Diagonalization. MATH 322, Linear Algebra I. J. Robert Buchanan. Spring Department of Mathematics

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1 Diagonalization MATH 322, Linear Algebra I J. Robert Buchanan Department of Mathematics Spring 2015

2 Motivation Today we consider two fundamental questions: Given an n n matrix A, does there exist a basis for R n consisting of eigenvectors of A? Given an n n matrix A, does there exist an invertible matrix P such that P 1 AP is a diagonal matrix?

3 Similarity Transformations Suppose A and P are n n matrices and suppose P is invertible. We will call the transformation A P 1 AP = B a similarity transformation.

4 Similarity Transformations Suppose A and P are n n matrices and suppose P is invertible. We will call the transformation A P 1 AP = B a similarity transformation. Matrices A and B share many properties in common. For instance det(b) = det(p 1 AP) = det(p 1 ) det(a) det(p) = det(a) det(p) det(p) = det(a).

5 Similarity Invariants Any property that is preserved by a similarity transformation is called a similarity invariant. Properties of a matrix which are invariant under similarity include: Determinant Invertibility Rank Nullity Trace Characteristic polynomial Eigenvalues Eigenspace dimension

6 Similar Matrices Definition If A and B are square matrices, then we say B is similar to A if there is an invertible matrix P such that B = P 1 AP. Remark: if B is similar to A then A is similar to B since B = P 1 AP PBP 1 = PP 1 APP 1 (P 1 ) 1 BP 1 = A where Q = P 1 is invertible. Q 1 BQ = A

7 Diagonalizable Matrices Definition A square matrix A is diagonalizable if there is an invertible matrix P such that P 1 AP is a diagonal matrix. The matrix P is said to diagonalize A.

8 Diagonalizable Matrices Definition A square matrix A is diagonalizable if there is an invertible matrix P such that P 1 AP is a diagonal matrix. The matrix P is said to diagonalize A. Theorem If A is an n n matrix, then the following statements are equivalent. 1. A is diagonalizable. 2. A has n linearly independent eigenvectors.

9 Proof (1 = 2) Suppose A is diagonalizable, then D = P 1 AP is a diagonal matrix. P D = A P

10 Proof (1 = 2) Suppose A is diagonalizable, then D = P 1 AP is a diagonal matrix. P D = A P Let the diagonal entries of D be λ 1, λ 2,..., λ n and let the columns of P be the vectors p 1, p 2,..., p n. P D = [ λ 1 p 1 λ 2 p 2 λ n p n ] = [ Ap1 Ap 2 Ap n ] = A P

11 Proof (1 = 2) Suppose A is diagonalizable, then D = P 1 AP is a diagonal matrix. P D = A P Let the diagonal entries of D be λ 1, λ 2,..., λ n and let the columns of P be the vectors p 1, p 2,..., p n. P D = [ λ 1 p 1 λ 2 p 2 λ n p n ] = [ Ap1 Ap 2 Ap n ] = A P Thus Ap 1 = λ 1 p 1, Ap 2 = λ 2 p 2,..., Ap n = λ 2 p n. Thus the columns of P are the eignenvectors of A.

12 Proof (1 = 2) Suppose A is diagonalizable, then D = P 1 AP is a diagonal matrix. P D = A P Let the diagonal entries of D be λ 1, λ 2,..., λ n and let the columns of P be the vectors p 1, p 2,..., p n. P D = [ λ 1 p 1 λ 2 p 2 λ n p n ] = [ Ap1 Ap 2 Ap n ] = A P Thus Ap 1 = λ 1 p 1, Ap 2 = λ 2 p 2,..., Ap n = λ 2 p n. Thus the columns of P are the eignenvectors of A. Since P is invertible, rank(p) = n which implies its columns (the eigenvectors of A) are linearly independent.

13 Proof (2 = 1) Suppose A has n linearly independent eigenvectors p 1, p 2,..., p n with corresponding eigenvalues λ 1, λ 2,..., λ n. Let the eigenvectors of A be the columns of matrix P and let D be the diagonal matrix with the eigenvalues of A on the diagonal. A P = [ Ap 1 Ap 2 Ap n ] = [ λ1 p 1 λ 2 p 2 λ n p n ] = P D

14 Proof (2 = 1) Suppose A has n linearly independent eigenvectors p 1, p 2,..., p n with corresponding eigenvalues λ 1, λ 2,..., λ n. Let the eigenvectors of A be the columns of matrix P and let D be the diagonal matrix with the eigenvalues of A on the diagonal. A P = [ Ap 1 Ap 2 Ap n ] = [ λ1 p 1 λ 2 p 2 λ n p n ] = P D Since the columns of P are linearly independent, P is invertible. Therefore D = P 1 AP.

15 Related Result Theorem If λ 1, λ 2,..., λ k are distinct eigenvalues of a matrix A and if v 1, v 2,..., v k are corresponding eigenvectors then {v 1, v 2,..., v k } is a linearly independent set. An n n matrix with n distinct eigenvalues is diagonalizable.

