x5.3 Diagonalization dfn: A square matrix A is diagonalizable if A is similar to a diagonal matrix. This means A = P DP −1 for some invertible P and diagonal D, with all matrices being n × n. EPIC FACT: If A = P DP −1 for some invertible P and diagonal D we can compute Ak without computing AA ··· A. In fact, Ak = PDkP −1. This is much less computation because | {z } k factors 2 3 2 k 3 λ1 0 0 ::: 0 λ1 0 0 ::: 0 k 6 0 λ2 0 ::: 0 7 6 0 λ2 0 ::: 0 7 6 7 6 k 7 6 0 0 λ3 ::: 0 7 k 6 0 0 λ ::: 0 7 if D = 6 7 then D = 6 3 7 6 . .. 7 6 . .. 7 4 . 5 4 . 5 k 0 0 0 : : : λn 0 0 0 : : : λn Theorem 5: The Diagonalization Theorem • An n × n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors. If so, these eigenvectors form a basis of Rn and this basis is called an eigenvector basis. • Assume A is diagonalizable with the eigenvalues λ1; λ2; : : : ; λn and corresponding eigenvector basis B = fv1; v2;:::; vng (Note that vj is an eigenvector associated with λj , and the λj do not need to be distinct numbers.) Then, A = P DP −1 2 3 λ1 0 0 ::: 0 6 0 λ2 0 ::: 0 7 6 7 6 0 0 λ3 ::: 0 7 where P = [v1 v2 ::: vn] and D = 6 7 6 . .. 7 4 . 5 0 0 0 : : : λn Theorem 6 An n × n matrix with n distinct eigenvalues is diagonalizable. (Because the n associated eigenvectors are always linearly independent.) Theorem 7 Let A be an n × n matrix whose distinct eigenvalues are λ1; : : : ; λp. a. For 1 ≤ k ≤ p, the dimension of the eigenspace for λk is less than or equal to the multiplicity of the eigenvalue λk. b. The matrix A is diagonalizable if and only if the sum of the dimensions of the distinct eigenspaces equals n, and this happens if and only if the dimension of the eigenspace for each λk equals the multiplicity of λk. c. If A is diagonalizable and Bk is a basis for the eigenspace corresponding to λk for each k, then the total n collection of vectors in the sets B1;:::; Bp forms an eigenvector basis for R . NB: • It's not necessary for an n × n matrix to have n distinct eigenvalues in order to be diagonalizable. What matters is having n linearly independent eigenvectors. • Two matrices with the same eigenvalues, with the same multiplicities, aren't necessarily both diagonal- izable, or both not diagonalizable. What matters is having enough linearly independent eigenvectors. • To summarize, an n × n matrix A is diagonalizable if and only if there are enough linearly independent eigenvectors to form a basis of Rn. This occurs precisely when the sum of the dimensions of the distinct eigenspaces = n. • If A is diagonalizable and λ1; λ2; : : : ; λn are its eigenvalues then { The λj do not all need to be distinct. { The eigenvector basis B = fv1;:::; vng must be in the same order as the eigenvalues in D (e.g., v1 must be an eigenvector for λ1). MORAL 1: Let A be an n × n matrix. To see if A is diagonalizable: 1. Find eigenvalues: that is, find roots, λ, to the characteristic equation det(A − λIn) = 0. Let λ be an eigenvalue of A. The eigenspace of A associated to λ is Nul (A − λIn), the solution space to (A − λIn)x = 0. 2. For each eigenvalue, λ, of A, find a basis for its eigenspace. (By Theorem 7, that the eigenspace associated to λ has dimension less than or equal to the multiplicity of λ as a root to the characteristic equation.) 3. The matrix A is diagonalizable if and only if the sum of the dimensions of the eigenspaces equals n. 4. If A is diagonalizable, use Theorem 5 (above) to find P and D so that A = P DP −1. x5.4 The Matrix of a Linear Transformation in General First we give a theorem that gives the matrix of a linear transformation. Theorem (see p. 289 of the book) Let V be a finite dimensional vector space with basis B = fv1; v2;:::; vng and let W be a finite dimensional vector space with basis C = fw1; w2;:::; wmg. Let T : V ! W be linear. Then there is a unique matrix C[T ]B = [[T (v1)]C; [T (v2)]C;::: [T (vn)]C] (the matrix formed by the coordinates of the T (vj ) in the C basis) which satisfies, for all v 2 V , [T v]C = C[T ]B[v]B: If T : V ! V is linear then we write [T ]B for the matrix B[T ]B. xDiagonalizing Linear Transformations from Rn to Rn Now we consider the case of linear transformations on Rn. Let T : Rn ! Rn be linear, and let A be the standard matrix of T . This means that n T (x) = Ax for all x 2 R : In class we will show that the standard matrix of T , A, is equal to the matrix of T in the standard basis E = fe1; e2;:::; eng: A = [T ]E . We say that the linear transformation T is a diagonalizable linear transformation if its standard matrix A = [T ]E is a diagonalizable matrix. Theorem 8. Suppose A is diagonalizable and A = P DP −1 where and D is diagonal. If B is the basis for Rn formed from the columns of P , then D is the B-matrix for the linear transformation x 7! Ax. Using the notation T (x) = Ax, we have [T ]B = D so n [T (x)]B = D[x]B for all x 2 R : MORAL 2: Let T : Rn ! Rn be linear. To see if T is a diagonalizable linear operator: 1. Find the standard matrix of T , A = [T ]E . 2. Determine if A is diagonalizable (see first part of these notes). (If not, T is not diagonalizable.) However, if A is diagonalizable, find eigenvalues λ1; λ2; : : : ; λn and an associated eigenvector basis, B = fv1; v2;:::; vng. 2 3 λ1 0 0 ::: 0 6 0 λ2 0 ::: 0 7 6 7 6 0 0 λ3 ::: 0 7 3. Then, the matrix of T in the B basis is [T ]B = D = 6 7 : 6 . .. 7 4 . 5 0 0 0 : : : λn 2.