Diagonalizability Consider the constant-coe±cient linear system y0 = Ay (1) where A is some n£n matrix. In general, this problem is not easy to solve since the equations in the system are usually coupled. · ¸ 2 1 Example 1. 3 If A = , the system y0 = Ay is 4 ¡1 0 y1 = 2y1 + y2 0 y2 = 4y1 ¡ y2 Because both y1 and y2 appear in both equations, we cannot solve these equations indepen- dently. 3 In some cases, however, the equations in the system are uncoupled. · ¸ 3 0 Example 2. 3 If A = , the system y0 = Ay is 0 ¡2 0 y1 = 3y1 0 y2 = ¡2y2: Since the ¯rst equation only contains y1 and the second equation only contains y2, we may solve the two equations independently of one another. The solution is 3t y1 = c1e ¡2t y2 = c2e or in vector form · ¸ · ¸ · ¸ 3t c1e 3t 1 ¡2t 0 y = ¡2t = c1e + c2e : c2e 0 1 3 The matrix A in this example is called a diagonal matrix, since its entries o® the main diagonal are all zero. In general, a linear system is uncoupled if and only if its coe±cient matrix is a diagonal matrix. Now suppose we are given a system y0 = Ay which is coupled. Is it possible to make a substitution which transforms the system into an uncoupled system? To answer this question, let's consider a substitution of the form y = Cx; 1 where C is some n £ n invertible matrix. This de¯nes a change of coordinates. The old variable y will be replaced by the new variable x. To see what a®ect this has on the system y0 = Ay, we substitute the change of coordinates into both sides. On the left side we have y0 = Cx0, and on the right side Ay = ACx, so Cx0 = ACx. Multiplying both sides by the inverse of C gives x0 = C¡1ACx: Thus the original system in y with coe±cient matrix A has been replaced by the new system in x with coe±cient matrix C¡1AC. In order for this new system to be uncoupled, we need this coe±cient matrix be a diagonal matrix. That is, we want C¡1AC = D for some diagonal matrix D. De¯nition 1. We say that a matrix A is diagonalizable if there exists and invertible matrix C and a diagonal matrix D such that C¡1AC = D. So a system y0 = Ay can be uncoupled if and only if A is a diagonalizable matrix. This conclusion leads to the following questions. ² Which matrices are diagonalizable? ² For those matrices A which are diagonalizable, how do we ¯nd C and D? To answer these questions, suppose that we have found matrices 2 ¯ ¯ ¯ 3 ¯ ¯ ¯ 2 3 ¯ ¯ ¯ ¸ 0 ¢ ¢ ¢ 0 6 ¯ ¯ ¯ 7 1 6 7 6 0 ¸ ¢ ¢ ¢ 0 7 6 7 6 2 7 C = 6v¯1 v¯2 ¢ ¢ ¢ v¯n7 and D = 6 . .. 7 4 ¯ ¯ ¯ 5 4 . 5 ¯ ¯ ¯ ¯ ¯ ¯ 0 0 ¢ ¢ ¢ ¸n such that C¡1AC = D. Then multiplying both sides by C gives AC = CD. But 2 ¯ ¯ ¯ 3 2 ¯ ¯ ¯ 3 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 6 ¯ ¯ ¯ 7 6 ¯ ¯ ¯ 7 6 7 6 7 6 7 6 7 AC = 6A¯v1 A¯v2 ¢ ¢ ¢ A¯vn7 and CD = 6¸1¯v1 ¸2¯v2 ¢ ¢ ¢ ¸n¯vn7 ; (2) 4 ¯ ¯ ¯ 5 4 ¯ ¯ ¯ 5 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ so if we equate the columns of these matrices we ¯nd that Av1 = ¸1v1 Av2 = ¸2v2 . Avn = ¸nvn: Motivated by this, we make the following de¯nition. 2 De¯nition 2. A nonzero vector v such that Av = ¸v for some scalar ¸ is called an eigenvector of A with eigenvalue ¸. Thus in order for A to be diagonalized by the matrices C and D, the columns of C must be eigenvectors whose eigenvalues are the diagonal entries of D. Since C is invertible, its columns must be linearly independent. Thus we have shown that if A is diagonalizable, then there exist n linearly independent eigenvectors of A. But conversely, if there exist n linearly independent eigenvectors of A, then placing these vectors into the columns of a matrix C, and placing their eigenvalues into the diagonal matrix D, it follows by the same calculations above that AC = CD, so C¡1AC = D and A is diagonalizable. Theorem 1. An n£n matrix A is diagonalizable if and only if there there exist n linearly independent eigenvectors of A. · ¸ · ¸ 2 1 1 Example 3. 3 Let A = and de¯ne v = . Then 4 ¡1 1 1 · ¸ · ¸ · ¸ 2 1 1 3 Av = = = 3v ; 1 4 ¡1 1 3 1 · ¸ ¡1 so v is an eigenvector with eigenvalue 3. Next, let v = . Then 1 2 4 · ¸ · ¸ · ¸ 2 1 ¡1 2 Av = = = ¡2v ; 2 4 ¡1 4 ¡8 2 so v2 is an eigenvector· ¸ with eigenvalue ¡2. Since v1 and v2 are linearly independent, the 1 ¡1 matrix C = is invertible, and by the reasoning above, 1 4 · ¸ 3 0 C¡1AC = D = : (Check this yourself!) 