Eigenvalues and Eigenvectors 7.2 Diagonalization November 8
Goals Suppose A is square matrix of order n. Provide necessary and sufficient condition when there is an invertible matrix P such that P 1 AP is a diagonal matrix.
Definitions Two square matrices A,B said to be similar, if there is an invertible matrix P, such that A = P 1 BP. A square matrix A said to be diagonalizable, if there is an invertible matrix P, such that P 1 AP is a diagonal matrix. That means, if A is similar to a diagonal matrix, we say that A is diagonalizable.
Theorem 7.3 Theorem 7.4 Suppose A,B are two similar matrices. Then, A and B have same eigenvalues. Proof. Write A = P 1 BP. Then λi A = λi P 1 BP = λ(p 1 P) P 1 BP = P 1 (λi B)P = P 1 λi B P = P 1 λi B P = λi B So, A and B has same characteristic polynomials. So, they have same eighenvalues. The proof is complete.
Conditions for Diagonalizability We ask, when a matrix is diagonalizable? Theorem 7.5 A square matrix A, of order n, is diagonalizable if and only if A has n linearly independent eigenvectors. Proof. Suppose A is diagonalizable. Then, P 1 AP = D for some invertible matrix P, where D is a diagonal matrix. So, Write D =. λ 1 0 0 0 λ 2 0 0 0 λ n AP = PD. and P = [ p 1 p 2 p n ].
Continued So, A [ ] p 1 p 2 p n = [ ] p 1 p 2 p n λ 1 0 0 0 λ 2 0 0 0 λ n. Or [ Ap1 Ap 2 Ap n ] = [ λ1 p 1 λ 2 p 2 λ n p n ]
Continued So, Ap i = λ i p i for i = 1,2,,n Since P is invertible, p i 0. So, p i are eigenvectors of A. Also, rank(p) = n. So, its columns {p 1,p 2,...,p n } are indpendent. So, it is established that if A is diagonilizable, then has n linearly independent eigenvectors.
Continued Now, we prove the converse. So, we assume A bas has n linearly independent eigenvectors: So, {p 1,p 2,...,p n } Ap 1 = λ 1 p 1,Ap 2 = λ 2 p 2,,Ap n = λ n p n for some λ i.
Continued Write, P = [ p 1 p 2 p n ] and D = λ 1 0 0 0 λ 2 0 0 0 λ n It follows from the equations Ap i = λ i p i that AP = PD. So, P 1 AP = D is diagonal. The proof is complete.
Steps for Diagonalizing Suppose A is a square matrix of order n. If A does not have n linearly independent eigenvectors, then A is not diagonalizable. When possible, find n linearly independent eigenvectors p 1,p 2,,p n for A with corresponding eigenvalues λ 1,λ 2,...,λ n. Then, write P = [ p 1 p 2 p n ] and D = We have D = P 1 AP is a diagonal matrix. λ 1 0 0 0 λ 2 0 0 0 λ n
With Distinct Eigenvalues Thoerem 7.6 Suppose a square matrix A, of order n, has n distinct eigenvalues. Then the corresponding eignevectors are linearly independent and A is diagonizable. Proof. The second statement follows from the first, by theorem 7.5. So, we prove the first statement only. Let λ 1,λ 2,...,λ n be distinct eigenvalues of A. So, for i = 1,2,...,n we have Ax i = λ i x i where x i 0 are eigenvectors.
Continued We are required to prove that x 1,x 2,...,x n are linearly independent. We prove by contra-positive argument. So, assume they are linearly dependent. There is an m < n such that x1,x 2,...,x m are mutually linearly independent and x m+1 is in can be written as a linear combination of {x 1,x 2,...,x m }. So, x m+1 = c 1 x 1 +c 2 x 2 + +c m x m Eqn I. Here, at least one c i 0. Re-labeling x i, if needed, we can assume c 1 0.
