final answers

7/29/2019 Final Answers

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Math 110 Final Exam with Answers

NAME (1 pt):

TA (1 pt):

Name of Neighbor to your left (1 pt):

Name of Neighbor to your right (1 pt):

Instructions : You are allowed to use 1 sheet (8 1/2 by 11 inches, both sides) of notes.Otherwise this is a closed book, closed notes, closed calculator, closed computer, closed PDA,closed cellphone, closed mp3 player, closed network, open brain exam.

You get one point each for lling in the 4 lines at the top of this page. All other questionsare worth 10 points.

Fill in the questions at the top of the page. Then stop and wait until we tell you toturn the page and start the rest of the exam. Do not start reading the rest of theexam until we tell you to start.

After you start, read all the questions on the exam before you answer any of them, so youdo the ones you nd easier rst.

Write all your answers on this exam. If you need scratch paper, ask for it, write your nameon each sheet, and attach it when you turn it in (we have a stapler).

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Question 1. (10 points) (version 1) Let P (R ) be the vector space of all real polynomials.Dene the linear map T : P (R ) P (R ) by T (f ) = f , the second derivative.

Part 1: Show that T is onto but not one-to-one.Answer: If g(x) = di=0 gix

i , then T (f ) = g where f (x) = f 0 + f 1x + di=0gi x i +2

(i+1)( i+2) (soT is onto) and f 0 and f 1 are arbitrary (so T is not one-to-one).

Part 2: Describe all eigenvectors of T , i.e. nonzero polynomials such that f (x) = f (x)for some .

Answer: Since the degree of f is less than the degree of f , f cannot be a nonzero multiple of f . Therefore = 0 and f (x) = 0 is satised by all linear polynomials f (x) = f 0 + f 1x.

Question 1. (10 points) (version 2) Let P (C ) be the vector space of all complex polyno-mials. Dene the linear map S : P (C ) P (C ) by S (g) = g , the second derivative.

Part 1: Show that S is onto but not one-to-one.Answer: If f (x) = di=0 f ix

i , then S (g) = f where g(x) = g0 + g1x + di=0f i x i +2

(i+1)( i+2) (soS is onto) and g0 and g1 are arbitrary (so S is not one-to-one).

Part 2: Describe all eigenvectors of S , i.e. nonzero polynomials such that g (x) = g(x)for some .

Answer: Since the degree of g is less than the degree of g, g cannot be a nonzero multiple of g. Therefore = 0 and g (x) = 0 is satised by all linear polynomials g(x) = g0 + g1x.

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Question 2. (10 points) (version 1) Let A be an m-by-n complex matrix, and let B bean n-by-m complex matrix. Show that I m + A B is invertible if and only if I n + B A isinvertible.

Answer: Solution 1: Suppose I m + A B is invertible, and (I n + B A) v = 0 ; we need toshow v = 0 . Multiply by A to get

A (I n + B A) v = A v + A B A v = ( I m + A B ) (A v)

so A v = 0 since I m + A B is invertible. But then v = B A v = 0 as desired. Thus I m + A B invertible implies I n + B A invertible. The converse follows by the same argument.

Solution 2: From the practice nal, we know A B and B A have the identical nonzeroeigenvalues. Therefore, 1 is an eigenvalue of A B if and only if it an eigenvalue of B A,implying 0 is an eigenvalue of I m + A B if and only if it is an eigenvalue of I n + B A, sothat I m + A B is singular if and only if I n + B A is singular.

Question 2. (10 points) (version 2) Let X be an m-by-n real matrix, and let Y be ann-by-m complex matrix. Show that I m + X Y is invertible if and only if I n + Y X is invertible.

Answer: Solution 1: Suppose I m + X Y is invertible, and (I n + Y X ) v = 0 ; we need toshow v = 0 . Multiply by X to get

X (I n + Y X ) v = X v + X Y X v = ( I m + X Y ) (X v)

so X v = 0 since I m + X Y is invertible. But then v = Y X v = 0 as desired. Thus I m + X Y invertible implies I n + Y X invertible. The converse follows by the same argument.

