Answer Keys For Math 225 Final Review Problem

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1 Answer Keys For Math Final Review Problem () For each of the following maps T, Determine whether T is a linear transformation. If T is a linear transformation, determine whether T is -, onto and/or bijective. You must justify your answer: (a) T : R R given by T (x, y, z) = (x y, y z, z x). (b) T : R R given by T (x, y, z) = (x + y + z, x + y + z). (c) T : Rx Rx given by T (f(x)) = f(x ) + x. (d) T : M (R) M (R) given by T (A) = A + A T. linear - onto bijective (a) Yes No No No Answer. (b) Yes No Yes No (c) No (d) Yes No No No (c) is not linear because T () = x. () Let M m n (R) be the vector space of m n real matrices and T : M (R) M (R) be the map given by T (A) = A 4 for all A M (R). (a) Show that T is a linear transformation. (b) Find the kernel, range and rank of T. (c) Is T onto, - and/or bijective? Justify your answer. (d) Find the characteristic polynomial, eigenvalues and eigenvectors of T. (e) Is T diagonalizable? If it is, find a basis B of M (R) such that T B B is diagonal. (f) Compute T Solution. Let D = Since. 4 T (A + cb) = D(A + cb) = DA + cdb = T (A) + ct (B) for all A, B M (R) and c R, T is a linear transformation.

2 The kernel, range and rank of T are K(T ) = {A M (R) : DA = } = { v v : D v v = } = { v v : Dv = Dv = } { { v = v : v, v Nul(D) = Span }} { } = Span, R(T ) = {DA : A M (R)} = {D v v : v v M (R)} = { Dv Dv : v v M (R)} { { u = u : u, u Col(D) = Span }} { } = Span, rank(t ) = dim R(T ) =. Since K(T ) {} and T is an endomorphism, T is not -, onto or bijective. Let A be an eigenvector of T corresponding to an eigenvalue λ. Then T (A) = λa DA = λa D v v = λ v v Dv = λv and Dv = λv v, v Nul(λI D) for A = v v. Since A, λ must be an eigenvalue of D. The characteristic polynomial of D is x x and it has eigenvalues and with eigenspaces { Nul(D I) = Nul(D) = Span } Nul(D I) = Nul 4 = Span { }.

3 Therefore, T has eigenvalues and with eigenspaces K(T I) = { v v : v, v Nul(D I) } { } = Span, K(T I) = { v v : v, v Nul(D I) } { } = Span,. Since dim K(T I) + dim K(T I) = 4 = dim M (R), T is diagonalizable and T B B = for B = {,,, }. It follows that the characteristic polynomial of T is Since = det(xi T B B ) = x (x ) we obtain T = T + T + T + T ( ) ( ) = + ( ) ( ) + + ( = ) ( ) ( ) (. )

4 4 () Let T : R R be a linear endormoprhism given by T (x, y, z) = (x + y + z, x + y + z, x + y + z). (a) Find the kernel, range and rank of T. (b) Is T onto, - and/or bijective? (c) Find the matrix T B B representing T under the standard basis B. (d) Find the matrix T C C representing T under the basis C =,,. (e) Find the eigenvalues, eigenvectors and characteristic polynomial of T. (f) Is T diagonalizable? If it is, find a basis D such that T D D is diagonal. Answer. K(T ) =, R(T ) = R and rank(t ) =. T is -, onto and bijective. T B B = T C C = = 4 The characteristic polynomial of T is (x 4)(x ). It has two eigenvalues 4 and with eigenspaces: K(T 4I) = Span K(T I) = Span,

5 Since dim K(T 4I) + dim K(T I) =, T is diagonalizable and for T D D = 4 D =,,. (4) Let P be the vector space of real polynomials of degree and T : P P be the map given by T (f(x)) = xf (x) + f(). (a) Show that T is a linear endomorphism. (b) Find the kernel, range and rank of T. (c) Find the characteristic polynomial, eigenvalues and eigenvectors of T. (d) Is T diagonalizable? If it is, find a basis B of P such that T B B is diagonal. Solution. Since T (f(x) + cg(x)) = x(f(x) + cg(x)) + (f() + cg()) = xf (x) + cxg (x) + f() + cg() = (xf (x) + f()) + c(xg (x) + g()) for all f(x), g(x) P and c R, T is a linear transformation. Let D = {, x, x, x } be the standard basis of P. Since we obtain T () = x() + = T (x) = x(x) + = + x T (x ) = x(x ) + = + x T (x ) = x(x ) + = + x T }{{ D D = }. A Since det(a), T is invertible. So K(T ) =, R(T ) = P and rank(t ) = 4.

