Diagonal the Matrix Remark: We learned in a previous section that in many cases, a matrix A can be displayed in a useful factorization of the form A = P DP −1 where D is a diagonal matrix. 7 2 Exercise 1. Let A = . Find a formula for Ak , given that A = P DP −1 , where −4 1 1 1 5 0 P = and D = . (Example 2) −1 −2 0 3 5 Class Exercise 1. Suppose P = 2 7 2 and D = 3 0 0 and A = P DP −1 . Compute A4 . (#1) 1 Definition: A square matrix A is said to be diagonalizable if A is similar to a diagonal matrix, that is, if A = P DP −1 for some invertible matrix P and some diagonal matrix D. Diagonalization Theorem: An n x n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors. In fact, A = P DP −1 , with D a diagonal matrix, if and only if the columns of P are n linearly independent eigenvectors of A. In this case, the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors in P . Definition: A is diagonalizable if and only if there are enough eigenvectors to form a basis of Rn . We call such a basis an eigenvector basis of Rn . Exercise 2. The matrix below is factored in the form P DP −1 . Use the Diagonalization Theorem to find the eigenvalues of A and a basis for each eigenspace. (#5)       1 1 2 1/4 2 2 1 5 0 0 1/4 1/2 1 3 1 =1 0 −1 0 1 0 1/4 1/2 −3/4 1 2 2 1 −1 0 0 0 1 1/4 −1/2 1/4   0 0 −2 Exercise 3. Find a matrix P that diagonalizes A = 1 2 1 . 1 0 3  1 Exercise 4. Show that the following matrix is not diagonalizable: A =  1 −3  0 0 2 0. 5 2 1 0 if possible. (In other words, find an invertible 6 −1 matrix P and a diagnol matrix D such that A = P DP −1 .) (#7) Class Exercise 2. Diagonalize the matrix Class Exercise 3. Diagonalize the matrix  −1 Exercise 5. Diagonalize the matrix −3 −3 λ = 1, 2, 3. (#11) 3 −1 if possible. (#9) 1 5  4 −2 4 0  if possible. The eigenvalues for the matrix are 1 3  2 2 −1 3 −1 if possible. The eigenvalues for the Class Exercise 4. Diagonalize the matrix  1 −1 −2 2 matrix are λ = 5, 1. (#13)  Theorem: An n x n matrix with n distinct eigenvalues is diagonalizable. Theorem: Let A be an n x 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 eigenspaces equals n, and this happens if and only if (i) the characteristic polynomial factors completely into linear factors and (ii) 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 collection of vectors in the sets B1 , . . . .., Bp forms an eigenvector basis for Rn .

Description

Related Documents

Free up your schedule!

Take 5 seconds to unlock