Continued Multiply Eqn-I by A on the left and use Ax i = λ i x i,: λ m+1 x m+1 = λ 1 c 1 x 1 +λ 2 c 2 x 2 + +λ m c m x m Eqn II. Also, multiply Eqn-I by λ m+1, we have λ m+1 x m+1 = λ m+1 c 1 x 1 +λ m+1 c 2 x 2 + +λ m+1 c m x m Eqn III. Subtract Eqn-II from Eqn -III: (λ m+1 λ 1 )c 1 x 1 +(λ m+1 λ 2 )c 2 x 2 + +(λ m+1 λ m )c m x m = 0.
Continued Since these vectors are linearly independent, (λ m+1 λ i )c i = 0 for i = 1,2,,n. Since c 1 0 we get λ m+1 λ 1 = 0 or λ m+1 = λ 1. This contradicts that λ i s are distinct. So, we conclude that x 1,x 2,...,x n are linearly independent. The proof is complete. Reading assignment: 7.2 Examples 1-7
Exercise 6 Let A = 2 3 1 0 1 2 0 0 3 and P = 1 1 5 0 1 1 0 0 2 Verify that A is diagonalizable, by computing P 1 AP. Solution: We do it in a two steps. 1. Use TI to compute P 1 = 1 1 3 0 1.5 0 0.5.So, P 1 AP = So, it is verified that P 1 AP is a diagonal matrix.. 2 0 0 0 1 0 0 0 3.
Exercise 10 [ Let A = 1.5 2 1 Show that A is not diagonalizable. Solution: Use theorem 7.5 and show that A does not have 2 linearly independent eigenvectors. To do this, we have find and count the dimensions of all the eigenspaces E(λ). We do it in a few steps. ]. First, find all the eigenvalues. To do this, we solve det(λi A) = λ 1.5 2 λ+1 = λ2 = 0. So, λ = 0 is the only eigenvalue of A.
Continued Now we compute the eigenspace E(0) of the eigenvalue λ = 0. We have E(0) is solution space of [ ] [ ] [ ][ ] [ ] x 0 1.5 x 0 (0I A) = or = y 0 2 1 y 0 Using TI (or by hand), a parametric solution of this system is given by x =.5t y = t. So E(0) = {(.5t,t) : t R} = R(.5,1). So, the (sum of) dimension(s) of the eigenspace(s) = dime(0) = 1 < 2. Therefore A is not diagonizable.
Exercise 14 Let A = 2 1 1 0 1 2 0 0 1 Show that A is not diagonalizable. Solution: Use theorem 7.5 and show that A does not have 2 linearly independent eigenvectors. To do this, we find and count the dimensions of all the eigenspaces E(λ). First, find all the eigenvalues. To do this, we solve λ 2 1 1 det(λi A) = 0 λ+1 2 0 0 λ+1 = (λ 2)(λ+1)2 = 0.. So, λ = 1,2 are the only eigenvalues of A.
Continued Now we compute the dimension dime( 1) of the eigenspace E( 1) of the eigenvalue λ = 1. We have E( 1) is solution space of Or ( I A) 3 1 +1 0 0 2 0 0 0 x y z = x y z 0 0 0 = (We will avoid solving this system.) The rank of the coefficient matrix is 2. So, dim(e( 1)) = nullity = 3 rank = 3 2 = 1. 0 0 0
Continued Now we compute the dimension dime(2) of eigenspace E(2) of the eigenvalue λ = 2. E(2) is the solution space of x 0 (2I A) y = 0 or z 0 0 1 +1 0 3 2 0 0 3 x y z = Use TI (or look at the columns) to see that rank of the coefficient matrix is 2. So, dim(e(2)) = nullity = 3 rank = 3 2 = 1. 0 0 0
Continued So, the sum of dimensions of the eigenspaces = dime( 1)+dimE(2) = 2 < 3. Therefore A is not diagonizable.
Exercise 20 Let A = 4 3 2 0 1 1 0 0 2 Find the eigenvalues of A and determine whether there is a sufficient number of them to guarantee that A is diagonalizable. Solution: First, find all the eigenvalues. To do this, we solve det(λi A) = λ 4 3 2 0 λ 1 1 0 0 λ+2. = (λ 4)(λ 1)(λ+2) = 0. So, A has three distinct eigenvalues λ = 4,1, 2. Since A is a 3 3 matrix, by theorem 7.6, A is diagonalizable.
: 7.2 See site