show that

1. A \ (B ∩ C) = (A \ B) ∪ (A \ C)

2. A \ (B \ C) = (A \ B) ∪ (A \ C ′)

hint/solution 1.6

1.

A \ (B ∩ C) = A ∩ (B ∩ C) ′

= A ∩ (B ′ ∪ C ′)= (A ∩ B ′) ∪ (A ∩ C ′)= (A \ B) ∪ (A \ C)

2.

A \ (B \ C) = A \ (B ∩ C ′)= A ∩ (B ∩ C ′) ′

= A ∩ (B ′ ∪ (C ′) ′)

= A ∩ (B ′ ∪ C)

= (A ∩ B ′) ∪ (A ∩ C)

= (A ∩ B ′) ∪ (A ∩ (C ′) ′)

(A \ B) ∪ (A \ C ′)

3.

A \ (B \A) = A \ (B ∩A ′)= A ∩ (B ∩A ′) ′

= A ∩ (B ′ ∪ (A ′) ′)

= A ∩ (B ′ ∪A)

= (A ∩ B ′) ∪ (AcapA)

= (A ∩ B ′) ∪A


done?problem 1.7 Show that

A ∪ (A ∩ B) = A.

You will need to argue by showing that the LHS is a subset of the RHS and

that the RHS is a subset of the LHS. Use set algebra to deduce that also

A ∩ (A ∪ B) = A.

These two results (which can not be deduced from other identities of set algebra

that we have previously stated) are called the ”axioms of absorption”. Use them

together with set algebra as before to show that A \ (B \A) = A.

hint/solution 1.7 Let x ∈ A ∩ (A ∪ B). Then x ∈ A and x ∈ A ∪ B. So

x ∈ A and so LHS ⊆ RHS. Now let x ∈ A. Then x ∈ A ∪ B and so x ∈ LHS.

Now,

A \ (B \A) = A \ (B ∩A ′)= A ∩ (B ∩A ′) ′

= A ∩ (B ′ ∪ (A ′) ′)

= A ∩ (B ′ ∪A)

= Ausing the second form of the axiom of absorption.

lecture 2 cartesian products and functions

problems

done?problem 2.1 Let A,B be finite sets. What is the size of A× B?

hint/solution 2.1 ](A× B) = ]A ]B.

done?problem 2.2 Show that A × (B ∩ C) = (A × B) ∩ (A × C). (Hint: to do

this carefully, show that the LHS is a subset of the RHS and that the RHS is a

subset of the LHS.)

hint/solution 2.2 Let x ∈ A × (B ∩ C). Then x = (a, u) with a ∈ A and

u ∈ B∩C. Hence u ∈ B and u ∈ C. Hence (a, u) ∈ A×B and (a, u) ∈ A×C.

Thus x = (a, u) ∈ (A× B) ∩ (A× C). (We have shown that LHS ⊆ RHS.)

Now let x ∈ (A × B) ∩ (A × C). Then x ∈ A × B and so x = (a, b) with

a ∈ A and b ∈ B. But also x = (a, b) ∈ A×C and so b ∈ C. Hence b ∈ B×Cand so x ∈ A× (B ∩ C). (So also RHS ⊆ LHS and the proof is complete.)


done?problem 2.3 Decide what relationship holds between the following pairs of

sets (one side is a subset of the other, or they are equal, or there is no relation).

Give a proof.

1. A× (B ∪ C) and (A× B) ∪ (A× C);

2. (A× B) ∩ (C×D) and (A ∩ C)× (B ∩D);

3. (A× B) ∪ (C×D) and (A ∪ C)× (B ∪D).

hint/solution 2.3

1. LHS = RHS

2. LHS = RHS

3. LHS ⊆ RHS but not generally equal. (You should find an example where

they are not equal.)

done?problem 2.4 Consider the function f : R→ R2 given by f : t 7→ (cos t, sin t).

The graph of f is a subset of R1 × R2 = R3. Sketch it.

hint/solution 2.4 The graph is {(t, cost, sin t) | t ∈ R} — a curve that screws

round the first axis direction. (It is a “helix”.)

done?problem 2.5 Consider the map f : R2 → R given by f : (x, y) 7→ √x2 + y2.

Describe each of the following using some combination of words, equations or

pictures (proofs not required):

1. f−1(2)

2. f−1([1, 2]) (here [1, 2] is the closed interval)

3. f−1(−1)

4. f(Z) where Z = {(x, y) | (x− 2)2 + (y− 2)2 = 2}

5. f(V) where V = {(x, y) |y > 0}

6. f(U) where U = {(x, y) | x2 − y2 = 1}

7. f(R2)


hint/solution 2.5

1. The circle of radius 2 centred at the origin.

2. The annulus bounded by the circles of radius 1 and 2 centred at the origin

(including the boundaries).

3. {}

4. [1, 3]

5. (0,∞)

6. [1,∞)

7. [0,∞)

done?problem 2.6 + Hand-in for tutorial Let f : X→ Y be a function and

let A,B be subsets of X. Show that

f(A ∩ B) ⊆ f(A) ∩ f(B).

(Hint: You proof should begin: “Let y ∈ f(A∩B)” and should finish with “and

hence y ∈ f(A)∩f(B)”.) Give an example to show that “⊆” can not be replace

with equality.

hint/solution 2.6 Let y ∈ f(A ∩ B). Then there exists x ∈ A ∩ B such

that y = f(x). So x ∈ A and x ∈ V = B. Hence y = f(x) ∈ f(A) and

y = f(x) ∈ f(B). Hence y ∈ f(A) ∩ f(B).

Consider the squaring function f : R→ R with f(x) = x2. Let A = [−2,−1]

and B = [1, 2]. Then A ∩ B = {} and so f(A ∩ B) = {}. On the other hand,

f(A) = f(B) = [1, 4] and so f(A) ∩ f(B) = [1, 4].

done?problem 2.7 let X and Y be finite non-empty sets. Write down a formula for

the number of different functions from X to Y. Now consider the case where

one of X and Y is empty. How many functions are there in that case? Are the

results consistent with your previous formula. (Hint: For the second part use

the definition of function in terms of a subset of the cartesian product.)


hint/solution 2.7 The number of functions is ] Y]X. If X is empty there is

just one function X → Y which has graph {} ⊆ X × Y = {}. Note that this

does (vacuously) satisfy the conditions for a subset of the product to define a

function. This agrees with the formula except where Y is also empty because

00 is generally taken to be undefined.

If Y is empty but X is not there are no functions X → Y since the only

subset of X× Y = {} is empty and that does not satisfy the condition to define

a function. This does agree with the formula.

lecture 3 more on functions

problems

done?problem 3.1 Give conditions on the size of the subsets f−1(y), y ∈ Y that

are characterize f being (a) injective; (b) surjective; (c) bijective.

hint/solution 3.1 The condition is that each set f−1(y) has: (a) no more

than one element; (b) at least one element; (c) exactly one element.

done?problem 3.2 + Hand-in for tutorial

1. Let g ◦ f be injective. Show that f is injective. (Hint: You may find it

best to show that if f is not injective then g ◦ f is not injective.

2. If g ◦ f is injective, does g have to be injective? Give a proof or a

counterexample.

3. What exactly can be deduced if we know that g ◦ f is surjective?

hint/solution 3.2 Suppose f : X→ Y and g : Y → Z.

1. Suppose f is not injective. Then there exist u 6= v in X such that f(u) =

f(v). Then

(g ◦ f)(u) = g(f(u)) = g(f(v)) = (g ◦ f)(v)

and so g ◦ f is not injective.

2. No. Consider f : [0, 1] → [−1, 1] where f : x 7→ x. Let g : [−1, 1] →[−1, 1] where g : x 7→ x2.

3. If g ◦ f is surjective then g is surjective. Proof: Let z ∈ Z. Since g ◦ f is

surjective there exists x ∈ X such that (g ◦ f)(x) = z. Now, f(x) ∈ Y and

g(f(x)) = z and so g is surjective.


done?problem 3.3 Let f : A→ A be a map and suppose that f◦ f = f. What extra

condition on f allows us to deduce that f is the identity map?

hint/solution 3.3 f ◦ f = f means that for all a ∈ A we have f(f(a)) = f(a)

and so we can deduce precisely that f acts as the identity on the image of f.

Thus the extra condition needed is that f is surjective.

done?problem 3.4

1. Let f : A → B be a map. Suppose there exists a map g : B → A such

that g ◦ f = IA : A→ A. Show that f is injective.

2. Suppose f : A→ B is injective. Deduce that there exists a map g : B→ A

such that g ◦ f = IA : A→ A.

3. Under what circumstances is the map g in the previous part unique?

hint/solution 3.4

1. Let f(u) = f(v). Then g(f(u)) = g(f(v)) and so (since g ◦ f = IA) we

have u = v and so f is injective.

2. Let f : A→ B be injective. Fix a ∈ A. Define a map g : B→ A by

g(b) =

{a if b 6∈ im f,

u if b = f(u)..

(We note that each b ∈ im f is of the form f(u) for one and only one

u ∈ A since f is injective.) Then g ◦ f(u) = g(f(u)) = u for all u ∈ Aand so g ◦ f = IA.

