5.3.5 The eigenvalues are 3, 2, 3 (i.e., the diagonal entries of D) with corresponding eigenvalues. Null(A 3I) = Null( ), 0 0

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1 535 The eigenvalues are 3,, 3 (ie, the diagonal entries of D) with corresponding eigenvalues,, 538 The matrix is upper triangular so the eigenvalues are simply the diagonal entries, namely 3, 3 The corresponding eigenspace is given by Null(A 3I) = Null( ), which is the span of e diagonalizable Since there is no basis of eigenvectors, this matrix is not 533 The eigenspaces are E = Null = Span{, } and 3 E 5 = Null = Span{ } 3 Since there are three linearly independent eigenvectors (since eigenvectors from distinct eigenspaces are always linearly independent), the matrix can be diagonalized by taking D with diagonal entries equal to,, 5 and P with columns as above, corresponding to the eigenvalues 534 The eigenspaces are E = Null = Span{ E 3 = Null = Span{, } Since we have a basis of three eigenvectors, the matrix is diagonalizable D containing the eigenvalues as diagonal entries and P containing the above eigenvectors as the corresponding columns 53 (a) False D must be a diagonal matrix (b) True See Theorem 5 (c) False Diagonalizability is about having enough linearly independent eigenvectors, not eigenvalues (d) False A is invertible iff A does not have zero as an eigenvalue, which has nothing to do with diagonalizability

2 537 Since A is diagonalizable, we can write A = P DP for some diagonal D whose entries are the eigenvalues of A Since A is invertible none of the eigenvalues are zero Thus, D has exactly n pivots so it is invertible, ie, D exists (alternatively, you can just consider the diagonal matrix with entries /λ i if D has entries λ i on the diagonal) Now observe that (P D P )A = P D P P DP = P D DP = P P = I, so A has a left inverse By the invertible matrix theorem (or by direct calculation) this implies that A also has a right inverse, and is invertible 533 Consider the matrix This matrix is already in REF and has a pivot in every row, so it is invertible Since it is upper triangular its eigenvalues are equal to its diagonal entries, so its only eigenvalue is However the corresponding eigenspace is E = Null = Span{ }, so the matrix does not have a basis of eigenvectors and is not diagonalizatable 533 Any diagonalizable matrix A with [ zero as ] an eigenvalue will do, since this implies A is not invertible Consider A = Since this matrix is upper triangular its eigenvalues are just,, and since they are distinct the matrix must be diagonalizable (One can also observe that the matrix is in REF and does not have a pivot in every row) 54 The B matrix of x Ax is given by where A is the standard matrix We have [A] B = P B E AP E B, P E B = P B = and so multiplying we obtain P B E = P B =, [A] B =

3 544 We compute the characteristic polynomial t 3 χ A (t) = det = ( t) 9 = t 4t 5 = (t 5)(t + ) 3 t Thus the eigenvalues are 5,, and since they are distinct the matrix must be diagonalizable The corresponding eigenvectors are given by solving: 3 3 v 3 3 = v = c, So [T ] B = 547 See answer key 5 in the basis B = { 543 Suppose Ax = λx We calculate 3 3 v 3 3 = v = c, } B(P x) = P AP P x = P Ax = P (λx) = λp x Moreover, P x since P is invertible and x Thus P x is an eigenvector of B with eigenvalue λ, as desired A 4 = P D 4 P = = See answer key 534 Noting that A k = P D k P and that D k = A k = 535 See answer key 3 ( 3) k ( ) k = 3 ( 3) k ( ) k, we obtain: [ 3( 3) k + 4( ) k 6( 3) k 6( ) k ( 3) k + ( ) k 4( 3) k 3( ) k 53 (a) False You need n linearly independent eigenvectors (b) False Consider the identity (c) True Let v,, v n be the columns of P Then the columns of AP are Av, Av n and the columns of P D are d i v i where d i is the i th diagonal entry of D Comparing columns we find that Av i = d i v i, which means that the v i are eigenvectors and the d i eigenvalues, whenever v i (d) False See No, it is not, since it has at most two linearly independent eigenvectors (one from each eigenspace) 3 ]

4 535 No it is not possible, ie, A must be diagonalizable Let the three eigenvalues be a, b, c with eigenspaces E a, E b, E c Assume E a has dimension and E b has dimension two Since c is an eigenvalue E c must have dimension at least one Thus the sum of the dimensions of the eigenspaces is 4, so by Theorem 7 the matrix must be diagonalizable 539 See answer key Basically, the idea is that you can switch the order of the eigenvalues and eigenvectors and get matrices P and D with correspondingly switched columns and rows Another way [ to get a different ] P would be to use some other scaling of the 3 eigenvectors (eg, P = ) 6 5Supp Assume ABx = λx Observe that BA(Bx) = BABx = B(λx) = λ(bx), so since Bx we conclude that Bx is an eigenvector of BA 5Supp9 Observe that if x is an eigenvector of A with eigenvalue λ, then Ax = λx, which is equivalent to x Ax = x λx which is the same as (I A)x = ( λ)x Thus the eigenvalues of I A are exactly λ i where λ i are the eigenvalues of A Since λ i < for all the eigenvalues, is not an eigenvalue of A, and consequently = is not an eigenvalue of I A, which means that I A is invertible 5Supp This question assumes quite a bit of familiarity with limits, which is why it was optional (and such questions will not be on the exam) The statement A k means that lim k (Ak )(i, j) ==, for every i, j n, ie, every entry of the sequence A, A, A 3, tends to zero Fix i, j and observe that the i, j entry of A k is equal to e T i A k e j Since A is diagonalizable, we have e T i A k e j = e T i P D k P e j = v T D k w, letting v = P T e i and w = P e j Suppose λ,, λ n are the diagonal entries of D (ie, the eigenvalues) and recall that λ i < We now have v T D k w = λ k v w +λ k v w + +λ k nv n w n λ k v w + λ k v w + + λ n k v n w n Since λ i k as k, the sum on the right hand side must tend to zero, so each entry of A k must also tend to zero, as desired 55 See answer key 554 To find the eigenvalues we first find the characteristic polynomial t χ A (t) = det = ( t)(3 t) + = t 4t + 5, 3 t 4

5 which has roots 4± 6 = ± i We find the corrresponding eigenvectors by row reduction; for λ = i + we have i i [ i ] i i i, since by complex arithmetic we have ( i)( i) = ( i ) =, which implies that = i, so the bottom right entry is zero Thus the second variable is free i and the nullspace is given by [ E +i = { x ] [ ] i : x x C} = Span{ i + i } = Span{ } + i Thus Av = ( + i)v where v = We could find the eigenvector corresponding to the other eigenvalue i by rowreduction, but a more conceptual way is to observe that Av = Av = ( + i)v = ( i)v, + i i so the other eigenvector must simply be v = v = = 558 Computing r = 3 + (3 3) = = 6, we have [ ] = [ ] / 3/ = cos π/3 sin( π/3 sin( π/3) cos(π/3), 3/ / 6 so the given matrix corresponds to a scale factor of r = 6 and a rotation by φ = π/3 55 Just as above, we calculate r = = 3 and cos φ = 3/( 3) = 3/, sin φ = 3/( 3) = /, yielding φ = π/6 553 See answer key 55 Assume Ax = λx and let µ C Observe that A(µx) = µax = µλx = λ(µx), since matrix vector multiplication has the same properties over C as over R, and complex number multiplication is commutative Since µ µx and we conclude that µx is an eigenvalue of A with eigenvalue λ 555 See answer key 5