Math 3191 Applied Linear Algebra
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1 Math 9 Applied Linear Algebra Lecture 9: Diagonalization Stephen Billups University of Colorado at Denver Math 9Applied Linear Algebra p./9
2 Section. Diagonalization The goal here is to develop a useful factorization A = P DP, when A is n n. We can use this to compute A k quickly for large k. The matrix D is a diagonal matrix (i.e. entries off the main diagonal are all zeros). D k is trivial to compute as the following example illustrates. Math 9Applied Linear Algebra p./9
3 Computing Powers of Diagonal Matrices EXAMPLE: Let D = where k is a positive integer?. Compute D and D. In general, what is D k, Solution: D = = D = D D = = and in general, D k = k k Math 9Applied Linear Algebra p./9
4 Powers of P DP EXAMPLE: Let A = where P =, D = Solution:. Find a formula for A k given that A = P DP and P = A = `P DP `P DP = P D `P P DP = P DDP = P D P. Math 9Applied Linear Algebra p./9
5 Powers of P DP, cont. Again, A = A A = `P D P `P DP = P D `P P DP = P D P In general, A k = P D k P = k k = k k k + k k k k + k. Math 9Applied Linear Algebra p./9
6 Diagonalizable Matrices A square matrix A is said to be diagonalizable if A is similar to a diagonal matrix, i.e. if A = P DP where P is invertible and D is a diagonal matrix. When is A diagonalizable? eigenvectors of A.) (The answer lies in examining the eigenvalues and In the previous example, note that the columns of P are eigenvectors of A: = and = Math 9Applied Linear Algebra p./9
7 All together = 8 Equivalently, = or = Math 9Applied Linear Algebra p./9
8 In general, h A and if h v v v n i = v v v n i h i v v v n {z } P h i v v v n is invertible, A equals λ λ..... {z λ n } D λ λ..... λ n h i v v v n {z } P Math 9Applied Linear Algebra p.8/9
9 THEOREM The Diagonalization Theorem An n n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors. In fact, A = P DP, 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. Math 9Applied Linear Algebra p.9/9
10 Diagonalizing a Matrix EXAMPLE: Diagonalize the following matrix, if possible. A = Step. Find the eigenvalues of A. λ det (A λi) = det λ λ = ( λ) ( λ) =. Eigenvalues of A: λ = and λ =. Math 9Applied Linear Algebra p./9
11 Diagonalizing a Matrix, cont. Step. Find three linearly independent eigenvectors of A. By solving (A λi) x =, for each value of λ, we obtain the following: Basis for λ = : v = Basis for λ = : v = v = Step : Construct P from the vectors in step. P = Math 9Applied Linear Algebra p./9
12 Diagonalizing a Matrix, cont. Step : Construct D from the corresponding eigenvalues. D = Step : Check your work by verifying that AP = P D AP = = P D = = Math 9Applied Linear Algebra p./9
13 Not Every Matrix is Diagonalizable! EXAMPLE: Diagonalize the following matrix, if possible. A =. Since this matrix is triangular, the eigenvalues are λ = and λ =. By solving (A λi) x = for each eigenvalue, we would find the following: Basis for λ = : v = Basis for λ = : v = Every eigenvector of A is a multiple of v or v which means there are not three linearly independent eigenvectors of A and by Theorem, A is not diagonalizable. Math 9Applied Linear Algebra p./9
14 Theorem EXAMPLE: Why is A = diagonalizable? Solution: Since A has three eigenvalues (λ =, λ =, λ = ) and since eigenvectors corresponding to distinct eigenvalues are linearly independent, A has three linearly independent eigenvectors and it is therefore diagonalizable. THEOREM An n n matrix with n distinct eigenvalues is diagonalizable. Math 9Applied Linear Algebra p./9
15 Motivation for Theorem EXAMPLE: Diagonalize the following matrix, if possible. A = Solution: Eigenvalues: and (each with multiplicity ). Solving (A λi) x = yields the following basis vectors for the eigenspaces of the two eigenvalues: λ = : v =, v = λ = : v =, v = Math 9Applied Linear Algebra p./9
16 {v, v, v, v } is linearly independent P = [v v v v ] is invertible. A = P DP, where P = and D =. Math 9Applied Linear Algebra p./9
17 THEOREM Let A be an n n matrix whose distinct eigenvalues are λ,..., λ p. a. For k p, the dimension of the eigenspace for λ k is less than or equal to the multiplicity of the eigenvalue λ k. 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 λ k equals the multiplicity of λ k. c. If A is diagonalizable and β k is a basis for the eigenspace corresponding to λ k for each k, then the total collection of vectors in the sets β,..., β p forms an eigenvector basis for R n. Math 9Applied Linear Algebra p./9
18 Section.: Eigenvectors and Linear Transformations Recall: A Linear Transformations T : V W can be represented by a matrix. How? Choose bases B and C for V and W respectively. Represent vectors using their coordinates relative to the bases. Construct the matrix A representing the transformation using the method we learned in Section.9. Then w = T (v) [w] C = A[v] B. Math 9Applied Linear Algebra p.8/9
19 Example Suppose V = IP and W = IP. Suppose that T : V W is defined by T (v) = v = dv dt. Let B = {b, b, b } = {, t, t } Let C = {c, c } = {, t}. T (b ) = d dt =, T (b ) = d dt t =, T (b ) = t. [T (b )] C = (, ), [T (b )] C = (, ), [T (b ] C = (, ). h i A = [T (b )] C [T (b )] C [T (b )] C =. Check: If v = a + a t + a t, then T (v) = a + a t. [v] B = (a, a, a ). [T (v)] C = (a, a ). And, A[v] B = (a, a ). Math 9Applied Linear Algebra p.9/9
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