6 The characteristic polynomial of T is det(xi A) = (x ) (x )(x ) So T has eigenvalues,,. Since the eigenspaces of A are Nul(A I) = Nul = Span Nul(A I) = Nul = Span Nul(A I) = Nul = Span Correspondingly, the eigenspaces of T are Since K(T I) = Span{} K(T I) = Span{ + x } K(T I) = Span{ + x }. dim K(T I) + dim K(T I) + dim K(T I) = < 4 = dim P, T is not diagonalizable. () Let B and C be two n n invertible matrices. Show that all three matrices CBC, BC and C B have the same characteristic polynomials. Proof. Since C (CBC)C = BC and B (BC )B = C B CBC BC C B. polynomials. () Let W be the subspace of R 4 given by So they have the same characteristic W = {(x, x, x, x 4 ) : x + x + x = x + x + x 4 = }. (a) Find an orthonormal basis for W.

7 7 (b) Find the projection of v = (,,, ) onto W. (c) Let T : R 4 R 4 be the linear transformation given by T (v) = proj W v. Find the kernel and range of T and find T B B under the standard basis B of R 4. Proof. We have W = Nul = Span{(,,, ) }{{}, (,,, ) }. }{{} u u Applying Gram-Schmidt to {u, u }, we obtain v = u = (,,, ) v = u u.v v v = (,,, ) Normalizing {v, v }, we obtain an orthonormal basis of W. So w = (,,, ) w = (,,, ) proj W v = (v.w )w + (v.w )w = (,,, ). For every w W, T (w) = proj W w = w. So W R(T ) and rank(t ) dim W =. For every v W, T (v) = proj W v =. So W K(T ) and hence dim K(T ) dim W =. By Rank Theorem, rank(t ) + dim K(T ) = 4. Combining rank(t ) + dim K(T ) = 4 rank(t ) dim K(T ) we conclude rank(t ) = dim K(T ) =. K(T ) = W. So R(T ) = W and

8 8 Since T (e ) = proj W e = (e.w )w + (e.w )w = (,,, ) T (e ) = proj W e = (e.w )w + (e.w )w = (,,, ) T (e ) = proj W e = (e.w )w + (e.w )w = (,,, ) T (e 4 ) = proj W e 4 = (e 4.w )w + (e 4.w )w = (,,, ) we obtain T B B =. (7) Let A be a real symmetric matrix whose characteristic polynomial is x 4 x +. (a) Show that A is invertible. (b) Find the characteristic polynomial of A + A. Solution. Since deg(x 4 x + 4) = 4, A is a 4 4 matrix. Therefore, det(a) = det( A) = det(i A) = 4 () + =. Since det(a), A is invertible. Since A is real symmetric, A is diagonalizable. And since x 4 x + = (x ) = (x + ) (x ) P AP = D = for an invertible matrix P. Then P (A + A )P = P AP + (P AP ) = D + D =.

9 So A + A and D + D are similar and they have the same characteristic polynomial det(xi (A + A )) = det(xi (D + D )) = (x + ) (x ). (8) Let T : V W and T : V W be two linear transformations between two vector spaces V and W. Show that K(T ) K(T T ) = K(T + T ) K(T ). 9 Proof. For every v K(T ) K(T T ), T (v) = (T T )(v) = T (v) = T (v) T (v) = T (v) = T (v) = T (v) + T (v) = T (v) = (T + T )(v) = T (v) = v K(T + T ) K(T ). So K(T ) K(T T ) K(T + T ) K(T ). On the other hand, for every v K(T + T ) K(T ), (T + T )(v) = T (v) = T (v) + T (v) = T (v) = T (v) = T (v) = T (v) = T (v) T (v) = T (v) = (T T )(v) = v K(T ) K(T T ). So K(T T ) K(T ) K(T ) K(T + T ). In conclusion, K(T ) K(T T ) = K(T + T ) K(T ). (9) Find all n n real matrices A with the property that every nonzero vector in R n is an eigenvector of A. Justify your answer.

10 Proof. Suppose that A has two distinct eigenvalues λ λ. Let v and v be two eigenvectors of A associated to λ and λ, respectively. Then Av = λ v and Av = λ v. Since v and v are eigenvectors associated to different eigenvalues, they are linearly independent. Since every nonzero vector is an eigenvector of A, v + v is also an eigenvector of A. So But on the other hand, So A(v + v ) = λ(v + v ). A(v + v ) = Av + Av = λ v + λ v. λ(v + v ) = λ v + λ v (λ λ )v + (λ λ )v =. Since λ λ, λ λ and λ λ cannot be both zero. So v and v are linearly dependent. Contradiction. So A has only one eigenvalue λ. Since e, e,..., e n are n linearly independent eigenvectors of A, P AP = λ λ for P = e e... e n = I. Since P = I, P AP = A. Hence A = λi. () Let {a n : n =,,,...} be a sequence satisfying... λ a n+ = a n+ a n + for all n and a = a =. Find a formula for a n. Answer. a n = + n n. () Let T : R R and T : R R be two linear transformations satisfying T T = T T =, T (, ) = (, ) and T (, ) = (, ). (a) Find T and T.