16 Diagonalizing a Matrix Given an n n matrix A, Steps: 1. Find n linearly independent eigenvectors of A, say p 1, p 2,..., p n. 2. Form the matrix P having p 1, p 2,..., p n as its columns. 3. The matrix P 1 AP will be diagonal with diagonal entries λ 1, λ 2,..., λ n, where λ i is the eigenvalue corresponding to p i for i = 1, 2,..., n.

17 Example Diagonalize the matrix below A =

18 Solution (1 of 2) 0 = det(λi A) = Eigenvalues: λ 1 = 0, λ 2 = λ 3 = 1. λ λ λ + 1 = λ(λ 1)2

19 Solution (1 of 2) 0 = det(λi A) = Eigenvalues: λ 1 = 0, λ 2 = λ 3 = 1. 1 Eigenvectors: p 1 = 1, p 2 = 1 λ λ λ , p 3 = = λ(λ 1)

20 Solution (1 of 2) 0 = det(λi A) = λ λ λ + 1 Eigenvalues: λ 1 = 0, λ 2 = λ 3 = Eigenvectors: p 1 = 1, p 2 = 2, p 3 = P = and P 1 = = λ(λ 1)

21 Solution (2 of 2) We can verify P 1 AP = = = D

22 Example Verify that the matrix below is not diagonalizable B =

23 Solution 0 = det(λi B) = λ λ λ + 4 = (λ 2)(λ 1)2 Eigenvalues: λ 1 = 2, λ 2 = λ 3 = 1.

24 Solution 0 = det(λi B) = λ λ λ + 4 = (λ 2)(λ 1)2 Eigenvalues: λ 1 = 2, λ 2 = λ 3 = 1. 3 Eigenvectors: p 1 = 3, p 2 = 4 Since there are only 2 linearly independent eigenvectors for this matrix, the matrix is not diagonalizable

25 Example Verify the following matrix is diagonalizable

26 Solution Since the matrix is triangular then the eigenvalues are λ 1 = 1, λ 2 = 5, λ 3 = 8, and λ 4 = 10. Since there are four distinct eigenvalues there will be four linearly independent eigenvectors and thus the matrix is diagonalizable.

27 Geometric and Algebraic Multiplicity Remark: an n n matrix may be diagonalizable even if it does not have n distinct eigenvalues. What is important is that it have n linearly independent eigenvectors.

28 Geometric and Algebraic Multiplicity Remark: an n n matrix may be diagonalizable even if it does not have n distinct eigenvalues. What is important is that it have n linearly independent eigenvectors. Suppose the characteristic polynomial of A can be factored as p(λ) = (λ λ 1 ) n 1 (λ λ 2 ) n2 (λ λ k ) n k

29 Geometric and Algebraic Multiplicity Remark: an n n matrix may be diagonalizable even if it does not have n distinct eigenvalues. What is important is that it have n linearly independent eigenvectors. Suppose the characteristic polynomial of A can be factored as p(λ) = (λ λ 1 ) n 1 (λ λ 2 ) n2 (λ λ k ) n k then the eigenvalues of A are λ 1, λ 2,..., λ k.

30 Geometric and Algebraic Multiplicity Remark: an n n matrix may be diagonalizable even if it does not have n distinct eigenvalues. What is important is that it have n linearly independent eigenvectors. Suppose the characteristic polynomial of A can be factored as p(λ) = (λ λ 1 ) n 1 (λ λ 2 ) n2 (λ λ k ) n k then the eigenvalues of A are λ 1, λ 2,..., λ k. n i is called the algebraic multiplicity of λ i. The dimension of the eigenspace corresponding to λ i is called the geometric multiplicity of λ i.

31 Geometric and Algebraic Multiplicity (continued) Theorem If A is a square matrix, then 1. The geometric multiplicity of every eigenvalue is less than or equal to its algebraic multiplicity. 2. A is diagonalizable if and only if the geometric multiplicity of every eigenvalue is equal to it algebraic multiplicity.

32 Powers of a Matrix If a square matrix A is diagonalizable, then powers of A are easy to compute. D = P 1 AP D k = (P 1 AP) k = (P 1 AP)(P 1 AP) (P 1 AP) }{{} k factors = P 1 A(PP 1 )A(P P 1 )AP = P 1 A k P PD k P 1 = A k

33 Example If A = then find A 10.

34 Example If A = then find A 10. We have already seen that if P = P 1 = then P 1 AP = D = with

35 Solution A 10 = PD 10 P = = =

36 Final Word The converse of the diagonalizability theorem is false. There are n n matrices with fewer than n distinct eigenvalues that are diagonalizable.

37 Homework Read Section 5.2 Exercises: 1, 3, 5, 10, 15, 20, 23, 24, 25.

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