Let A [ 2 2 2 Compute the eigenvalues and eigenspaces of A We start b computing the characteristic polnomial of A as [ 2 λ 2 det (A λi) det 2 λ ( 2 λ)( λ) (2)(2) λ 2 + λ 2 4 λ 2 + λ 6 (λ 2)(λ + 3) The eigenvalues are the roots of the characteristic polnomial, λ 2 and λ 3 To find the eigenspace associated with each, we set (A λi)x 0 and solve for x This is a homogeneous sstem of linear equations, so we put A λi in row echelon form [ 4 2 For λ 2, we have A 2I Add half of the top row to [ 2 4 2 the bottom to get Then x 0 0 2 t is a free variable, and we have 4x [ + 2t 0, or x 2t The eigenspace corresponding to λ 2 is the span [ of 2, or equivalentl of 2 [ 2 For λ 3, we have A + 3I Subtract two times the top row [ 2 4 2 from the bottom to get Then x 0 0 2 t is a free variable, and we have x + 2t 0, or x 2t The eigenspace corresponding to λ 3 is the span of [ 2
[ 8 6 2 Let A 8 3 [ Given that 2 of A, find a matrix B such that B 2 A and [ 2 3 are eigenvectors We can compute the eigenvalues corresponding to the given eigenvectors of A [ [ [ [ 8 6 4 4 8 3 2 8 2 [ [ [ [ 8 6 2 2 2 8 3 3 3 3 [ 4 0 Thus, the eigenvalues are 4 and If we let D and P [ 0 2, then b the wa that we diagonalize matrices, A P DP [ 2 0 If we let E, then D E 0 2 If we let B P EP, then B 2 P EP P EP P E 2 P P DP A We can [ compute det P ( )(3) ( 2)(2) We can compute P 3 2 We can compute det P CT 2 B [ P EP 2 [ 2 2 4 3 [ 2 2 6 5 [ [ 2 0 3 2 0 2 [ 3 2 2 Other possible choices for B (depending on the order of the eigenvalues on the diagonal of D and whether [ the positive or negative square root of the 0 6 eigenvalues are taken) include or the negative of either of the 8 matrices found so far
3 Let A be a 7 7 matrix that is not diagonalizable Suppose that nullit (A + I), nullit (A + 2I) 2, and nullit (A + 3I) 3 Determine, with proof, the nullit of (A + 4I) We are given that λ, λ 2, and λ 3 are eigenvalues of A, and that their associated eigenspaces have dimensions, 2, and 3, respectivel The eigenspaces thus have, 2, and 3 linearl independent eigenvectors, respectivel Suppose that λ 4 is an eigenvalue of A Then it has an eigenvector Since eigenvectors associated with different eigenvalues are linearl independent, A has at least seven linearl independent eigenvectors Since A is 7 7, A is diagonalizable, a contradiction Therefore, λ 4 is not an eigenvalue of A Thus, nullit (A + 4I) 0
4 Determine whether the following matrix is diagonalizable 3 2 0 2 0 4 0 0 0 0 0 Be sure to justif our answer (A correct solution does not require computing the inverse of an matrix) The matrix is 4 4, so it is diagonalizable if and onl if it has a set of four linearl independent eigenvectors Since the matrix is upper triangular, we can read the eigenvalues off as the numbers on the diagonal We have λ 3 and λ, each occurring with multiplicit The other eigenvalue is λ 2, which occurs with multiplicit 2 From the multiplicit of the eigenvalues, we know that λ 3 and λ each have one-dimensional eigenspaces λ 2 could have an eigenspace of dimension one or two To determine which, we can compute its eigenspace b subtracting 2I from the matrix This gives us 5 2 0 0 0 4 0 0 0 3 0 0 0 3 This matrix isn t quite in row echelon form, but it is close enough that we can tell that the second and third columns are not pivot columns This matrix thus has a null space of dimension two, so the eigenspace of the original matrix corresponding to λ 2 has dimension two Therefore, there are two linearl independent eigenvectors corresponding to λ 2 Eigenvectors corresponding to distinct eigenvalues are linearl independent, so these two eigenvectors together with one for λ 3 and one for λ are a set of four linearl independent eigenvectors for the matrix Therefore, the matrix is diagonalizable
([ x 5 Let T ) We are given that T T ( ( ([ x T T [ 2x + x 2 ([ x ))) ) ( ( ([ x Find a formula for T T T [ 2 2 [ x As such, [ [ [ [ 2 2 2 x 2 2 2 [ [ [ 5 0 2 x 0 5 2 [ [ 0 5 x 5 0 [ 0x + 5 5x 0 ))) ( ( ([ ))) ([ ) x x Oddl enough, T T T 5T That was an unintentional fluke of the problem, and noticing it was unnecessar
([ 2x + 3 6 Let T be a linear transformation such that T 2x + ([ ) x Find a formula for T ) [ x Since T is a linear transformation ([ on a) finite dimensional [ vector space, it x x must have a formula of the form T A for some matrix A Then we can compute [ x ([ ) 2x + 3 T 2x + [ 2x + 3 A 2x + [ [ x A 2 Since this must hold for all possible choices of x and, we must have [ [ [ A I, and so A We can compute det 4, 2 2 2 [ ([ ) [ 3 x from which A 4 Thus, T 4 x + 3 4 2 2 2 x 2