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 : 3 0 3 4 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
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
4 0 Find the eigenvalues of A 3 5. 6 0 Solution. First we find the characteristic equation : 4 λ 0 det(a λi) 3 λ 5 6 0 λ 3 5 0 6 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
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 3+5. 4 / 4
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 0 0 0 x 0 0 x 0 x x x Which means that the eigenspace corresponding to λ is one-dimensional. 5 / 4 x
3 Let A 0 0. Calculate the eigenvalues of A and their 5 5 4 multiplicities. 3 λ 0 0 λ 3 A λi 0 0 0 λ 0 0 λ 0. 5 5 4 0 0 λ 5 5 4 λ We use the cofactor expansion across the second row : λ 3 0 λ 0 3 0 5 5 4 λ 5 4 λ + ( λ) λ 5 4 λ 0 λ 3 5 5 ( 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
Recall : diagonal matrices An n n matrix is called a diagonal matrix if every entry outside the diagonal of the matrix is zero.. 0 0 0 0 3 0 0 0 0 0 0 0 0 0 is a diagonal matrix. 0 0 0 0 0 0 But is NOT a diagonal matrix. 0 0 0 0 0 0 7 / 4
Powers of matrices - the diagonal case 0 0. Let A 0 0. Calculate A0. 0 0 Solution. It is easily seen that 0 0 0 0 () 0 0 A A A 0 0 0 0 0 0 0 0 0 0 0 0 ( ) Similarly () 3 0 0 A 3 A A A 0 3 0 0 0 ( )3 We have : () 0 0 0 0 0 A 0 0 0 0 0 048576 0 0 0 ( )0 0 0 048576 The above example shows that we can easily compute any power of a diagonal matrix. 8 / 4
Powers of matrices - the general case 5 4 Let A. Calculate A 8 7 5. Solution. Suppose we can find an invertible matrix P such that P AP is diagonal. For example, consider P P and 5 4 0 P AP D 8 7 0 3 Then and P AP D A PDP A 5 (PDP )(PDP )(PDP )(PDP )(PDP ) PD 5 P 9 / 4
A 5 (PDP )(PDP )(PDP )(PDP )(PDP ) PD 5 P 0 0 0 0 0 A 5 0 3 0 3 0 3 0 3 0 3 5 0 0 0 ( 3) 5 0 43 43 45 44 486 488 487 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
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
How to find D and P? 7 9 Is A diagonalizable? If it is, use diagonalization to evaluate A 6 8 0. 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
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 6 8 0 λ 6 8 λ 7 λ 9 6 8 λ ( 7 λ)(8 λ) ( 9)6 λ λ (λ + )(λ ) λ () ± (4)( ) { 3 / 4
(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. 6 9 6 9 [ 6 9 0 6 9 0 ] Row reduction 6 9 x 6 9 [ 3 0 0 0 0 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
(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 9 9 6 6 [ 9 9 6 6 ] x x 0 0 [ 9 9 0 6 6 0 ] Row reduction x : basic; x : free. {x x x x x x x [ 0 0 0 0 and a basis for the eigenspace corresponding to λ is ] { } 5 / 4
(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 0 3 6 8 0 6 / 4
(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 0 0 0 A 0 P P 0 3 () 0 0 0 0 3 3 0 045 3069 0 04 3 046 3070 7 / 4
Another example 3 3 Let A 0. Is A diagonalizable? 0 0 3 Solution. The matrix is triangular and its eigenvalues are 3 and -. We find a basis for each eigenspaces : λ 3 3 0 0 4 3 A ()I 0 0 0 0 0 0 0 3 0 0 0 0 4 4 3 0 0 0 0 0 0 4 0 x x x 3 3 x 4 x 0 Row Reduction 3 4 0 0 0 0 0 0 0 0 0 4 3 x 0 0 0 x 0 0 0 4 x 3 0 { x 3 4 x 0 x 3 0 3 3 x 4 4 Basis for eigenspace : 0 0. 8 / 4
Another example (cont.) 3 3 Let A 0. Is A diagonalizable? 0 0 3 λ 3 3 3 3 0 0 0 3 A 3I 0 0 3 0 0 4 0 0 3 0 0 3 0 0 0 0 3 0 0 4 0 0 0 0 0 Row Reduction 0 0 0 0 0 0 0 0 0 0 0 3 x 0 0 4 x 0 0 0 0 x 3 0 { x 0 x 3 0 x x x 0 x 0 Basis for eigenspace : 0 x 3 0 0 0. 3 4 We now list all of the vectors we have obtained : v, v 0 0 0 We only have vectors, but A is 3 3. Therefore A is NOT diagonalizable. 9 / 4
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. 3 0 0 Is A diagonalizable? 0 0 5 0 0 0 5 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
5 0 3 Let A 6 3. We know that the eigenvalues of A are and -. Is A 6 0 4 diagonalizable? Solution. First we find a basis for each eigenspace : λ 5 0 3 0 0 6 0 3 A I 6 3 0 0 6 0 3 6 0 4 0 0 6 0 3 6 0 3 0 6 0 3 0 6 0 3 0 R R R R 3 R 3 +R 6 0 3 0 0 0 0 0 0 0 0 0 6 0 3 6 0 3 6 0 3 R 6 R x x x 3 0 0 0 0 0 0 0 0 0 0 0 0 0 x : basic; x, x 3 : free; x + x 3 0 x x x 3 0 x x + x 3 0 x 3 x 3 0 0 Basis for the eigenspace is, 0 0 / 4
5 0 3 Let A 6 3. We know that the eigenvalues of A are and -. Is A 6 0 4 diagonalizable? λ 5 0 3 0 0 3 0 3 A ( )I 6 3 0 0 6 3 3 6 0 4 0 0 6 0 6 3 0 3 x 0 6 3 3 x 0 6 0 6 0 x 3 3 0 3 0 6 3 3 0 6 0 6 0 x, x : basic; x 3 : free; R R R R 3 R 3 +R { x + x 3 0 x + x 3 0 3 0 3 0 0 3 3 0 0 0 0 0 R 3 R R 3 R x x 3 x x 3 x 3 x 3 x 3 0 0 0 0 0 0 0 0 0 Our list of vectors : v, v 0, v 3 Therefore A is diagonalizable. 0 / 4
5 0 3 Let A 6 3. We know that the eigenvalues of A are and -. Is A 6 0 4 diagonalizable? Our list of vectors : 0 0 v, v 0, v 3 P 0, P 0 0 0 0 and we have 5 0 3 0 0 0 P AP 0 6 3 0 0 0 0 6 0 4 0 0 0 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 0 0 5 0 3 0 0 0 P AP 0 6 3 0 0 0 6 0 4 0 0 0 3 / 4
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