3. The choice of a makes g non-unique unless f is surjective and hence a

bijection, in which case g has to be the usual inverse.

done?problem 3.5 State and prove results analogous to the previous exercise that

involve a map h : B→ A such that f ◦ h = IB?

done?problem 3.6 Show that there exists an injection A → B if and only if there

exists a surjection B→ A.


done?problem 3.7 (Harder!) Let S be a set. Show that there can not exist a

surjection f : S → P(S). You might proceed as follows. Suppose there is such

a surjection f. Now consider the subset A ⊆ S defined by

A = {x ∈ S | x 6∈ f(x)}.

Deduce that A itself is not in the image of f for a contradiction.

done?problem 3.8 (Optional!) The Schroder-Bernstein Theorem states that if there

are injections A → B and B → A then there exists a bijection A → B. It is

not entirely trivial. Find a proof (Halmos’s “Naive Set Theory” or perhaps the

web) and understand it!

lecture 4 relations and quotients

problems

done?problem 4.1 + Hand-in for tutorial Show that u ∼ v if and only if

u−v ∈ Z defines an equivalence relation on R. Describe [x] for this equivalence

relation and give a set of representatives.

hint/solution 4.1

1. x− x = 0 ∈ Z and so x ∼ x.

2. Let x ∼ y. Then x− y = k ∈ Z. Then y− x = −k ∈ Z and so y ∼ x.

3. Let x ∼ y and y ∼ z. Then there exist k, l ∈ Z such that x − y = k and

y− z = l. Then x− z = (x− y) + (y− z) = k+ l ∈ Z and so x ∼ z.

So ∼ is an equivalence relation. The equivalence class of x is

[x] = {x+ k |k ∈ Z}.

A set of representatives is [0, 1).

done?problem 4.2 Does a ∼ b ⇐⇒ a + 2b = 3k where k ∈ Z define an

equivalence relation on Z? (Check carefully and investigate if you are not sure

— don’t just guess!)


hint/solution 4.2 Yes it does. To check the three axioms:

1. a+ 2a = 3a and so a ∼ a.

2. Let a ∼ b. Then there exists k such that a + 2b = 3k and so b + 2a =

3(a+ b) − (a+ 2b) = 3(a+ b+ k) and so b ∼ a.

3. Let a ∼ b and b ∼ c. Then there exists k, l such that a + 2b = 3k and

b+ 2c = 3l. So a+ 2c = (a+ 2b) + (b+ 2c) − 3b = 3(k+ l− b) and

so a ∼ c.

Alternatively, notice that a+ 2b is a multiple of 3 iff a−b is and so this is just

a familiar example in disguise.

done?problem 4.3 + Hand-in for tutorial Let a be the vector (1, 1) ∈ R2.Show that

x ∼ y ⇐⇒ x− y = λa for some λ ∈ R

defines an equivalence relation on R2. Sketch the equivalence classes and show

that R = {(x, y) | x+ y = 0} is a set of representatives.

hint/solution 4.3

• x− x = 0 = 0a and so x ∼ x

• Let x ∼ y. Then x − y = λa for some λ ∈ R. Then y − x = (−λ)a and

so y ∼ x.

• Let x ∼ y and y ∼ z. Then x−y = λa and y−z = µa for some λ, ν ∈ R.

Then x− z = (x− y) + (y− z) = (λ+ µ)a and so x ∼ z.

Thus ∼ is an equivalence relation. The equivalence classes are all lines parallel

to x = y. The line R intersects every such line in a single point and so is a set

of representatives.


done?problem 4.4

(a) Show that if x ∈ R2 is non-zero then there exists an invertible 2× 2 matrix

A such that Ae1 = x where e1 is the first standard basis vector in R2.

(b) Use the above to show that given two non-zero vectors x, y ∈ R2 there

exists an invertible 2 × 2 matrix P such that y = Px. (Hint: take x to e1and then e1 to y.)

(c) Let x ∼ y iff there exists an invertible 2 × 2 matrix A such that y = Ax.

Show that this defines an equivalence relation on R2.

(d) What are the equivalence classes for this equivalence relation? Give a set

of representatives. How many elements does R2/ ∼ have?

hint/solution 4.4

(a) Let z be any vector such that x, z are linearly independent. Then let A be

the matrix with x, z as columns. Then Ae1 = x and A is invertible.

(b) Let A be as above and let Be1 = y with B invertible. Then y = Be1 =

BA−1x and BA−1 is invertible with inverse AB−1.

(c) (1) x = Ix and so x ∼ x

(2) Let x ∼ y then there exists invertible A with y = Ax. Then x = A−1y

and so x ∼ y.

(3) Let x ∼ y and y ∼ z then there exists invertible A,B with y = Ax and

z = By. Then z = (BA)x and of course BA is invertible (with inverse

A−1B−1). Thus x ∼ z.

(d) There are thus two equivalence classes: all the non-zero vectors in one and

just the zero vector in the other. A set of representatives could be {e1, 0}.

The quotient has two elements.

done?problem 4.5 Consider the set X = {(a, b) |a, b ∈ Z and b 6= 0} (so an

element of X is a pair of integers with the second one non-zero). Show from

the definition that

(a, b) ∼ (k, l) ⇐⇒ al = bk

defines an equivalence relation on X.


hint/solution 4.5

• (a, b) ∼ (a, b) since ab = ab.

• The relation is obviously symmetric under exchange of (a, b) and (c, d).

• Suppose (a, b) ∼ (c, d) and (c, d) ∼ (e, f). Then ad = bc and cf = de.

Multiply the first through by ef to get

afde = cfbe.

Now, assuming cf = de 6= 0 we can cancel to get af = be and so

(a, b) ∼ (e, f). If on the other hand cf = de = 0 then a, c, e are all zero

(remembering that b, d, f are all nonzero) and the result follows in this

case too.

lecture 5 the first isomorphism theorem (FIT) for sets

problems

done?problem 5.1 Prove that the operation of multiplication is well-defined in Zn.

hint/solution 5.1 Taking the equivalence relation ∼ on Z so that Zn = Z/ ∼,

let a ∼ a ′ and b ∼ b ′. Then there exist k, l ∈ Z such that a ′ − a = kn and

b ′ − b = ln. Then a ′b ′ = (a+ kn)(b+ ln) = ab+n(bk+ la+ kln) and so

ab ∼ a ′b ′.

done?problem 5.2 + Hand-in for tutorial Consider the equivalence relation

x ∼ y ⇐⇒ |x| = |y| on Z. Use the absolute value function (i.e. the modulus

function) and FIT for sets to deduce that the quotient Z/ ∼ can be identified

with N ∪ {0}.

hint/solution 5.2 Consider the surjection f : Z→ N ∪ {0} where f : n 7→ |n|.

Then x ∼ y iff f(x) = f(y). FIT for sets implies there is a canonical bijection

Z/ ∼→ N ∪ {0}.

done?problem 5.3 Consider the equivalence relation ∼ on the set X = {(a, b) |a, b ∈Z and b 6= 0} as in the problem for lecture 4. Show that setting

[(a, b)] + [(c, d)] = [(ad+ bc, bd)], [(a, b)] ∗ [(c, d)] = [(ac, bd)]

is a well-defined “addition” and “multiplication” on X/ ∼.


hint/solution 5.3 Suppose that [(a, b)] = [(a ′, b ′)] and [(c, d)] = [(c ′, d ′)].

Then ab ′ = a ′b and cd ′ = c ′d. Then

(ad+ bc)b ′d ′ = adb ′d ′ + bcb ′d ′ = a ′bdd ′ + bc ′b ′d = bd(a ′d ′ + b ′c ′)

and so [(ad + bc, bd)] = [(ad + bc, bd)] and so addition is well defined. The

case of multiplication is similar (but easier).

done?problem 5.4 This continues from the previous problem. Define a map f :

X → Q (where Q is the rational numbers) by f : (a, b) 7→ a/b. Use FIT

for sets to deduce that X/ ∼ can be identified with Q. Note by the way that

the previous exercise can be taken to be a definition of Q and its arithmetic

operations which does not use the idea of fractions or real numbers.

hint/solution 5.4 Consider the map f : X→ Q defined by f : (a, b) 7→ a/b.

Then

f(a, b) = f(k, l) ⇐⇒ a/b = k/l ⇐⇒ (a, b) ∼ (k, l).

The map is clearly surjective too and so by FIT for sets we can identify the set

of equivalence classes with Q.

done?problem 5.5 Let S1 denote the unit circle, thought of as the unit-modulus

complex numbers. Consider the map f : R → S1 defined by f : x 7→ e2πxi.

Show that the equivalence relation on R defined by u ∼ v ⇐⇒ f(u) = f(v) is

that u and v are equivalent if and only if they differ by an integer. Use FIT for

sets to show that R/ ∼ can be identified with S1.

lecture 6 fields and n-dimensional space

problems

done?problem 6.1 + Hand-in for tutorial Find the multiplicative inverses

of the non-zero elements in Z7. (Just experimenting is probably easier than

using the Euclidean algorithm.)

hint/solution 6.1 By experiment, 2.4 = 1, 3.5 = 1 6.6 = 1 and so 2, 4 are

mutually inverse as are 3, 5. The element 6 is its own inverse.

done?problem 6.2 Show that if L ⊆ K is a subfield then 1, 1+ 1, 1+ 1+ 1, . . . are

all elements of L. (It is tempting to call these 1, 2, 3, . . . but note that (e.g. in

Zp) that they are not necessarily all distinct.) Deduce that Zp does not have

any subfields (other than itself).