Solution 2: From the practice nal, we know X Y and Y X have the identical nonzeroeigenvalues. Therefore, 1 is an eigenvalue of X Y if and only if it an eigenvalue of Y X ,implying 0 is an eigenvalue of I m + X Y if and only if it is an eigenvalue of I n + Y X , sothat I m + X Y is singular if and only if I n + Y X is singular.

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Question 3. (10 points) (version 1) Let A = P R L U P C be an LU decompositionof the m-by-n real matrix A of rank r > 0. Thus P R and P C are permutation matrices, Lis m-by-r and unit lower triangular, and U is r -by-n and upper triangular with U ii nonzero.Show how to express an LU decomposition of At using simple modications of the parts of this LU decomposition of A.

Answer: Write U = D U , where D is r -by-r and diagonal with D ii = U ii , so U is unit upper triangular. Then A = P R L U P C = P R L D U P C so At = P tC U t (D L t ) P tR .This is an LU decomposition of At .

Question 3. (10 points) (version 2) Let B = P R L U P C be an LU decomposition of the m-by-n complex matrix B of rank r > 0. Thus P R and P C are permutation matrices, Lis m-by-r and unit lower triangular, and U is r -by-n and upper triangular with U ii nonzero.Show how to express an LU decomposition of B using simple modications of the parts of this LU decomposition of B .

Answer: Write U = D U , where D is r -by-r and diagonal with D ii = U ii , so U is unit upper triangular. Then B = P R L U P C = P R L D U P C so B

= P C U (D L) P R .

This is an LU decomposition of B .

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Question 4. (10 points) (version 1) Let T : V V be a linear operator. Supposethat T (vi) = ivi for i = 1 ,...m , and all i are distinct. If W is an invariant subspace of T and includes the vector mi=1 a i vi , where all the a i = 0, then prove that W contains vi fori = 1 ,...,m .

Answer: Let w1 = mi=1 a i vi . Since w1 W , so are w2 = T w1 through wn = T n 1 w1

since W is invariant. Let V = [v1 ,...,v m ] and W = [w1,...,w m ]. Let Aij = a i j 1

i be m-by-m. Then we can express the dependence of all the wi on all the vi by W = V A. Now A = diag( a1, a2 ,...,a n ) B , where B ij = j

1i . Thus B is the m by m Vandermonde matrix

with distinct i , and so nonsingular by homework question 4.3.22(c). Thus A is the product of nonsingular matrices and also nonsingular. Thus W A 1 = V , so all the columns of V ,namely the vi , are linear combinations of the wi , and so within W as desired.

Question 4. (10 points) (version 2) Let S : X X be a linear operator. Supposethat S (x i ) = ixi for i = 1 ,...m , and all i are distinct. If Y is an invariant subspace of S and includes the vector mi=1 bi x i , where all the bi = 0, then prove that Y contains x i fori = 1 ,...,m .

Answer: Let y1 = mi=1 bi x i . Since y1 Y , so are y2 = S y1 through yn = S n 1 y1

since Y is invariant. Let X = [x1,...,x m ] and Y = [y1,...,y m ]. Let Aij = bi j 1i be m-by-m. Then we can express the dependence of all the yi on all the x i by Y = X A. Now A = diag( b1, b2,...,bn ) B , where B ij = j

1i . Thus B is the m by m Vandermonde matrix

with distinct i , and so nonsingular by homework question 4.3.22(c). Thus A is the product of nonsingular matrices and also nonsingular. Thus Y A 1 = X , so all the columns of X ,namely the x i , are linear combinations of the yi , and so within Y as desired.

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Question 5. (10 points) (version 1) Let A = X X 1 be diagonalizable ( is diagonal).Let X = QR be the QR decomposition of X , so that Q is unitary and R upper triangular.Show that T = Q A Q is upper triangular. What is the name we gave to the matrixfactorization A = Q T Q?