11 (b) Find the kernels and ranges of T and T. Solution. Since T T = T T =, we conclude that = T T (, ) = T (T (, )) = T (, ) = T T (, ) = T (T (, )) = T (, ). Suppose that T x y = A x y and T x y = A x y for some matrices A and A. Then A = and A =. So Therefore, A = = A = =. T (x, y) = (y, y) T (x, y) = ( x + y, ) K(T ) = Span{(, )} K(T ) = Span{(, )} R(T ) = Span{(, )} R(T ) = Span{(, )}. () Construct the following examples (You must justify your examples): (a) Three vectors v, v, v in R such that each pair of the three are linearly independent but v, v, v are linearly dependent. (b) Two linear transformations T : R R and T : R R such that rank(t T ) = and rank(t T ) =. (c) Two matrices with the same characteristic polynomials but not similar. (d) A linear endomorphism T : R R such that the characteristic polynomial of T is x and T.

12 (e) Two matrices A and B such that both A and B are diagonalizable but A + B is not diagonalizable. (f) Two matrices A and B such that both A and B are diagonalizable but AB is not diagonalizable. (g) Four matrices A, A, B, B such that A B and A B but A + A B + B. (h) Four matrices A, A, B, B such that A B and A B but A A B B. Solution. (a) Let v = (,, ), v = (,, ) and v = (,, ). Then v and v are linearly independent, v and v are linearly independent, and v and v are linearly independent since the matrices, and all have rank. But v, v and v are linearly dependent since the matrix has rank <. (b) Let T (x, y) = (, x, y) and T (x, y, z) = (x, ). Then has rank and (c) Let T T (x, y, z) = T (x, ) = (, x, ) T T (x, y) = T (, x, y) = (, ). A = and B =. Then both A and B have the same characteristic (x ). Since dim Nul(A I) = <, A is not diagonalizable. But B is diagonal. So A B. (d) Let T x = x. x x } {{ } A x x

13 Then det(xi T ) = det(xi A) = x. But A = and hence T. (e) Let A = and B =. Then both A and B are diagonalizable since they have three distinct eigenvalues. But A + B = is not diagonalizable since it has one eigenvalue 4 and dim Nul(A + B 4I) = <. (f) Let A = and B =. Then both A and B are diagonalizable since they have three distinct eigenvalues. But AB = is not diagonalizable since it has one eigenvalue and dim Nul(AB I) = <. (g) Let A = A = B = and B = Then A B since they are diagonal with the same characteristic and A B since A = B. But A + B = = A 4 + B since they have different characteristic polynomials.

14 4 (h) Let A = A = B = and B = Then A B since they are diagonal with the same characteristic and A B since A = B. But A B = = A 4 B since they have different characteristic polynomials. () Show that two real symmetric matrices with the same characteristic polynomials must be similar. Proof. Every real symmetric matrix is diagonalizable. Let A and B be two symmetric matrices with characteristic polynomial f(x) = (x λ )(x λ )...(x λ n ). Then there exist nonsingular matrices P and Q such that λ λ P AP = Q BQ =.... λ n Therefore, A = P Q BQP = (QP ) B(QP ) and hence A and B are similar. (4) Determine whether the following matrices A and B are similar. Justify your answers. (a) A = and B = (b) A = and B = 4 (c) A = and B = 4

15 Proof. (a) Since A has two eigenvalues and and dim Nul(A I) + dim Nul(A I) = + = <, A is not diagonalizable. But B is diagonal. So A B. (b) Since A has two eigenvalues and and dim Nul(A I) + dim Nul(A I) = + =, A is diagonalizable and A. Since B has two eigenvalues and and dim Nul(B I) + dim Nul(B I) = + =, B is diagonalizable and B. Therefore, A B. (c) Since both A and B have three distinct eigenvalues,,, both A and B are diagonalizable and A B. () Orthogonally diagonalize and find spectral decompositions of the following symmetric matrices: 4 a) b) 4 Answer. a) Q T AQ = for Q = and A =.

16 b) and A = Q T AQ = + ( ) for + ( ) () Solve the following ODEs: a) dx dt = x + x + x dx dt = x + x + x dx dt = x + x + x with b) dx dt dx + x = with x() =. dt x () = x () = x () = Answer. a) x x = e t + e 4(t ) x b) x = ce (t ) + ( c)e (t ) for c R. (7) Let A be an n n symmetric matrix with characteristic polynomial (x ) n (x + ) and Nul(A + I) = Span..

17 7 Find A. Solution. Since A has two eigenvalues and and A is real symmetric, Nul(A + I) and Nul(A I) are orthogonal complements to each other. Normalizing (,,..., ), we obtain w = n.. Let w, w,..., w n be an orthonormal basis of Nul(A I). Then Q T AQ = D =...

18 8 for Q = w w... w n. Then A = QDQ T = Q... QT = Q I +... = QQ T + Q... QT = I w w T = I n n n n =... n n n n n... n