What do you think the smallest subfield of R is?


hint/solution 6.2 We know 1 ∈ L and so all the given elements are as they

are all obtained by repeated addition. If K = Zp then this exhausts all the

elements of Zp so every subfield contains (and therefore is) the whole field.

By the above argument, every subfield of R contains N and hence (closure

under additive inverses) Z. Then consider closure under division — we deduce

every subfield contains all of Q. Thus Q is the smallest subfield of R.

done?problem 6.3 + Hand-in for tutorial Do the following equations have

solutions in the fields C,R,Q,Z3,Z2?

x2 + 1 = 0, x2 − x− 1 = 0

Note: what this means in each case is this: is there an element of the given

field such that if you substitute it in to this equation and do all the arithmetic

in that field then you get zero? The answers for the first three fields should be

easy from elementary background knowledge. The last two fields have very few

elements and so you can just experiment.

hint/solution 6.3 The first equation has solutions in C but not R or Q. In

Z3 there are no solutions but in Z2 then x = 1 is a solution since 12+1 = 2 = 0

in Z2.The second equation has solutions in C and R but not Q. It has no solutions

in the two finite fields.

done?problem 6.4 Show that Q[√2] (definition in notes) is a subfield of R.

hint/solution 6.4 Q[√2] := {a+ b

√2 |a, b ∈ Q} ⊆ R. Clearly 0, 1 ∈ Q[

√2]

and if u, v ∈ Q[√2] then so are −u, u + v, uv (the latter by multiplying out).

Also, if u = a+b√2 then so is 1/(a+b

√2) as can be seen by multiplying top

and bottom by a− b√2.

done?problem 6.5 Find all the vectors in Z23 that are scalar multiples of

a =

(1

2

).

hint/solution 6.5

0a =

(0

0

), 1a = a =

(1

2

), 2a =

(2

1

),

done?problem 6.6 Find the vectors in the subspaces in §6.4.3.


hint/solution 6.6

• The given 1-dim subspace of Z23 contains just the vectors(0

0

),

(1

1

),

(2

2

).

• The given 2-dim subspace of Z32 contains just the vectors

0a+0b =

000

, 1a+0b =

110

, 0a+1b =

111

, 1a+1b =

001

.lecture 7 vector spaces — revision

7.1 setting

Vector spaces over a general field F.

problems

done?problem 7.1 + Hand-in for tutorial In Z32 find all the vectors in

Span(x, y) where

x =

110

, y =

011

hint/solution 7.1 Since there are only two scalars, there are just four vectors:

0 = 0x+ 0y, x = 1x+ 0y, y = 0x+ 1y, x+ y = 1x = 1y, the last is101

.done?problem 7.2 Find all the vectors in the subspace V ⊆ Z32 given by V = {x ∈

Z32 | x1 + x2 + x3 = 0}.

hint/solution 7.2 You should find four vectors - it’s the zero vector together

with every vector containing two 1’s and one 0.


done?problem 7.3 Show that in Z35 the vectors113

,202

,430

are linearly dependent.

hint/solution 7.3 Various methods: you could calculate the determinant of

the matrix with these columns, either by row expansion or Gauusian elimination,

and show that it is zero. Alternatively, hunt around for a linear dependency: in

fact the last column is the first plus twice the second.

done?problem 7.4 Give a basis of the subspace of Z35 defined by the equation

x1 + x2 + x3 = 0. What is the dimension of this subspace? How many vectors

are there in this subspace? Find the coordinate matrix of the vector

v =

433

in you chosen basis.

hint/solution 7.4 In this subspace, the general element is x1

x2

−(x1 + x2)

= x1

104

+ x2

014

and so the two vectors on the right-hand side are a basis. The dimension is

therefore 2. The coordinate matrix of v in this basis is(4

3

).

The subspace has 25 vectors: since it is two-dimensional an element is specified

by choosing two scalars.

done?problem 7.5 + Hand-in for tutorial Complete the following sentence

without using any terms from linear algebra. “If R were finite-dimensional as a

vector space over Q it would mean that there exist a finite number r1, . . . , rnof . . . such that every . . . could be written as . . . ”


hint/solution 7.5 If R were finite-dimensional as a vector space over Q it

would mean that there exist a finite number r1, . . . , rn of real numbers such

that every real number r could be written as

r = q1r1 + · · ·+ qnrn

with the qk all rational.

done?problem 7.6 + Hand-in for tutorial There are seven different non-

zero vectors in Z32 and hence (since the only scalars are {0, 1}) there are seven

different 1-dimensional subspaces. There are also seven different linear equations

of the form

λ1x1 + λ2x2 + λ3x3 = 0, λj ∈ Z2

apart from the trivial one with all the λj being zero. Each

of these defines a different 2-dimensional subspace of Z32.In the picture there are seven blobs, and seven lines

(six straight together with a circle). Label each blob

with a different 1-dimensional subspace and each line

with a 2-dimensional subspace in such a way that a

blob is on a line iff the 1-dimensional subspace lies in-

side the 2-dimensional one (i.e. iff the vector satisfies

the equation). (Note by the way that this config-

uration has the property that through every pair of

points there is a unique line and every pair of lines

meet in precisely one point. It is an example of a

”finite projective plane”.)

hint/solution 7.6 One solution is to set the vertices of the triangle equal to

the standard basis vectors and the edges can then be given by the equations

x = 0, y = 0, z = 0. The centre of the triangle can be (111)T and the other

three non-zero vectors can be put in the middle of the sides. The circle has

equation x+ y+ z = 0.

done?problem 7.7 u Challenge How many 2-dimensional subspaces does Z42have? You might want to think along the lines of defining such a subspace

by choosing a non-zero vector and then choosing another vector that is not a

multiple of it — their span determines a subspace. Count how many ways there

are of doing this and then work out how many times each subspace has been

counted.


lecture 8 linear maps — revision

8.1 setting

Linear maps T : U → V where U,V are finite-dimensional vector spaces over

the same field F.

problems

done?problem 8.1 + Hand-in for tutorial Consider the linear map T :

Z32 → Z32 with matrix 1 1 0

1 0 1

0 1 1

.Find all the vectors in ker T and im T .

hint/solution 8.1 The determinant is zero. There are two linearly indepen-

dent colums and so the rank is 2. The image is therefore all the vectors that

are linear combinations (with Z2 coefficients) of (say) the first two columns. In

fact, the image is precisely the three vectors that are the columns.

By the rank theorem, the kernel is 1-dimensional. A vector in the kernel is111

.Thus the kernel is this vector and the zero vector.

done?problem 8.2 How many linear maps T : Z22 → Z22 are there? How many of

them are invertible? (Hint: equivalently, how many 2 × 2 matrices are there

with entries in Z2 and how many have inverses? Remember that a 2× 2 matrix

has an inverse if its first column is non-zero and the second column is not a

multiple of the first.)

Let A be such an invertible matrix and let

e1 =

(1

0

), e2 =

(0

1

), e3 =

(1

1

).

Show that Aej 6= Aek unless j = k and that Aej 6= 0. Deduce that multi-

plication by A permutes the three non-zero vectors in Z22. Find explicitly the

permutation given by each of the six invertible matrices.


hint/solution 8.2 There are two choices for each entry and so there are

24 = 16 linear maps. To count the invertible ones, there are 3 choices for a

non-zero first column. Then for each such choice there are 2 non-zero vectors

that are not multiples of the first for the second column. Hence there are

2× 3 = 6 such matrices.

The linear map given by an invertible A is injective and so the given con-

ditions hold and thus A permutes the three vectors. The images of e1, e2 are

the columns of the matrix and so we see easily that permutations are (in cycle

notation)(1 0

0 1

)! identity perm,

(0 1

1 1

)! (123),

(1 1

1 0

)! (132)(

0 1

1 0

)! (12),

(1 0

1 1

)! (13),

(1 1

0 1

)! (23)

done?problem 8.3 Let Pj[Z3] denote the vector space of polynomials of degree ≤ jwith coefficients in Z3.

1. State the dimension of Pj[Z3].

2. Show that 1+ x, x+ x2, x2 is a basis for P2[Z3].

3. Show that T : P2[Z3] → P3[Z3] where T : p(x) 7→ (x + 2)p(x) is a linear

map.

4. Calculate the matrix of T with respect to the basis above for P2[Z3] and

the basis 1, x, x2, x3 for P2[Z3].


hint/solution 8.3

1. The dimension id j+ 1.

2. Suppose

λ1(1+ x) + λ2(x+ x2) + λ3x2 = 0.

The constant term implies λ1 = 0, then the linear term implies λ2 = 0

and the quadratic term then implies λ3 = 0. So the vectors are linearly

independent and since the space is 3-dimensional they form a basis. (Al-

ternatively, show that the vectors span and then appeal to dimension, or

show that the span and are linearly independent (and omit the argument

by dimension.)

