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Generalized Eigenvectors and Jordan Form We have seen that an n £ n matrix A is diagonalizable precisely when the dimensions of its eigenspaces sum to n. So if A is not diagonalizable, there is at least one eigenvalue with a geometric multiplicity (dimension of its eigenspace) which is strictly less than its algebraic multiplicity. In this handout, we will discuss how one can make up for this de¯ciency of eigenvectors by ¯nding what are called generalized eigenvectors, which can in turn be used to ¯nd the Jordan form of the matrix A. First consider the following non-diagonalizable system. Example 1. 3 The matrix · ¸ 3 1 A = 0 3 2 has characteristic polynomial (¸ ¡ 3)µ·, so¸¶ it has only one eigenvalue ¸ = 3, and the cor- 1 responding eigenspace is E = span . Since dim(E ) = 1 < 3, the matrix A is not 3 0 3 diagonalizable. Nevertheless, it is still possible to solve the system dy = Ay dt without much di±culty. Writing out the two equations dy 1 = 3y + y dt 1 2 dy 2 = 3y dt 2 3t we see that the second equation has general solution y2 = c1e . Plugging this into the ¯rst equation gives dy 1 = 3y + c e3t dt 1 1 This is a ¯rst order linear equation that can be solved by the method of integrating factors. 3t 3t Its solution is y1 = c1te + c2e . Thus the general solution of the system is · ¸ · ¸ µ · ¸ · ¸¶ · ¸ 3t 3t y1 c1te + c2e 3t 1 0 3t 1 y = = 3t = c1e t + + c2e y2 c2e 0 1 0 3 The matrix A in the previous example is said to be in Jordan form. De¯nition 1. A 2 £ 2 matrix of the form · ¸ ¸ 1 J = 0 ¸ for some number ¸ is called a Jordan form matrix. 1 The following theorem generalizes the calculations of the previous example. Theorem 1. The general solution of the system · ¸ dy ¸ 1 = Jy;J = dt 0 ¸ is µ · ¸ · ¸¶ · ¸ 1 0 1 y = c e¸t t + + c e¸t 1 0 1 2 0 In general, suppose A is a 2£2 matrix with a single repeated eigenvalue ¸ with dim(E¸) = 1. Then, as we have seen before, making the change of variables y = Cx transforms the system dy = Ay dt into the system dx = C¡1ACx: dt We now want to ¯nd a matrix C such that the coe±cient matrix C¡1AC for the new system is a Jordan form matrix J. That is, we want AC = CJ. Writing 2 3 j j · ¸ ¸ 1 C = 4v v 5 ;J = ; 1 2 0 ¸ j j we have 2 3 2 3 j j j j AC = 4Av1 Av25 ;CJ = 4¸v1 ¸v2 + v15 j j j j Therefore the columns of C must satisfy Av1 = ¸v1 Av2 = ¸v2 + v1 Thus the vector v1 is an eigenvector with eigenvalue ¸. Rewriting these equations (A ¡ ¸I)v1 = 0 (A ¡ ¸I)v2 = v1 2 2 it follows that (A ¡ ¸I) v2 = (A ¡ ¸I)v1 = 0. Thus the vector v2 must be in ker(A ¡ ¸I) . 2 De¯nition 2. A nonzero vector v which satis¯es (A¡¸I)pv = 0 for some positive integer p is called a generalized eigenvector of A with eigenvalue ¸. The vectors v1 and v2 form a generalized eigenvector chain, as the following diagram illustrates: v2 ¡! v1 ¡! 0 A¡¸I A¡¸I Therefore, to ¯nd the columns of the matrix C that puts A in Jordan form, we must ¯nd a chain of generalized eigenvectors, as follows: 2 ² Find a nonzero vector v2 in ker(A ¡ ¸I) that is not in ker(A ¡ ¸I). ² Set v1 = (A ¡ ¸I)v2. dy Having found such vectors, we can then write down the solution of the system dt = Ay. dx From above, we know that the solution of the system dt = Jx is µ · ¸ · ¸¶ · ¸ 1 0 1 x = c e¸t t + + c e¸t 1 0 1 2 0 so multiplying this by C we get ¸t ¸t y = Cx = c1e (tv1 + v2) + c2e v1: In summary, we have the following result. Theorem 2. Let A be a 2 £ 2 matrix and suppose v2 ! v1 ! 0 is a chain of generalized dy eigenvectors of A with eigenvalue ¸. Then the general solution of the system dt = Ay is ¸t ¸t y(t) = c1e (tv1 + v2) + c2e v1: Example 2. 