Lecture 11: Eigenvalues and Eigenvectors

Transcription

1 Lecture : Eigenvalues and Eigenvectors De nition.. Let A be a square matrix (or linear transformation). A number λ is called an eigenvalue of A if there exists a non-zero vector u such that A u λ u. () In the above de nition, the vector u is called an eigenvector associated with this eigenvalue λ. The set of all eigenvectors associated with λforms a subspace, and is called the eigenspace associated with λ. Geometrically, if we view any n n matrix A as a linear transformation T. Then the fact that u is an eigenvector associated with an eigenvalue λ means u is an invariant direction under T. In other words, the linear transformation T does not change the direction of u : u and T u either have the same direction (λ > ) or opposite direction (λ< ). The eigenvalue is the factor of contraction (jλj < ) or extension (jλj > ). Remarks. () u 6 is crucial, since u always satis es Equ (). () If u is an eigenvector for λ, then so is c u for any constant c. () Geometrically, in D, eigenvectors of A are directions that are unchanged under linear transformation A. We observe from Equ () that λ is an eigenvalue i Equ () has a non-trivial solution. Since Equ () can be written as (A λi) u A u λ u, () it follows λ is an eigenvalue i Equ () has a non-trivial solution. By the inverse matrix theorem, Equ () has a non-trivial solution i det (A λi). () We conclude that λ is an eigenvalue i Equ () holds. We call Equ () "Characteristic Equation" of A. The eigenspace, the subspace of all eigenvectors associated with λ, is eigenspace Null (A λi). ² Finding eigenvalues and all independent eigenvectors: Step. Solve Characteristic Equ () for λ. Step. For each λ, nd a basis for the eigenspace Null (A λi) (i.e., solution set of Equ ()). Example.. Find all eigenvalues and their eigenspace for A. Solution: A λi λ λ λ. λ λ

2 The characteristic equation is We nd eigenvalues det (A λi) ( λ) ( λ) ( ), λ λ+, (λ ) (λ ). λ, λ. We next nd eigenvectors associated with each eigenvalue. For λ, x x (A λ I) x, x x or x x. The parametric vector form of solution set for (A λ I) x : x x x x x x. basis of Null (A I) :. This is only (linearly independent) eigenvector for λ. The last step can be done slightly di erently as follows. From solutions (for (A λ I) x ) x x, we know there is only one free variable x. Therefore, there is only one vector in any basis. To nd it, we take x to be any nonzero number, for instance, x, and compute x x. We obtain x λ, u. x For λ, we nd or (A λ I) x x x x x. x, x To nd a basis, we take x. Then x, and a pair of eigenvalue and eigenvector λ, u.

3 Example.. Given that is an eigenvalue for A Find a basis of its eigenspace. Solution: A I ! Therefore, (A I) x becomes x x + 6x, or x x + 6x, (4) where we select x and x as free variables only to avoid fractions. Solution set in parametric form is x 4 x x x 5 4x + 6x 5 x x A basis for the eigenspace: x x u 45 and u 465. Another way of nding a basis for Null (A λi) Null (A I) may be a little easier. From Equ (4), we know that x an x are free variables. Choosing (x, x ) (, ) and (, ), respectively, we nd x, x ) x ) u 4 5 x, x ) x 6 ) u Example.. Find eigenvalues: (a) 6 λ 6 A 4 65, A λi 4 λ 6 5. λ det (A λi) ( λ) ( λ) ( λ)

4 The eigenvalues are,,, exactly the diagonal elements. (b) B 4 4 5, B λi 4 4 λ λ λ det (B λi) (4 λ) ( λ). The eigenvalues are 4,, 4 (4 is a double root), exactly the diagonal elements. Theorem.. () The eigenvalues of a triangle matrix are its diagonal elements. () Eigenvectors for di erent eigenvalues are linearly independent. More precisely, suppose that λ, λ,..., λ p are p di erent eigenvalues of a matrix A. Then, the set consisting of a basis of Null (A λ I), a basis of Null (A λ I),..., a basis of Null (A λ p I) is linearly independent. Proof. () For simplicity, we assume p : λ 6 λ are two di erent eigenvalues. Suppose that u is an eigenvector of λ and u is an eigenvector of λ To show independence, we need to show that the only solution to x u + x u is x x. Indeed, if x 6, then We now apply A to the above equation. It leads to u x x u. (5) A u x x A u ) λ u x x λ u. (6) Equ (5) and Equ (6) are contradictory to each other: by Equ (5), or Equ (5) ) λ u x x λ u Equ (6) ) λ u x x λ u, x x λ u λ u x x λ u. Therefor λ λ, a contradiction to the assumption that they are di erent eigenvalues. ² Characteristic Polynomials 4