3.

T(p(x)+q(x)) = (x+2)(p(x)+q(x)) = (x+2)p(x)+(x+2)q(x) = T(p(x))+T(q(x))

and also

T(λp(x)) = (x+ 2)λp(x) = λ(x+ 2)p(x) = λT(p(x)).

4. T(1+x) = (x+2)(x+1) = x2+2 and T(x+x2) = (x+2)(x2+x) = x3+2x

and T(x2) = (x+ 2)x2 = x3 + 2x2. So taking the coordinates of each in

the given basis we get that the matrix is2 0 0

0 2 0

1 0 2

0 1 1

.lecture 9 invariant subspaces and block matrices

9.1 setting

Linear maps V → V where V is a (normally finite-dimensional) vector space

over a field F.

problems

done?problem 9.1 Let T : V → V be linear. Prove ker T is an invariant subspace.

hint/solution 9.1 Let v ∈ ker T . Then Tv = 0 ∈ ker T . So ker T is invariant.


done?problem 9.2 Consider rotations (about the origin) and reflections (in lines

through the origin) in the plane. Describe all 1-dimensional invariant subspaces.

hint/solution 9.2 For rotations there are none unless θ = 0, Pi in both of

which cases every 1-dimensional subspace is invariant.

done?problem 9.3 Show that every 1-dimensional invariant subspace is the span of

an eigenvector.

hint/solution 9.3 Let U ⊆ V be invariant under T and 1-dimensional. Then

U = Span(u) for some u. Then Tu ∈ U and so Tu = λu for some λ and so u

is an eigenvector.

done?problem 9.4 Consider matrix multiplication of block upper-triangular matri-

ces. Using the notation of §9.3.1, show that M is invertible if and only if the

blocks A and D are invertible.

hint/solution 9.4 The block upper-triangular case is

MM ′ =

(A B

0 D

)(A ′ B ′

0 D ′

)=

(AA ′ AB ′ + BD ′

0 CB ′ +DD ′

).

So for MM ′ = I we need AA ′ = I and DD ′ = I and so A and D must

be invertible. Conversely, suppose A,D are invertible. Then if we set B ′ =

−A−1BD ′ then we will have MM ′ = I and so M is invertible.

done?problem 9.5 + Hand-in for tutorial Show that T : V → V has an

invariant subspace of dimension l if and only if V has a basis with respect to

which the matrix of T is block lower-triangular.

Deduce that if the matrix M is n×n block upper-triangular and P is n×nsuch that Pij = 1 when i+ j = n+ 1 and zero otherwise, then P−1MP is block

lower-triangular.

hint/solution 9.5 If a linear map has block lower-triangular matrix with re-

spect to a basis, and the second diagonal block is l× l then the span of the last

l basis vectors is invariant. The proof is the same as the block upper-triangular

case.

The matrix P is the change of basis matrix from a basis to the same basis

in reverse order. Thus changing basis with it will convert block upper-triangular

to block lower-triangular and vice versa.


done?problem 9.6 Let T : V → V be a linear map and suppose there is a flag {Vk}

in V such that T(Vk) ⊆ Vk−1 for k = 1, . . . , n. Show that Tn = 0.

Show that there exists such a flag for T if and only if there exists a basis

for V such that the matrix of T is strictly upper-triangular (meaning i ≥ j =⇒Tij = 0).

hint/solution 9.6 Suppose there exists a basis such that the matrix of T is

strictly upper triangular. Then setting Vk to be the span of the first k basis

vectors we have such a flag. Conversely, given such a flag, construct a basis

such that Vk is the span of the first k basis vectors. Then the matrix is strictly

upper-triangular.

lecture 10 quotients and the 1st isomorphism theo-rem

10.1 setting

Vector spaces over a field F, which may be assumed to be finite-dimensional

(and needs to be when we make statements about dimension).

problems

done?problem 10.1 Let V ⊆ X be a subspace. Check that x ∼ y ⇐⇒ x− y ∈ Vdoes define an equivalence relation on X

hint/solution 10.1 Just check the three axioms.

done?problem 10.2 Check that the addition and scalar multiplication defined in

§10.3.3 is well-defined.

hint/solution 10.2 For addition, let [x] = [x ′] and [y] = [y ′]. Then there

exist u, v ∈ V such that x− x ′ = u and y− y ′ = v. Then

(x ′ + y ′) − (x+ y) = u+ v ∈ V

and so [x + y] = [x ′ + y ′] and the addition is well-defined. The argument for

scalar multiplication is similar.

done?problem 10.3 Check that the distributive law (λ(x + y) = λx + λy for all

vectors x, y and a scalar λ) holds in X/V.


hint/solution 10.3

λ([x] + [y]) = λ([x+ y])

= [λ(x+ y)]

= [λx+ λy]

= [λx] + [λy]

= λ[x] + λ[y]

done?problem 10.4 Write out a careful proof of the fact that P : X→ X/V where

P : x 7→ [x] is linear, surjective and has kernel V.

hint/solution 10.4

• It is linear because

P(λx+ µy) = [λx+ µy] = λ[x] + µ[y] = λPx+ µPy.

• It is surjective because if [x] ∈ X/V then P(x) = [x].

•Px = 0 = P0 ⇐⇒ [x] = [0] ⇐⇒ x ∼ 0 ⇐⇒ x− 0 = x ∈ V.

So ker P = V.

done?problem 10.5 + Hand-in for tutorial Suppose T : X→ Y is a linear

map and that V ⊆ X is a subspace such that V ⊆ ker T . Define a linear map

T : X/V → Y (checking that the map you have defined is linear) such that the

following diagram commutes. (The vertical map is the usual one.) Find the

dimension of the kernel of T in terms of the dimensions of V and ker T . (Hint:

apply the rank theorem to T and T .)

XT

−→ Y↓ ↗ eTX/V


hint/solution 10.5 Define a map T : X/V → Y by T : [x] 7→ Tx. This is

well-defined since if [x] = [x ′] then x−x ′ = v ∈ V and then T(x−x ′) = 0 since

v ∈ ker T . The map is linear since

T(λ[x] + µ[y]) = T([λx+ µy]) = T(λx+ µy) = λTx+ µTy = λT [x] + µT [y].

(N.B. If you’re reading this solution you need to understand why each one of

that chain of equalities is true.)

The triangle commutes since if x ∈ X then the vertical map takes you to [x]

and T([x]) = T(x) by definition of T .

To find the dimension, the rank theorem for T gives

dimX− dim ker T = dim im T

and that for T gives

dimX/V − dim ker T = dim im T .

Now clearly im T = im T and so subtracting and using dimX/V = dimX−dimV

we get dim ker T = dim ker T − dimV.

done?problem 10.6 (Harder!) Let U ⊆ V ⊆ X be subspaces of X.

(a) Show that there is a canonical linear map S : V/U→ X/U.

(b) Show that S is injective.

(c) Show that there is a canonical linear map T : X/U→ X/V. Show that T is

surjective.

(d) Show that ker T = imS.

So, if we identify V/U with its image in X/U (reasonable, since S is injective)

then we can deduce that there is an isomorphism

X/U

V/U→ X/V.

(Notation suggestion: write x ∼U y if x − y ∈ U and write [x]U for the

equivalence class under this relation. Similarly for V.)


hint/solution 10.6

(a) Since if v ∈ V then v ∈ X we can define S by S : [v]U 7→ [v]U.

(b) Let S([v]U) = S([v ′]U). Then v− v ′ ∈ U and so [v]U = [v ′]U.

(c) Define T by T : [x]U 7→ [x]V . This is well-defined because if [x]U = [x ′]Uthen x − x ′ = u ∈ U ⊆ V. So [x]V = [x] ′V . Clearly every [x]V is obtained

in this way so T is surjective.

(d) T [x]V = 0 iff x ∈ V and so ker T = imS.


lecture 11 quotients and linear maps

11.1 setting

Finite-dimensional vector spaces over a field F. For the result that for every

linear map T : V → V there exists a basis with respect to which the matrix of

T is upper-triangular, the field is assumed to be C.

problems

done?problem 11.1 + Hand-in for tutorial The 3× 3 real matrix

A =

1 0 −1

0 −2 0

1 0 0

has a single real eigenvalue. Find a real invertible matrix P such that P−1AP is

block upper-triangular.

hint/solution 11.1 The eigenvalue is −2 with eigenvector the second stan-

dard basis vector. Thus P can be any invertible 3 × 3 matrix with that as its

first column. For example,

P =

0 1 0

1 0 0

0 0 1

.done?problem 11.2 Consider the differentiation map D : P3 → P3 where as usual

Pn is the vector space of polynomials of degree ≤ n in a variable x. Show

that D gives rise to a linear map D : P3/V → P3/V where V is the subspace

of constant polynomials. What is the matrix of D with respect to the basis

[x], [x2], [x3] of P3/V?

hint/solution 11.2 The map D is such that D(V) = {0} ⊆ V and so there is

an induced linear map D : P3/V → P3/V . The matrix of D with respect to the

basis 1, x, x2, x3 is 0 1 0 0

0 0 2 0

0 0 0 3

0 0 0 0

and so the bottom-right 3 × 3 square is the matrix of D with respect to the

given basis.


done?problem 11.3 + Hand-in for tutorial Let J : R2 → R2 be rotation

anticlockwise by a rightangle. Does there exist a basis for R2 with respect to

which the matrix of J is upper-triangular? If not, explain where the proof we

gave for complex vector spaces breaks down.

hint/solution 11.3 No. The reason is that there are no real eigenvectors,

which in turn is because not all real polynomials have real roots.

done?problem 11.4 The aim of this question is to prove that if T : V → V is

nilpotent (so that Tk = 0 for some k), then there exits a basis for V such that

the matrix of T is strictly upper-triangular. (The proof in the notes applies only

if V is complex.) Let k be the least k such that Tk = 0.