0 ¡2 Thus we have found matrices C and D which diagonalize A. Therefore, if we now make the change of variable y = Cx, then the system y0 = Ay is transformed into the uncoupled system x0 = Dx: 0 x1 = 3x1 0 x2 = ¡2x2: 3 3t ¡2t The general solution is x1 = c1e , x2 = c2e , or · ¸ 3t c1e x = ¡2t : c2e Therefore the general solution of the system y0 = Ay is · ¸ · ¸ · ¸ · ¸ 3t 1 ¡1 c1e 3t 1 ¡2t ¡1 y = Cx = ¡2t = c1e + c2e : 1 4 c2e 1 4 3 The last statement in this example can be generalized as follows. Theorem 2. If fv1; v2;:::; vng are linearly independent eigenvectors of A with eigenval- 0 ues ¸1; ¸2; : : : ; ¸n, respectively, then the general solution of y = Ay is ¸1t ¸2t ¸nt y = c1e v1 + c2e v2 + ¢ ¢ ¢ + cne vn: Proof. Let C be the matrix with columns v1 through vn and D the diagonal matrix with diagonal entries ¸1 through ¸n. Then using equations (2), we have AC = CD and therefore C¡1AC = D. So, as above, the substitution y = Cx transforms y0 = Ay into x0 = Dx. The solution of this uncoupled system is 2 3 ¸1t c1e 6 ¸2t 7 6c2e 7 x = 6 . 7 4 . 5 ¸nt cne so ¸1t ¸2t ¸nt y = Cx = c1e v1 + c2e v2 + ¢ ¢ ¢ + cne vn: We now focus on the task of ¯nding the eigenvectors and eigenvalues of a matrix. We begin with the eigenvalues. Finding Eigenvalues Suppose that v is an eigenvector of A with eigenvalue ¸. Then Av = ¸v: We can rewrite this equation as Av ¡ ¸v = 0; or, since Iv = v for any vector v, Av ¡ ¸Iv = 0: 4 This may be written (A ¡ ¸I)v = 0; which means that v 2 N(A ¡ ¸I): Since eigenvectors are nonzero by de¯nition, this implies that the null space of A ¡ ¸I is non-trivial. This in turn implies that A ¡ ¸I is not invertible, and therefore det(A ¡ ¸I) = 0: Each step in this chain of implications can be reversed, so we have the following test for eigenvalues. Theorem 3. Let A be an n £ n matrix. Then ¸ is an eigenvalue of A if and only if det(A ¡ ¸I) = 0. Example 4. 3 Let · ¸ 1 2 A = : 4 3 Then · ¸ · ¸ · ¸ 1 2 ¸ 0 1 ¡ ¸ 2 A ¡ ¸I = ¡ = 4 3 0 ¸ 4 3 ¡ ¸ so det(A ¡ ¸I) = (1 ¡ ¸)(3 ¡ ¸) ¡ 8 = ¸2 ¡ 4¸ ¡ 5 = (¸ ¡ 5)(¸ + 1): This expression is zero when ¸ = 5 or ¸ = ¡1, so these are the only eigenvalues of A. 3 The expression pA(¸) = det(A ¡ ¸I) is in general a polynomial of degree n, called the characteristic polynomial of A. The eigenvalues of A are the roots of this polynomial. Next let's turn to the task of ¯nding the eigenvectors. Finding Eigenvectors Once an eigenvalue ¸ of a matrix A is known, the associated eigenvectors are the nonzero solutions of Av = ¸v. Equivalently, they are the nonzero elements of the null space of A ¡ ¸I. We call E¸ = N(A ¡ ¸I) the eigenspace associated with the eigenvalue ¸. Every nonzero vector in E¸ is an eigen- vector with eigenvalue ¸. 5 · ¸ 1 2 Example 5. 3 Let A = . In the previous example we found that the eigenvalues of 4 3 A were 5 and ¡1. For ¸ = 5, · ¸ · ¸ µ· ¸¶ 1 ¡4 2 1 ¡ 2 1 A ¡ 5I = ¡! ; =) E5 = N(A ¡ 5I) = span : 4 ¡2 rref 0 0 2 | {z } v1 For ¸ = ¡1, · ¸ · ¸ µ· ¸¶ 2 2 1 1 ¡1 A ¡ (¡1)I = ¡! ; =) E¡1 = N(A ¡ (¡1)I) = span : 4 4 rref 0 0 1 | {z } v2 Thus v1 and all its nonzero multiples are eigenvectors with eigenvalue 5, and v2 and all of its nonzero multiples are eigenvectors· ¸ with eigenvalue ¡1. Since v1 and v2 are linearly 1 ¡1 independent, the matrix C = is invertible, and we have 2 1 · ¸ 5 0 C¡1AC = D = ; (Check!) 0 ¡1 so the matrix A is diagonalizable. 3 Here is an example of a non-diagonalizable matrix. · ¸ 2 1 Example 6. 3 Let A = . The characteristic polynomial of A is p (¸) = ¸2 ¡ 6¸ + ¡1 4 A 9 = (¸ ¡ 3)2, so ¸ = 3 is the only eigenvalue of A. Since · ¸ · ¸ µ· ¸¶ ¡1 1 1 ¡1 1 A ¡ 3I = ¡! ; =) E3 = N(A ¡ 3I) = span ; ¡1 1 rref 0 0 1 | {z } v the only eigenvectors of A are multiples of v. Thus there do not exist two linearly independent eigenvectors of A, so by Theorem 1, A is not diagonalizable. 3 Part of the problem is this example was that the 2 £ 2 matrix A only had one eigenvalue. That is, the number of di®erent eigenvalues of A was less than the size of A. It turns out that if we avoid this situation, then we are guaranteed diagonalizability.