Recall : Eigenvalues and Eigenvectors

Transcription

1 Recall : Eigenvalues and Eigenvectors Let A be an n n matrix. If a nonzero vector x in R n satisfies Ax λx for a scalar λ, then : The scalar λ is called an eigenvalue of A. The vector x is called an eigenvector of A with eigenvalue λ. 0. is an eigenvector of with eigenvalue λ : Eigenspaces Let A be an n n matrix. If λ is an eigenvalue of A, the set of vectors x in R n satisfying Ax λx forms a subspace of R n. It is called the eigenspace of A corresponding to λ. The eigenspace corresponding to λ is identical to Nul(A λi). Ax λx (A λi)x 0 / 4

2 Finding Eigenvalues and Eigenvectors Let A be an n n matrix. To find the eigenvalues of A, we solve the equation det(a λi) 0 for λ. For each eigenvalue λ, we find a basis for Nul(A λi). The characteristic equation Let A be an n n matrix. The equation det(a λi) 0 is called the characteristic equation of A. The characteristic equation has the following form : ±λ n + c n λ n + c n λ n + + c λ + c 0 0 / 4

3 4 0 Find the eigenvalues of A Solution. First we find the characteristic equation : 4 λ 0 det(a λi) 3 λ λ λ + ( λ) 4 λ 6 λ 0 4 λ 3 5 ( λ) ( ( 4 λ)( λ) ()( 6) ) Because ( λ)(λ + 3λ + ) ( λ)(λ + )(λ + ) λ + 3λ + 0 λ 3 ± 9 8 Therefore the eigenvalues of A are,,. { 3 / 4

4 Multiplicity of an eigenvalue Sometimes the characteristic equation may have repeated roots. The number of times an eigenvalue λ appears as a root of the characteristic equation is called its multiplicity. If we take multiplicities into account, the characteristic equation of an n n matrix has n roots. Suppose A is a 5 5 matrix whose characteristic equation is λ 5 λ 4 λ 3 0. Find the eigenvalues of A and their multiplicities. Solution. λ 5 λ 4 λ 3 λ 3 (λ + λ + ) λ 3 (λ + ) Therefore the eigenvalues are : 0 with multiplicity 3 with mutiplicity You can see that the total sum of multiplicities is / 4

5 Multiplicity and dimension of eigenspace Theorem Let A be a square matrix an λ an eigenvalue of A. In general we have : Dimension of the eigenspace Multiplicity of λ corresponding to λ The inequality can be strict. 0. Let A. Then : λ 0 det(a λi) λ ( λ) Therefore λ is an eigenvalue with multiplicity. 0 0 x 0 However, Ax x (A I)x x 0 0 x 0 x x x Which means that the eigenspace corresponding to λ is one-dimensional. 5 / 4 x

6 3 Let A 0 0. Calculate the eigenvalues of A and their multiplicities. 3 λ 0 0 λ 3 A λi λ 0 0 λ λ λ We use the cofactor expansion across the second row : λ 3 0 λ λ 5 4 λ + ( λ) λ 5 4 λ 0 λ ( λ) ( ( λ)( 4 λ) ( 5) ) ( λ)(λ + λ 3) { For the second factor we have : λ ± (4)()( 3) 3 It follows that ( λ)(λ λ 3) (λ )(λ )(λ ( 3)) And the eigenvalues of A are : with multiplicity, -3 with multiplicity. 6 / 4

7 Recall : diagonal matrices An n n matrix is called a diagonal matrix if every entry outside the diagonal of the matrix is zero is a diagonal matrix But is NOT a diagonal matrix / 4

8 Powers of matrices - the diagonal case 0 0. Let A 0 0. Calculate A Solution. It is easily seen that () 0 0 A A A ( ) Similarly () A 3 A A A ( )3 We have : () A ( ) The above example shows that we can easily compute any power of a diagonal matrix. 8 / 4

9 Powers of matrices - the general case 5 4 Let A. Calculate A Solution. Suppose we can find an invertible matrix P such that P AP is diagonal. For example, consider P P and P AP D Then and P AP D A PDP A 5 (PDP )(PDP )(PDP )(PDP )(PDP ) PD 5 P 9 / 4

10 A 5 (PDP )(PDP )(PDP )(PDP )(PDP ) PD 5 P A ( 3) Calculating A r Let A be an n n matrix. We want to calculate A r. Suppose we can find an invertible n n matrix P and a diagonal matrix D such that A PDP. Then: But how do we find such a matrix P? A r PD r P 0 / 4

11 Diagonalizable matrices An n n matrix A is called diagonalizable if A PDP where an invertible n n matrix P, and a diagonal n n matrix D. If A is diagonalizable, then we can calculate powers of A quickly using the following method : Calculating A r Suppose A is diagonalizable, i.e., A PDP with D diagonal. Then: Reason : A r A A A A } {{ } r times A r PD r P (PDP )(PDP ) (PDP ) } {{ } r times P D D D P PD r P } {{ } r times / 4