1. Show that if T is nilpotent then T has zero as an eigenvalue.

2. Let U ⊆ V be an invariant subspace for nilpotent T . Show that T :

V/U→ V/U is also nilpotent.

3. Now argue analogously to the proof of the main theorem in the notes for

this lecture.

hint/solution 11.4

1. Since Tk−1 6= 0 there must be a vector v ∈ V such that u := Tk−1v 6= 0.

Then Tu = 0 and this is an eigenvector with eigenvalue 0.

2. Suppose T : V → V and Tk = 0. Let [v] ∈ V/U. Then Tk[v] = [Tkv] =

[0] = 0. So Tk = 0.

3. The proof now is completely analogous to that in the notes.

lecture 12 linear maps V → V — eigenspaces

12.1 setting

Linear maps T : V → V where V is finite-dimensional over a field F.

problems

done?problem 12.1 Write out the details of the proof of §12.4.3.


hint/solution 12.1 By definition of the sum, if v ∈ V then we can write

v = u1 + · · ·+ uk, uj ∈ Uj.

Suppose that also

v = u ′1 + · · ·+ u ′k, u ′j ∈ Uj.

Then subtracting we get

0 = (u1 − u ′1) + · · ·+ (uk − u ′k)

and so by the definition of “direct sum” uj − u′j = 0 for all j.

done?problem 12.2 + Hand-in for tutorial Let x1, . . . , xk be non-zero

vectors and let Uj = Span(xj). Show that the sum of the subspaces Uj is direct

if and only if the vectors xj are linearly independent.

hint/solution 12.2 Note that Span(xj) = {αxj |α ∈ F}. Now suppose that

the xj are linearly independent. Let

u1 + · · ·+ uk = 0, uj ∈ Uj.

Then for each j we can write uj = αjxj where αj ∈ F. So

α1x1 + · · ·+ αkxk = 0

and so by linear independence all the αj are zero and so all the uj are zero.

Conversely, suppose the sum is direct and let

α1x1 + · · ·+ αkxk = 0.

Then setting uj = αjxj we have

u1 + · · ·+ uk = 0, uj ∈ Uj

and so by the sum being direct we have uj = 0 for all j. Hence λj = 0 (given

that the xj are nonzero) for all j and so the xj are linearly independent.

done?problem 12.3 Let V =⊕j=1,...,kUj be a direct sum of subspaces. Suppose

that we are given a basis for each subspace Uj. Show that the totality of all

these basis vectors forms a basis for V.


hint/solution 12.3 Clearly they span. Now suppose that vj,i, i = 1, . . . , dj is

a basis for the dj-dimensional subspace Uj. Let

k∑j=1

dj∑i=1

αj,ivj,i = 0.

Then since the sum is direct and

uj :=

dj∑i=1

αj,ivj,i ∈ Uj

we deduce that uj = 0 for all j. But vj,i, i = 1, . . . , dj is a basis for Uj and so

linearly independent. So all the αj,i are zero.

done?problem 12.4 Write out the details of the proof of §12.4.4. In particular,

provide more detail on the final sentence.

In the following problems, the idea is to calculate and spot the pattern. Don’t

get too hung-up on proofs!

In the following problems we write Jn(α) for the Jordan matrix which is the

n × n matrix whose i, j-th entry is: α if i = j; 1 if j = i + 1 and 0 otherwise.

Thus for example

J3(−5) =

−5 1 0

0 −5 1

0 0 −5

.done?problem 12.5 + Hand-in for tutorial What is the characteristic

polynomial of J3(α)? Find its eigenvalues, and their algebraic and geometric

multiplicity. How does this generalize to Jn(α)?

hint/solution 12.5 The characteristic polynomial is (λ−α)3. The only eigen-

value is α and the eigenspace is the span of the first basis vector. It has algebraic

multiplicity 3 and geometric multiplicity 1.

The general result is obtained by replacing “3” with “n” in the previous

paragraph.

done?problem 12.6 Consider D : P3 → P3 (where D denotes differentiation and

Pn is the vector space of real polynomials in “x” of degree n or less). Find a

basis with respect to which D has matrix J4(0).

hint/solution 12.6 The easiest example is 1, x, x2/2!, x3/3!.


done?problem 12.7 For k ≥ 1, the k-th generalized eigenspace of T : V → V with

eigenvalue λ is

Eλ,k := {v ∈ V | (T − λI)kv = 0}.

So, for k = 1 the generalized eigenspace is just the eigenspace in the usual

sense.

1. Show that if k ≤ l then Eλ,k ⊆ Eλ,l.

2. Let A = J3(α). Show that Eα,k = V for k ≥ 3. Describe Eα,2 and give

its dimension.

3. In general, what is dimEα,k for the Jordan matrix Jn(α)?

hint/solution 12.7

1. Clearly if (T − λI)kv = 0 then (T − λI)lv = 0 for all l ≥ k and so

Eλ,k ⊆ Eλ,l.

2. Computing,

(J3(α) − αI)2 =

0 0 1

0 0 0

0 0 0

The generalized eigenspace is thus 2-dimensional and it is the span of the

first 2 standard basis vectors.

3. A bit more experiment shows that generally the k-th generalized

eigenspace is k-dimensional and is the span of the first k standard ba-

sis vectors.

lecture 13 the Cayley-Hamilton theorem and the min-imal polynomial

13.1 setting

T : V → V is a linear map and V is a finite-dimensional vector space over a field

F. Its characteristic polynomial is cT (x) and its minimal polynomial is mT (x).

(We will usually write the characteristic polynomial using x as the variable in

place of the more familiar λ.)

We will write λ1, . . . , λk for the distinct eigenvalues of T or A but µ1, . . . , µlfor the eigenvalues listed with multiplicity.


problems

In the problems for this lecture the idea is to calculate and spot the pattern.

Don’t get hung-up on proofs!

In the following problems we write Jn(α) for the Jordan matrix which is the

n × n matrix whose i, j-th entry is: α if i = j; 1 if j = i + 1 and 0 otherwise.

Thus for example

J3(−5) =

−5 1 0

0 −5 1

0 0 −5

.done?problem 13.1 What is the minimal polynomial of J3(α)? How does this

generalize to Jn(α)?

hint/solution 13.1 The minimal polynomial is (λ− α)3 (by experiment - no

lower power of (λ− α) annihilates J3(α)).

The general result is obtained by replacing “3” with “n”.


done?problem 13.2 + Hand-in for tutorial Consider ODEs of the form

x ′ = Ax where x(t) =

(u(t)

v(t)

)and A is a constant 2 × 2 real matrix. Re-

call from CVD that one can solve such systems when A is real-diagonalizable

(“nodes”,“saddles” and “stars”) and complex-diagonalizable (“foci” and “cen-

tres”).

1. Let A be a real 2× 2 matrix that is not diagonalizable by real or complex

P. Show that the minimal polynomial of A is equal to the characteristic

polynomial which is of the form (x− λ)2, where λ is the eigenvalue.

2. Deduce (Cayley-Hamilton) that if f2 is not an eigenvector then f1 :=

(A− λI)f2 is.

3. Show that the matrix of x 7→ Ax in a basis f1, f2 (as in the previous part)

is J2(λ) and hence that A = PJ2(λ)P−1 where P has f1, f2 as columns.

4. Show that

exp(tJ2(λ)) =

(etλ tetλ

0 etλ

).

5. Show that exp(tA) = P exp(tJ2(λ))P−1 where P is the matrix with

columns f1, f2.

6. Solve

u ′ = −3u− v, v ′ = 4u+ v.

(Recall from CVD that the general solution of x ′ = Ax is x(t) =

exp(tA)

(C1

C2

).)

It is purely optional for this course, but you might like to figure out what the

phase portraits for these systems look like.

Remark: this approach to the solution obscures the relationship of the so-

lutions to the basis. Alternatively we can observe that

eλtf1, eλt(tf1 + f2)

are two independent solutions.


hint/solution 13.2

1. For A not to be diagonalizable it must not have two distinct eigenvalues.

So its characteristic polynomial is (x− λ)2. Its minimal polynomial can’t

be (x − λ) otherwise A = λI which is diagonalizable. Hence its minimal

polynomial is (x− λ)2.

2. Let v2 be such that v1 := (A − λI)v2 6= 0. Then (A − λ)v1 = 0 by

Cayley-Hamilton and so v1 is an eigenvector.