Answer: A = X X 1 = Q R R 1 Q , since Q 1 = Q , and so Q 1 A Q =R R 1 = T . Since R is upper triangular and nonsingular, so is R 1, and so the product R R

1 is a product of upper triangular matrices and so upper triangular as desired. This is the Schur decomposition.

Question 5. (10 points) (version 2) Let B = Z Z 1 be diagonalizable ( is diagonal).Let Z = QR be the QR decomposition of Z , so that Q is unitary and R upper triangular.Show that T = Q B Q is upper triangular. What is the name we gave to the matrixfactorization B = Q T Q?

Answer: B = Z Z 1 = Q R R 1 Q , since Q 1 = Q , and so Q 1 B Q =R R 1 = T . Since R is upper triangular and nonsingular, so is R 1, and so the product R R 1 is a product of upper triangular matrices and so upper triangular as desired. This is the Schur Factorization.

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Question 6. (10 points) (version 1) Let < x, y > = mi=1 x iyi = xt y be the standard

dot product on R m . An m-by-m real symmetric matrix T = T t is called positive denite if < T x, x > is positive for all nonzero vectors x R m .

We dene the function < , > T : R m R m R by

< x, y > T = < Tx,y > .

Show that < x, y > T is an inner product on R m if and only if T is symmetric and positivedenite.

Answer: First assume T is symmetric and positive denite: We need to conrm that < , > T satises the axioms of an inner product:

< 1x1 + 2x2, y > T = < T (1x1 + 2x2), y >= < 1T x1 + 2T x2, y >= 1 < T x1, y > + 2 < T x2, y >= 1 < x 1, y > T + 2 < x 2, y > T

as desired. Next

< x, y > T = xt

T y = ( xt

T y)t

= yt

T t

x = yt

T x = < y, x > T Finally, x = 0 implies < x, x > T = < T x,x > > 0 as desired.

Now assume < x, y > T = yt T x is an inner product. Then

T ij = eti T e j = < e i , e j > T = < e j , ei > T = et j T ei = T ji

so T is symmetric. Also x = 0 implies 0 < < x, x > T = x t T x, so T is positive denite.

Question 6. (10 points) (version 2) Let < u, v > = ni=1 u ivi = ut v be the standard

dot product on R n . An n-by-n real symmetric matrix X = X t is called positive denite if < X u,u > is positive for all nonzero vectors u R n .

We dene the function < , > X : R n R n R by< u, v > X = < Xu, v >

Show that < u, v > X is an inner product on R n if and only if X is symmetric and positivedenite.

Answer: First assume X is symmetric and positive denite: We need to conrm that < , > X satises the axioms of an inner product:

< 1u1 + 2u2, v > X = < X (1u1 + 2u2), v >= < 1X u1 + 2Xu 2, v >= 1 < X u1, v > + 2 < X u2, v >

= 1 < u 1 , v > X + 2 < u 2, v > X as desired. Next

< u, v > X = u t X v = ( u t X v)t = vt X t u = vt X u = < v, u > X

Finally, u = 0 implies < u, u > X = < X u, u > > 0 as desired.Now assume < u, v > X = vt X u is an inner product. Then

X ij = eti X e j = < e i , e j > X = < e j , ei > X = et j X ei = X ji

so X is symmetric. Also u = 0 implies 0 < < u, u > X = u t X u, so X is positive denite.

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Question 7. (10 points) (version 1) Find a matrix B such that B 2 = A =

10 6 66 10 90 0 1

.

What are all possible sets of eigenvalues of B ?Answer: The squares of the eigenvalues of B are the eigenvalues of A, so they can be

{ 1, 2, 3}, where each sign can be chosen independently. Thus there are 23 = 8 possible

triples in all for all possible choices of signs.

One square root of A, with positive eigenvalues, is B =

3 1 11 3 20 0 1

. We can compute it

in (at least) two ways.

First way: Diagonalize A = V V 1 with = diag(16 , 4, 1), V =

1 1 01 1 10 0 1

and

V 1 = 12

1 1 1 1 1 10 0 2

. Then B = V 1/ 2 V 1 = V diag(4 , 2, 1) V 1.