12 How to find D and P? 7 9 Is A diagonalizable? If it is, use diagonalization to evaluate A Diagonalization Algorithm To diagonalize an n n matrix A: Find the eigenvalues of A. Find a basis for each eigenspace. 3 make a list of all of the vectors you obtain in this way : v,..., v k If k < n, then A is NOT diagonalizable. If k n, then A is diagonalizable : take P [ v v v k ] and the diagonal entries of D are the eigenvalues of A with multiplicities. The case k > n is impossible. / 4

13 7 9 Let A. Is A diagonalizable? If so, use diagonalization to calculate 6 8 A 0. Solution. Finding the eigenvalues of A : 7 9 λ 0 7 λ 9 A λi λ 6 8 λ 7 λ λ ( 7 λ)(8 λ) ( 9)6 λ λ (λ + )(λ ) λ () ± (4)( ) { 3 / 4

14 (cont.) 7 9 Let A. Is A diagonalizable? If so, use diagonalization to calculate 6 8 A 0. Solution. Finding a basis for each eigenspace : λ A ()I x : basic; x : free [ ] Row reduction 6 9 x 6 9 [ x ] 0 0 {x 3 x [ x 3 x ] x x 3 x and a basis for the eigenspace corresponding to λ is { } 3 4 / 4

15 (cont.) 7 9 Let A. Is A diagonalizable? If so, use diagonalization to calculate 6 8 A 0. Finding a basis for each eigenspace : λ A I [ ] x x 0 0 [ ] Row reduction x : basic; x : free. {x x x x x x x [ and a basis for the eigenspace corresponding to λ is ] { } 5 / 4

16 (cont.) 7 9 Let A. Is A diagonalizable? If so, use diagonalization to calculate 6 8 A 0. 3 Listing all of the vectors we obtained : v, v A is and the list contains two vectors. Therefore A is diagonalizable. 3 P P 3 3 We have : 7 9 P 3 AP / 4

17 (cont.) 7 9 Let A. Is A diagonalizable? If so, use diagonalization to calculate 6 8 A 0. Finally, we calculate A 0 : 3 P P AP 0 0 A P 0 P A 0 P P 0 3 () / 4

18 Another example 3 3 Let A 0. Is A diagonalizable? Solution. The matrix is triangular and its eigenvalues are 3 and -. We find a basis for each eigenspaces : λ A ()I x x x 3 3 x 4 x 0 Row Reduction x x x 3 0 { x 3 4 x 0 x x 4 4 Basis for eigenspace : / 4

19 Another example (cont.) 3 3 Let A 0. Is A diagonalizable? λ A 3I Row Reduction x x x 3 0 { x 0 x 3 0 x x x 0 x 0 Basis for eigenspace : 0 x We now list all of the vectors we have obtained : v, v We only have vectors, but A is 3 3. Therefore A is NOT diagonalizable. 9 / 4

20 Theorem Let A be an n n matrix. If A has exactly n distinct eigenvalues, then A is diaonalizable. Reason : For each eigenvalue, we obtain one eigenvector, and therefore the list of vectors v,.., v k that we obtain has exactly n elements, i.e., k n Is A diagonalizable? Solution. Yes! In fact since A is triangular, its eigenvalues are 3,,-,5. A is 4 4 and has exactly 4 distinct eigenvalues. Therefore A is diagonalizable. 0 / 4

21 5 0 3 Let A 6 3. We know that the eigenvalues of A are and -. Is A diagonalizable? Solution. First we find a basis for each eigenspace : λ A I R R R R 3 R 3 +R R 6 R x x x x : basic; x, x 3 : free; x + x 3 0 x x x 3 0 x x + x 3 0 x 3 x Basis for the eigenspace is, 0 0 / 4

22 5 0 3 Let A 6 3. We know that the eigenvalues of A are and -. Is A diagonalizable? λ A ( )I x x x x, x : basic; x 3 : free; R R R R 3 R 3 +R { x + x 3 0 x + x R 3 R R 3 R x x 3 x x 3 x 3 x 3 x Our list of vectors : v, v 0, v 3 Therefore A is diagonalizable. 0 / 4

23 5 0 3 Let A 6 3. We know that the eigenvalues of A are and -. Is A diagonalizable? Our list of vectors : 0 0 v, v 0, v 3 P 0, P and we have P AP Question. What happens if we change the order of v, v, and v 3? Answer. The order of diagonal entries of D changes. 0 0 For example, if P [v v 3 v ] 0 then P 0 and P AP / 4

24 Summary and final remarks Let A be an n n matrix. Method for calculating A r : If A PDP, then A r PD r P. Having A, how can we find P and D? Find the eigenvalues of A. Find a basis for each eigenspace. 3 make a list of all of the vectors you obtain in this way : v,..., v k If k < n, then A is NOT diagonalizable. If k n, then A is diagonalizable : We take P [ v v v k ] and the diagonal entries of D are the eigenvalues of A with multiplicities. The case k > n is impossible. 4 / 4