3. Follows from Tf1 = λf1 and Tf2 = f1 + λf2 and the change of basis

formula.

4. Multiplying a couple of times one finds that

(tJ2(λ))k =

(tkλk ktkλk−1

0 tkλk

)and summing, the result follows. (Alternatively, use the clever trick of

writing J2(λ) as λI plus a matrix with an entry only in the top corner and

use the fact that for commuting matrices it is true that exp(A + B) =

exp(A) exp(B).

5. We know A = PJ2(λ)P−1 and so Ak = P(J2(λ))

kP−1 and sum.

6. The matrix of the system is (−3 −1

4 1

).

The characteristic equation has the single root λ = −1. Set f2 =

(0

1

)which is not an eigenvector. Then f1 = (A− λI)f2 =

(−1

2

).

So

exp(tA) = P exp(tJ2(−1))P−1

=

(−1 0

2 1

)(etλ tetλ

0 etλ

)(−1 0

2 1

)= e−t

(1− 2t −t

4t 2t+ 1

)So the general solution is

x(t) = exp(tA)

(C1

C2

)= e−t

(C1(1− 2t) − tC24tC1 + (2t+ 1)C2

)


done?problem 13.3 A matrix is said to be in Jordan form if it is block-diagonal

with each diagonal block being a Jordan matrix. (There may be more than one

block with a given parameter value α.) So for example the 7× 7 matrixJ3(5) 0 0

0 J1(5) 0

0 0 J3(−2)

is in Jordan form. (Note that J1(α) is the 1 × 1 matrix (a.k.a. “number”) α

and so a diagonal matrix is an example of Jordan form where all the blocks are

of size 1.)

1. Consider a matrix A in Jordan form with just two Jordan blocks

Jp(α), Jq(β) where α 6= β. Find the characteristic polynomial, eigenval-

ues, minimal polynomial, and dimensions of the generalized eigenspaces

of A (definition is in a problem for lecture 12).

2. The same, only now assume that α = β.

3. Conjecture how this generalizes to a general matrix in Jordan form.

hint/solution 13.3

1. The characteristic and minimal polynomials are the products of those for

the blocks separately. The eigenspaces are of the same dimension as for

the blocks separately.

2. The characteristic polynomial is still the product. The minimal polynomial

is (λ−α)max(p,q). The eigenspace is 2-dimensional (corresponding to the

top entry in each block). The generalized eigenspaces are just the direct

sum of those for the blocks separately.

3. For each eigenvalue λ the characteristic polynomial has a factor (x− λ)p

where p is the sum of the size of all blocks with that λ. The minimal

polynomial has a factor (x−λ)q where q is the maximum size of all blocks

with that λ. The generalized eigenspaces just add, as before.

done?problem 13.4 Find two matrices A,B in Jordan form which have the same

minimal polynomial, characteristic polynomial and dimensions of the (ordinary,

not generalized) eigenspaces and such that A 6= B (and neither do A,B differ

only by a change in the order of the blocks down the diagonal).


hint/solution 13.4 I think the smallest example is the two 7 × 7 matrices,

each made up of three Jordan blocks with the same value of α. In one the sizes

are 3, 3, 1 and in the other 3, 2, 2.

13.1.1 something worth knowing In fact, every complex square matrix is sim-

ilar to one in Jordan form. Some thought shows that the Jordan form is de-

termined (up to choosing the order of the blocks) by the dimensions of all

the generalized eigenspaces and so this provides a solution to the classification

problem for complex square matrices: two complex matrices are similar iff they

have the same eigenvalues and the same dimension for all the corresponding

generalized eigenspaces.

Knowing the characteristic and minimal polynomials is not enough, as ex-

ercise 4 demonstrates.

lecture 14 a diagonalizability theorem

14.1 setting

T : V → V is a linear map and V is a finite-dimensional vector space over a field

F. Its characteristic polynomial is cT (x) and its minimal polynomial is mT (x).

(We usually write the characteristic polynomial using x as the variable in place

of the more familiar λ.)

We will write λ1, . . . , λk for the distinct eigenvalues of T or A but µ1, . . . , µlfor the eigenvalues listed with multiplicity.

problems

done?problem 14.1 Use the process described in §14.3.1 to find quadratic polyno-

mials p1, p2, p3 such that for every quadratic polynomial q we have

q(x) = q(0)p1(x) + q(1)p2(x) + q(2)p3(x).

hint/solution 14.1 Following the formulae in §14.3.1 the polynomials are

p1(x) =1

2(x− 1)(x− 2), p2(x) = −x(x− 2), p3(x) =

1

2x(x− 1).


done?problem 14.2 For what values of k is the matrix

M :=

1 1 2

0 −2 k

0 0 1

diagonalizable?

hint/solution 14.2 Clearly, cT (x) = −(x − 1)2(x + 2) and so by the main

theorem of this lecture, M is diagonalizable if and only ifmT (x) = (x−1)(x+2).

Computing, (M− I)(M+ 2I) has all entries zero except for one which is k+ 6.

So M is diagonalizable iff k = −6.

done?problem 14.3 A is a 2× 2 matrix and λ is an eigenvalue of A. Also

A− λI =

(2 −3

−4 6

).

Find a basis for R2 consisting of eigenvectors of A.

hint/solution 14.3 Solve (A − λI)v = 0 to get an eigenvector v1 and using

the previous part we know that the columns of (A−λI) are eigenvectors for the

other eigenvalue. Thus

v1 =

(3

2

), v2 =

(2

−4

)is a basis of eigenvectors.

done?problem 14.4 Let

Rθ =

(cos θ − sin θ

sin θ cos θ

), θ ∈ [0, 2π].

If you calculate you will find that the different Rθ have different (complex)

eigenvalues but the same (complex) eigenvectors. How does this relate to the

theory in this lecture?

hint/solution 14.4 Since the Rθ are rotation matrices we have RθRφ = RφRθ.

Thus, (thought of as complex matrices) if they are diagonalizable (and they all

in fact are) then they are simultaneously diagonalizable - and for one dimensional

eigenspaces that just means they have the same eigenvectors.


done?problem 14.5 + Hand-in for tutorial Use the main theorem to check

that

A =

−2 −4 2

3 6 −1

−6 −4 6

is diagonalizable. Find a basis v1, v2, v3 with respect to which the matrix of

u 7→ Au is diagonal and hence write down a matrix P such that P−1AP is

diagonal. You are given that cA(x) = (x − 2)(x − 4)2. (You can (and should)

do all this without explicitly solving (A− λI)v = 0 for eigenvectors.)

hint/solution 14.5 Just check that (A− 2I)(A− 4I) = 0, which indeed it is.

Now,

A− 4I =

−6 −4 2

3 2 −1

−6 −4 2

, A− 2I =

−4 −4 2

3 4 −1

−6 −4 4

and so we can take one column from the first matrix to span the λ = 2

eigenspace (the right-most, say) and two from the second (one quarter of the

second and the last, say) to span the λ = 4 eigenspace. So set

P =

2 −1 2

−1 1 −1

2 −1 4

done?problem 14.6 Continuing the previous exercise, let

B =

−3 −4 3

5 6 −3

−5 −4 5

Check that AB = BA. Find a basis that diagonalizes both A and B. A

suggested strategy is as follows. Use your change of basis matrix P from the

previous exercise (that diagonalizes A) to find the matrix of x 7→ Bx in the

basis v1, v2, v3 (Maple, perhaps). Now you should discover that B is block-

diagonal and you just need to change basis again within the 2-dimensional

eigenspace. (There are other ways: you could think about the intersection of

the 2-dimensional eigenspaces of the two matrices, for example.)


hint/solution 14.6 Using Maple to change basis for A and B (the former just

as a check) we get

P−1AP =

2 0 0

0 4 0

0 0 4

, P−1BP =

2 0 0

0 4 −6

0 0 2

.So we need to diagonalise the submatrix

S :=

(4 −6

0 2

).

This has eigenvalues 4, 2 and assembling eigenvectors as usual we discover that

Q =

(1 3

0 1

)diagonalizes S. Combining the two changes of basis we get that

R :=

2 −1 2

−1 1 −1

2 −1 4

1 0 0

0 1 3

0 0 1

=

2 −1 −1

−1 1 2

2 −1 1

diagonalizes both. It’s columns are the required basis.

There are other ways of approaching this. One way is to observe that the

basis has to consist of a vector in A’s 1-dimensional eigenspace, a vector in B’s

1-dimensional eigenspace and a vector in the intersection of the two matrices

2-dimensional eigenspaces.

lecture 15 bilinear and quadratic forms on R-vectorspaces

15.1 setting

Finite-dimensional vector spaces over R. If b is a symmetric bilinear form (SBF)

we write β for the associated quadratic form and B for the matrix of b with

respect to some basis.

problems

done?problem 15.1 Prove that an SBF is linear in the second entry (§15.3.3).

done?problem 15.2 Check the claim in §15.4.1 that if B is a symmetric n×n matrix

then b(x, y) = xTBy defines an SBF on Rn, were x, y are column vectors as

usual. (Hint: for the symmetric part, take the transpose of the 1 × 1 matrix

xTBY.)


hint/solution 15.2 Firstly,

b(λx+ µy, z) = (λx+ µy)TBz = λxtBz+ µyTBz = λb(x, z) + µb(y, z).