5 We know that the key to nd eigenvalues and eigenvectors is to solve the Characteristic Equation () det (A λi). For matrix, So a λ b A λi. c d λ det (A λi) (a λ) (d λ) bc λ + ( a d) λ+ (ad bc) is a quadratic function (i.e., a polynomial of degree ). In general, for any n n matrix A, a λ a a n A λi 6 a a λ a n 4 5. a n a n a nn λ We may expand the determinant along the rst row to get det (A λi) (a λ) det 4 a λ a n a n a nn λ By induction, we see that det (A λi) is a polynomial of degree n.we called this polynomial the characteristic polynomial of A. Consequently, there are total of n (the number of rows in the matrix A) eigenvalues (real or complex, after taking account for multiplicity). Finding roots for higher order polynomials may be very challenging. Example.4. Find all eigenvalues for Solution: A λi 5 6 A λ 6 6 λ λ 4 5, λ λ 8 det (A λi) (5 λ) det 4 5 λ 4 5 λ 5 λ 4 (5 λ) ( λ) det λ (5 λ) ( λ) [(5 λ) ( λ) 4]. 5

6 There are 4 roots: (5 λ) ) λ 5 ( λ) ) λ (5 λ) ( λ) 4 ) λ 6λ+ ) λ 6 p 6 4 p. We know that we can computer determinants using elementary row operations. One may ask: Can we use elementary row operations to nd eigenvalues? More speci cally, we have Question: Suppose that B is obtained from A by elementary row operations. Do A and B has the same eigenvalues? (Ans: No) Example.5. R A +R!R! B A has eigenvalues and. For B, the characteristic equation is λ det (B λi) λ The eigenvalues are ( λ) ( λ) λ 4λ+. λ 4 p p 8 p. This example show that row operation may completely change eigenvalues. De nition.. Two n n matrices A and B are called similar, and is denoted as A» B, if there exists an invertible matrix P such that A P BP. Theorem.. If A and B are similar, then they have exact the same characteristic polynomial and consequently the same eigenvalues. Indeed, if A P BP, then P (B λi) P P BP λp IP (A λi). Therefore, det (A λi) det P (B λi) P det (P ) det (B λi) det P det (B λi). Example.6. Find eigenvalues of A if 5 6 A» B Solution: Eigenvalues of B are λ 5,, 5, 4. These are also the eigenvalues of A. 6

7 Caution: If A» B, and if λ is an eigenvalue for A and B, then an corresponding eigenvector for A may not be an eigenvector for B. In other words, two similar matrices A and B have the same eigenvalues but di erent eigenvectors. Example.. Though row operation alone will not preserve eigenvalues, a pair of row and column operation do maintain similarity. We rst observe that if P is a type elementary matrix (row replacement), ar P +R!R Ã, a then its inverse P is a type (column) elementary matrix obtained from the identity matrix by an elementary column operation that is of the same kind with "opposite sign" to the previous row operation, i.e., P C ac!c Ã. a We call the column operation is "inverse" to the row operation C ac! C R + ar! R. Now we perform a row operation on A followed immediately by the column operation inverse to the row operation R A +R!R! (left multiply by P ) C C!C! B (right multiply by P.) We can verify that A and B are similar through P (with a ) P AP. Now, λ is an eigenvalue. Then, (A ) u ) u is an eigenvector for A.

8 But In fact, (B ) u ) u is NOT an eigenvector for B. (B ) v. So, v is an eigenvector for B. This example shows that. Row operation alone will not preserve eigenvalues.. Two similar matrices share the same characteristics polynomial and same eigenvalues. But they have di erent eigenvectors. ² Homework #.. Find eigenvalues if 8 (a) A» (b) B» Find eigenvalues and a basis of each eigenspace. 4 (a) A. 9 (b) B

9 . Find a basis of the eigenspace associated with eigenvalue λ for 4 A Determine true or false. Reason your answers. (a) If A x λ x, then λ is an eigenvalue of A. (b) A is invertible i is not an eigenvalue. (c) If A» B, then A and B has the same eigenvalues and eigenspaces. (d) If A and B have the same eigenvalues, then they have the same characteristic polynomial. (e) If det A det B, then A is similar to B. 9