3 Let · ¸ 2 1 A = ¡1 4 The characteristic polynomial of A is ¸2 ¡ 6¸ + 9 = (¸ ¡ 3)2, so ¸ = 3 is the only eigenvalue of A. Next, we compute · ¸ · ¸ ¡1 1 0 0 A ¡ 3I = ; (A ¡ 3I)2 = ¡1 1 0 0 2 Now we choose· v¸2 to be any vector in ker(A ¡ 3I) that is not in ker(·A¸¡ 3I). One such 0 1 vector is v = . With this choice, we than have v = (A ¡ 3I)v = . The solution of 2 1 1 2 1 dy the system dt = Ay is therefore µ · ¸ · ¸¶ · ¸ 1 0 1 y = c e3t t + + c e3t : 1 1 1 2 1 3 3 For larger systems it turns out that any chain of generalized eigenvectors of the coe±cient matrix can be used to write down a linearly independent set of solutions. Theorem 3. Suppose fv1; v2;:::; vkg is a chain of generalized eigenvectors of A with eigenvalue ¸. That is, vk ¡! vk¡1 ¡! vk¡2 ¡! ¢ ¢ ¢ ¡! v2 ¡! v1 ¡! 0: A¡¸I A¡¸I A¡¸I A¡¸I A¡¸I A¡¸I Then ¸t y1(t) = e v1 ¸t y2(t) = e (v2 + tv1) · ¸ t2 y (t) = e¸t v + tv + v 3 3 2 2! 1 . · ¸ t2 tk¡2 tk¡1 y (t) = e¸t v + tv + v + ¢ ¢ ¢ + v + v k k k¡1 2! k¡2 (k ¡ 2)! 2 (k ¡ 1)! 1 dy is a linearly independent set of solutions of dt = Ay. Proof. We show that yk is a solution. The proof that the others are solutions is similar. Di®erentiating gives · dy t2 k = e¸t (¸v + v ) + t(¸v + v ) + (¸v + v ) + ¢ ¢ ¢ dt k k¡1 k¡1 k¡2 2! k¡2 k¡3 ¸ tk¡2 tk¡1 + (¸v + v ) + (¸v ) : (k ¡ 2)! 2 1 (k ¡ 1)! 1 Multiplying by A gives · ¸ t2 tk¡2 tk¡1 Ay = e¸t (Av ) + t(Av ) + (Av ) + ¢ ¢ ¢ + (Av ) + (Av ) : k k k¡1 2! k¡2 (k ¡ 2)! 2 (k ¡ 1)! 1 The two expressions above are equal, since Av1 = ¸v1, Av2 = ¸v2 + v1, and so on. Next we show that the chain fv1; v2;:::; vkg is necessarily linearly independent. Suppose c1v1 + c2v2 + ¢ ¢ ¢ + ckvk = 0: (1) k¡1 The multiplying by (A ¡ ¸I) sends v1 through vk to zero, and vk to v1, so we are left with ckv1 = 0. Since v1 is nonzero, this implies ck = 0. Similar reasoning shows that the remaining coe±cients must also be zero. Thus the chain of generalized eigenvectors is linearly independent. Now suppose c1y1(t) + c2y2(t) + ¢ ¢ ¢ + cnyn(t) = 0 for all t. Then since yj(0) = vj, at t = 0 this equation is precisely equation (1) above, which we just showed implies c1 = c2 = ¢ ¢ ¢ = ck = 0. 4 The next theorem guarantees that we can always ¯nd enough solutions of this form to generate a fundamental set of solutions. Theorem 4. Let A be an n£n matrix, and suppose ¸ is an eigenvalue of A with algebraic multiplicity m. Then there is some integer p · m such that dim(ker(A ¡ ¸I)p) = m. Example 3. 3 Let 2 3 5 1 ¡4 A = 44 3 ¡55 3 1 ¡2 3 The characteristic polynomial of A is pA(¸) = (¸ ¡ 2) . Since 2 3 2 3 3 1 ¡4 1 0 ¡1 A ¡ 2I = 44 1 ¡55 ¡! 40 1 ¡15 ; 3 1 ¡4 rref 0 0 0 the eigenspace E2 = ker(A ¡ 2I) has dimension 1. Next, since 2 3 2 3 1 0 ¡1 1 0 ¡1 (A ¡ 2I)2 = 41 0 ¡15 ¡! 40 0 0 5 ; 1 0 ¡1 rref 0 0 0 the space ker(A ¡ 2I)2 has dimension 2. Finally, since 2 3 0 0 0 (A ¡ 2I)3 = 40 0 05 ; 0 0 0 the space ker(A ¡ 2I)3 has dimension 3, which matches the algebraic multiplicity of the eigenvalue ¸ = 2. We can now form a chain of 3 generalized eigenvectors by choosing a vector v3 in ker(A¡ 3 2 2I) and de¯ning v2 = (A ¡ 2I)v3 and v1 = (A ¡ 2I)v2 = (A ¡ 2I) v3. To ensure that v2 2 and v1 are both non-zero, we need v3 to not be in ker(A ¡ 2I) (which in turn implies that 3 3 v3 is not in ker(A ¡ 2I)). Since ker(A ¡ 2I) = R , we can choose v3 to be any vector not in ker(A ¡ 2I)2. Since the ¯rst column of (A ¡ 2I)2 is non-zero, we may choose 2 3 2 3 2 3 1 3 1 v3 = 405 =) v2 = (A ¡ 2I)v3 = 445 =) v1 = (A ¡ 2I)v2 = 415 : 0 3 1 5 Then 2 3 1 2t y1(t) = e 415 1 0 2 3 2 31 1 3 2t y2(t) = e @t 415 + 445A 1 3 0 2 3 2 3 2 31 1 3 1 t2 y (t) = e2t @ 415 + t 445 + 405A 3 2! 1 3 0 dy is a fundamental set of solutions of dt = Ay. 3 There may in general be more than one chain of generalized eigenvectors corresponding to a given eigenvalue. Since the last vector in each chain is an eigenvector, the number of chains corresponding to an eigenvalue ¸ is equal to the dimension of the eigenspace E¸.
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