Secondly

b(y, x) = yTBx = (yTBx)T = xTBT (yT )T = xTBy = b(x, y).

Note the trick at the start where we take the transpose of a 1× 1 matrix (a.k.a

number).

done?problem 15.3 Consider the SBF b on R2 with matrix

B =

(2 3

3 1

).

Find a vector v 6= 0 in R2 such that b(v, v) = 0. Give a sketch showing the

regions in the plane where b(x, x) is positive, negative and zero.

hint/solution 15.3 b(v, v) = 2v21 + 6v1v2 + v22. Set t = v2/v1 so that

b(v, v) = 0 iff t2 + 6t + 2 = 0. This has solns t = −3 ±√7. So e.g.

(1 − 3+√7)t is such a vector.

The sketch should show the two lines v2 = (−3 ±√7)v1 with the sign of

b(v, v) alternating in sign as you go round the four regions in such a way that

it is positive on the coordinate axes. b(v, v) = 0 on the two lines.

done?problem 15.4 Consider the SBF on R3 with matrix

B =

1 0 0

0 1 0

0 0 −1

.Sketch the regions in R3 where b(x, x) is positive, negative and zero.

hint/solution 15.4 The sketch should show a double cone with b(x, x) = 0

on the cone and negative in the two components of its interior.


done?problem 15.5 + Hand-in for tutorial Let V be the vector space of

2 × 2 matrices with real entries and trace zero. Consider the SBF (known, by

the way, as the “trace form”) b(X, Y) = Trace(XY) on V. Find the matrix of b

with respect to the basis(0 1

0 0

),

(0 0

1 0

),

(1 0

0 −1

)of V.

hint/solution 15.5 The associated quadratic form is q(X) = Trace(X2). La-

beling the basis vectors f1, f2, f3 in the order given and calculating, we get

b(f1, f1) = b(f2, f2) = b(f1, f3) = b(f2, f3) = 0, b(f1, f2) = 1, b(f3, f3) = 2

and so the matrix is 0 1 0

1 0 0

0 0 2

.lecture 16 diagonalizing SBF’s

16.1 setting




problems

done?problem 16.1 Show that Nb := {x ∈ V |b(x, v) = 0 for all v ∈ V is a

subspace of V. (i.e. give the proof of §16.3.2.)

hint/solution 16.1 Standard subspace exercise.

done?problem 16.2 Suppose that the matrix of b in a basis is the standard form

as in §16.4.2. Identify a p-dimensional subspace on which b is positive definite.

Identify also an (n − p)-dimensional subspace on which b is “negative semi-

definite” (meaning that b(v, v) ≤ 0 for all v in the subspace). By considering

intersections, deduce that there is no subspace of dimension larger than p on

which b is positive definite.


hint/solution 16.2 It is positive definite on U := span of the first p basis

vectors. It is negative semi-definite on the span W of the remaining n−p basis

vectors. Suppose there is a subspace Y of dimension > p on which b is positive

definite. Then by the dimension theorem W ∩ Y has dimension > 0 which is a

contradiction.

done?problem 16.3 + Hand-in for tutorial

1. Find a 2× 2 orthogonal matrix P such that PtSP is diagonal where

S =

(7 −6

−6 −2

).

2. Find also a matrix P such that PtSP is diagonal with diagonal entries ±1or 0.

3. What is the type of the SBF on R2 given by the matrix S? What is its

rank and what is its signature?

hint/solution 16.3

1. The matrix S =

(7 −6

−6 −2

)has eigenvalues 10 and -5. Forming the

matrix P with the corresponding unit eigenvectors as columns we get

P =

(2/√5 1/

√5

−1/√5 2/

√5

)(Note that either column could be multiplied by -1. Also the columns

would be swapped if you decided to take the eigenvalues in the reverse

order.) Then PtSP = diag(10,−5) as one can check.

2. Dividing the columns by√

|λ| we get

P =

( √2/5 1/5

−√2/10 2/5

)

3. Rank is 2, signature is zero.


done?problem 16.4 + Hand-in for tutorial Let P2 denote the real vector

space of polynomials of degree ≤ 2 in a variable x.

1. Show that

b(p(x), q(x)) :=

∫10p ′(x)q ′(x)dx

defines an SBF on P2 (the “dashes” indicate derivatives).

2. Find a nonzero element of Nb and hence deduce that b is degenerate.

3. Find the matrix of b with respect to the basis {x2, x, 1} of P2 and hence

find the rank of b.

hint/solution 16.4

1. Clearly b(p(x), q(x)) = b(q(x), p(x)). Also

b(λp(x) + µq(x), r(x)) =

∫(λp(x) + µq(x)) ′r ′(x)dx

= λ

∫p ′(x)r ′(x)dx+ µ

∫q ′(x)r ′(x)dx

= λb(p(x), r(x)) + µb(q(x), r(x))

2. 1 ∈ Nb and so dimNb > 0 and so rankb < dimP2 = 3.

3. The matrix is 4/3 1 0

1 1 0

0 0 0

.This clearly has two linearly independent columns and so the rank is 2.

done?problem 16.5 Quick questions

1. True or false: An SBF is non-degenerate iff its matrix does not have zero

as an eigenvalue.

2. What are the possible types of an SBF on R3 if there exists a 2-dimensional

subspace on which it is negative-definite?

3. An SBF on Rn has type (p, q). What is the largest possible dimension

for a subspace V such that b(v, v) < 0 for all non-zero v ∈ V?

4. Same as above but now b(v, v) ≤ 0 for all non-zero v ∈ V.


hint/solution 16.5

1. True

2. It could be any of the types (0, 3), (1, 2), (0, 2).

3. q

4. Slightly harder. Let V have dimension n. Then in standard form, b clearly

satisfies the condition on the (n − p)-dimensional subspace spanned by

the last n − p basis vectors. This is the largest possible dimension since

any subspace of larger dimension would have non-trivial intersection with

the subspace spanned by the first p basis vectors (on which b is positive-

definite).

done?problem 16.6 + Hand-in for tutorial Let V denote the vector space

of n× n real matrices.

1. What is the dimension of V and of the subspace of symmetric matrices

and of the subspace of antisymmetric matrices?

2. Show that b(X, Y) = Trace(XY) defines an SBF on V.

3. Show that

b(X, Y) =

n∑j=1

n∑k=1

XjkYkj.

4. Show that b is positive-definite on the subspace of symmetric matrices

and negative-definite on the subspace of antisymmetric matrices.

5. Find the type, rank and signature of b.


hint/solution 16.6

1. n2, n(n+ 1)/2, n(n− 1)/2.

2. Firstly,

b(λX+µY, Z) = Trace((λX+µY)Z) = λTrace(XZ)+µTrace(YZ) = λb(X,Z)+µb(Y, Z).

Secondly

b(Y, X) = TraceXY = TraceYX = b(y, x)

because in general for matrices TraceAB = TraceBA. (Alternatively, the

properties follow from the explicit formula of the next part.)

3. The matrix product is given by

(XY)jl =

n∑k=1

XjkYkl

and so taking the trace

TraceXY =

n∑j=1

(XY)jj =

n∑j=1

n∑k=1

XjkYkj.

4. On symmetric matrices

b(X,X) =

n∑j=1

n∑k=1

XjkXkj =

n∑j=1

n∑k=1

XjkXjk

which is the sum of the squares of the entries and hence non-negative and

zero only for the zero matrix. Similarly for the antisymmetric, except that

there is a minus sign.

5. So (since V is the direct sum of the two subspaces) the type is (n(n +

1)/2, n(n− 1)/2), the rank is n2 and the signature n.


done?problem 16.7 Let b be the SBF on R4 given by the matrix B given in block

form as

B =

(I2 0

0 −I2

).

Let A be a fixed 2× 2 matrix. Define

U :=

{x ∈ R4

∣∣∣∣ x =

(Av

v

), v ∈ R2

}.

(We are using “block form” notation above.) Show that U is a subspace of R4

and state its dimension.

Show that b is identically zero on U if and only if A is an orthogonal matrix

(i.e. iff AtA = I).

hint/solution 16.7 Let x, y ∈ U so that x =

(Au

u

)and y =

(Av

v

)for

some u, v ∈ R2. Then

λx+ µy =

(A(λu+ µv)

λu+ µv

)∈ U

and so U is a subspace. It has dimension 2 since a basis is obtained by taking

v to be each standard basis vector inR2.Now consider b on U. We have (taking x, y as in the proof above)

b(x, y) = xtBy =(utAt ut

)(I2 0

0 −I2

)(Av v

)= utAtAv−Utv = ut(AtA−I2)v.

This is identically zero iff AtA = I2, that is, iff A is orthogonal.

done?problem 16.8 Suppose b is a non-degenerate SBF on V. Can there exist

a subspace U of V such that b restricted to U is degenerate? What if b is

assumed to be positive-definite?

hint/solution 16.8 Yes — the SBF on R2 with matrix

(0 1

1 0

)is non-

degenerate, but it is degenerate (in fact, zero) when restricted to the span

of the first basis vector.