Second way: Write A = A 690 1

and B = B b13b230 b33

where A and B are

2-by-2 submatrices, and then note that

B B =

B 2 (B + b33 I 2) b13b23

0 b233

Thus we see that B 2 = A. We solve for B by diagonalizing A = U U 1 where U =1 11 1 and = diag(16 , 4), so B = U diag(4 , 2) U

1 = 3 11 3 . Then b233 = 1 so we

take b33 = 1 . Finally, we solve (B + I ) b13b23= 69 for b13 = 1 and b23 = 2 as desired.

The set of all possible answers consists of

3 1 11 3 20 0 1

,

3 1 11 3 40 0 1

,

1 3 33 1 20 0 1

,

1 3 33 1 00 0 1

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Question 7. (10 points) (version 2) Find a matrix X such that X 2 = Y =

13 12 712 13 70 0 4

.

What are all possible sets of eigenvalues of X ?Answer: The squares of the eigenvalues of X are the eigenvalues of Y , so they can be

{ 1, 2, 5}, where each sign can be chosen independently. Thus there are 23 = 8 possible

triples in all for all possible choices of signs.

One square root, with positive eigenvalues, is X =

3 2 12 3 10 0 2

. We can compute it in (at

least) two ways.

First way: Diagonalize Y = V V 1 with = diag(25 , 1, 4), V =

1 1 11 1 10 0 3

, and

V 1 =

12

12

13

1212 0

0 0 13

. Then X = V 1/ 2 V 1 = V diag(5 , 1, 2) V 1.

Second way: Write Y = Y 770 4

and X = X x13x230 x33

where Y and X are

2-by-2 submatrices, and then note that

X X =

X 2 (X + x33 I 2) x13x23

0 x233

Thus we see that X 2 = Y . We solve for X by diagonalizing Y = U U 1 where U =1 11 1 and = diag(25 , 1), so X = U diag(5 , 1) U

1 = 13 1212 13 . Then x233 = 4 so

we take x33 = 2 . Finally, we solve (X + 2 I ) x13x23= 77 for x13 = 1 and x23 = 1 as

desired.The set of all possible answers consists of

3 2 12 3 10 0 2

,

2 3 13 2 10 0 2

,

3 2 7/ 32 3 7/ 30 0 2

,

2 3 7/ 33 2 7/ 30 0 2

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Question 8. (10 points) (version 1) Determine all possible Jordan canonical forms formatrices with the characteristic polynomial ( x 1)3(x 2)2. In other words, list the sets of Jordan blocks that could appear in each Jordan canonical form.

Answer: Let J i () denote an i-by-i Jordan block with eigenvalue . For the triple eigen-value at 1, the possible Jordan structures are {J 3(1), (J 2(1), J 1(1)) , (J 1(1), J 1(1), J 1(1)) }. for the double eigenvalue at 2, the possible Jordan structures are {J 2(2), (J 1(2), J 1(2)) }. Since the Jordan structures for the 2 eigenvalues can be chosen independently, and the order in which they appear does not matter, there are 6 possible structures in all, for all choices from the rst set and the second set.

Question 8. (10 points) (version 2) Determine all possible Jordan canonical forms formatrices with the characteristic polynomial ( x 3)2(x 4)3. In other words, list the sets of Jordan blocks that could appear in each Jordan canonical form.

Answer: Let J i () denote a i-by-i Jordan block with eigenvalue . For the triple eigenvalue at 4, the possible Jordan structures are {J 3(4), (J 2 (4), J 1(4)) , (J 1(4), J 1(4), J 1(4)) }. for the double eigenvalue at 3, the possible Jordan structures are {J 2(3), (J 1(3), J 1(3)) }. Since the Jordan structures for the 2 eigenvalues can be chosen independently, and the order in which they appear does not matter, there are 6 possible structures in all, for all choices from the rst set and the second set.

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