If however b is positive-definite then it is positive-definite on every subspace

and hence non-degenerate on every subspace.

done?problem 16.9 True or False: If an SBF is positive definite on subspaces

U,U ′ ⊆ V then it is positive definite on their sum U+U ′. Explain your answer.


hint/solution 16.9 It is false. It is not hard to find a counterexample.

lecture 17 determining type — applications

17.1 setting




problems

done?problem 17.1 Consider the quadratic form

β = 2x2 + 3y2 + z2 − 4xy+ 2xz+ 2yz.

1. Write down the matrix B of this quadratic form.

2. By evaluating determinants only determine the type of this form.

3. Find the eigenvalues and eigenvectors of B and check that the signs of

the eigenvalues are consistent with derivation of the type in the previ-

ous part. (You might want to use Maple to find the eigenvalues — use

“evalf(LinearAlgebra[Eigenvalues](B))”.)

hint/solution 17.1

1.

B =

2 −2 1

−2 3 1

1 1 1

2. Here detB1 = 2 (top-left 1 × 1) and detB2 = 2 (top-left 2 × 2). Here

detB = −7 (row-expansion or row operations). So β is positive-definite

on a 2-dim subspace (the (x, y)-plane) but has rank three (since detB 6=0). Hence it has type (2, 1).

3. “evalf( Eigenvalues(A) )” in Maple gives eigenvalues 4.6,−0.71, 2.14

(with a tiny imaginary part that is numerical error). Two positive and

one negative, as expected.


done?problem 17.2 + Hand-in for tutorial Show that the origin is a critical

point of

f(x, y, z) = 2x2 + y siny+ z2 + 2(y+ z) sin x− 2ky sin z

(where k is a constant). What can you say about the nature of the critical point

for different values of k?

hint/solution 17.2 The derivatives are

fx = 4x+2(y+z) cos x, fy = y cosy+siny+2 sin x, fz = 2z+2 sin x−2ky cos z.

These all vanish when x = y = z = 0 and so the origin is a critical point.

Taking the 2nd derivatives at zero the Hessian matrix is4 2 2

2 2 −2k

2 −2k 2

.The determinants of the top-left square submatrices are 4, 4,−16k(1+ k) (the

last after some calculation). This last number is positive for k ∈ (−1, 0) and

negative outside of [−1, 0]. Thus the critical point is a local min for k ∈ (−1, 0)

and a ”generalized saddle” for k < −1 and k > 0. One would need to look at

higher derivatives to determine what happens at k = 0 and k = −1.

done?problem 17.3 What is the type of the SBF with matrix−3 12 −7

12 4 2

−7 2 2

You do not need to evaluate a 3×3 determinant (or compute eigenvalues (don’t

even think of it) ).

hint/solution 17.3 This is positive-definite on the span of the last two basis

vectors and so has type (2, 0), (3, 0) or (2, 1). But it is negative definite on the

span of the first basis vector, which rules out the first two possibilities.

done?problem 17.4 What is the type of the SBF in problem 5 of lecture 15?

lecture 18 SBF’s on inner-product spaces


18.1 setting


we write B for the matrix of b with respect to some basis.

problems

done?problem 18.1 Classify the following quadrics.

1. x2 + 2y2 + 3z2 + 2xy+ 2xz = 1

2. 2xy+ 2xz+ 2yz = 1

3. x2 + 3y2 + 2xz+ 2yz− 6z2 = 1

You should be able to do all of these using “determinants” analysis if you think

carefully. You can always check your answer by asking Maple for the eigenvalues.

hint/solution 18.1

1. The matrix is

S =

1 1 1

1 2 0

1 0 3

.Calculating, detS = 1 and the top-left 1 × 1 and 2 × 2 submatrices are

also positive so S has type (3, 0) and this is an ellipsoid.

2. The matrix is

S =

0 1 1

1 0 1

1 1 0

.Here detS = 1 and so S has type (3, 0) or (1, 2). But it cannot be the

former since the first standard basis vector satisfies et1Se1 = 0. Thus this

is a hyperboloid of two sheets.

3. The matrix is

S =

1 0 1

0 3 1

1 1 −6

.Here, detS < 0 and so S has type (2, 1) and this is a hyperboloid of one

sheet.


done?problem 18.2 Let β(x) be a quadratic form on Rn given by a symmetric

matrix S. How are the maximum and minimum values of β(x) on the unit

sphere xtx = 1 related to the eigenvalues of S. (Hint: orthogonal change of

coordinates to standard form.)

hint/solution 18.2 Let λ1, . . . , λn be the eigenvalues with multiplicity in non-

increasing order. Then there exists an orthogonal change of coordinates y =

P−1x such that

β = λ1y21 + · · ·+ λny2n.

Since we have changed coordinates orthogonally, the sphere will still be given

as

y21 + · · ·+ y2n = 1.

I claim that the max value of β on the sphere is λ1 and it occurs at (1, 0, . . . , 0).

Clearly that value is obtained there and clearly also for all x on the sphere

λ1y21 + λ2y

22 + · · ·+ λny2n ≤ λ1y21 + λ1y

22 + · · ·+ λ1y2n = λ1.

Similarly, the min value is λn.

done?problem 18.3 u Challenge Continuing the previous exercise, use La-

grange multipliers to find the critical points of xtSx subject to the constraint

xtx = 1.

lecture 19 simultaneous diagonalization

19.1 setting


we write B for the matrix of b with respect to some basis.

problems

done?problem 19.1 + Hand-in for tutorial Consider the SBF’s on R2

given (with respect to the standard basis) by the matrices

B =

(1 3

3 3

), A =

(2 1

1 1

).

Show that one of these is positive definite and hence find a basis for R2 with

respect to which the matrices of both the SBF’s are diagonal (with the positive-

definite one having the identity as its matrix). Write down the change of basis

matrix that diagonalises both forms.


hint/solution 19.1 The 2nd matrix is positive-definite since its top-left entry

is positive as is its determinant. The relative eigenvalues are the solutions of

det(B− λA) =

∣∣∣∣1− 2λ 3− λ

3− λ 3− λ

∣∣∣∣ = (3− λ)(−2− λ) = 0.

Thus the relative eigenvalues are 3 and −2.

Solving (B− λA)v = 0 in each case we get relative eigenvectors(0

1

),

(1

−1

)respectively. Finally we must scale these to be of unit length w.r.t. the positive-

definite form. Checking, in fact these already satisfy this condition. Thus they

form the required basis and the change of basis matrix is the matrix with these

as columns.

done?problem 19.2

1. Let

A =

(1 3

3 2

).

Check explicitly that 〈Ax, y〉 = 〈x,Ay〉 for all x, y ∈ R2 where the inner

product is the standard one on R2.

2. Let

A =

(1 3

−1 2

).

Find vectors x, y ∈ R2 such that 〈Ax, y〉 6= 〈x,Ay〉 where the inner

product is the standard one on R2.

hint/solution 19.2 Just calculate along the lines for the first of observing

that

Ax =

(x1 + 3x23x1 + 2x2

)and so

〈Ax, y〉 = (x1 + 3x2)y1 + (3x1 + 2x2)y2

etc.


done?problem 19.3 + Hand-in for tutorial Let n ∈ N and consider

Tn =

{a0 +

n∑k=1

ak cos kx+ bk sinkx |ak, bk ∈ R

}

with the inner product

〈p(x), q(x)〉 :=

∫2π0p(x)q(x)dx.

Consider the linear map −D2 : Tn → Tn where −D2 : p(x) 7→ −p ′′(x).

1. Show that −D2 is self-adjoint. (Hint: integration by parts.)

2. What are the eigenvalues of −D2 : Tn → Tn and what is the multiplicity

of each eigenvalue? (Think ODE’s — no clever theory required.)

3. Show that the SBF associated to −D2 is b(p, q) =∫2π0 p ′(x)q ′(x)dx.

4. What is the type of the SBF just found? Relate that to the eigenvalues

of −D2.

hint/solution 19.3 Note that all the integrands appearing below are periodic

and so when we integrate by parts the boundary term will always be zero. All

integrals are from 0 to 2π.

1. Integrating by parts twice

〈p,−D2q〉 = −

∫p(x)q ′′(x)dx = −

∫p ′′(x)q(x)dx = 〈−D2p, q〉.

2. We are looking for solutions to −p ′′(x) = λp(x). So the eigenvalues are

λ = 0 with a 1-dim eigenspace consisting of the constants and λ = k2

where k > 0 with multiplicity 2 — the 2-dim eigenspace being spanned

by cos(kx) and sin(kx).

3. Integrating by parts

〈p,−D2q〉 = −

∫p(x)q ′′(x)dx =

∫p ′(x)q ′(x)dx.

4. This SBF is positive-definite on the 2n-dimensional subspace spanned by

cos(kx), sin(kx). The SBF is zero on the 1-dim subspace of constant

functions. Thus it has type (2n, 0). This agrees with −D2 having 2n

positive eigenvalues (counted with multiplicity) and one zero one.

mat3-alg algebra 2008-2009 | toby bailey lecture 0 preamblechris/alg/course_2013probs.pdf ·...

Documents