MATH 374 Lecture 24 Repeated Eigenvalues 8 6

MATH 374 Lecture 24 Repeated Eigenvalues

8. 6: Repeated Eigenvalues l 2 For the problem X’ = AX (1) what happens if some of the eigenvalues of A are repeated?

Repeated Eigenvalues 3

X’ = AX Theorem 8. 9 4 (1)

Example 1 5

Example 1 6

Example 1 7

Example 1 8

Example 1 9

Example 1 10

X’ = AX (1) One Eigenvector Case l l 11 Let’s consider the case when m is an eigenvalue of A of multiplicity two and there is only one eigenvector C associated with m. Then one solution to (1) is X 1(t) = Cemt and Theorem 8. 9 says a second linearly independent solution of the form X 2(t) = K 21 temt + K 22 emt (5) can be found (here, K 21 and K 22 are vectors).

X’ = AX (1) X 2(t) = K 21 temt + K 22 emt (5) One Eigenvector Case We will now figure out what K 21 and K 22 can be to make (5) a solution of (1). l Substituting (5) into (1): (K 21 temt + K 22 emt)’ = A(K 21 temt + K 22 emt) ) K 21 emt + K 21 mtemt + K 22 memt = A K 21 temt + AK 22 emt ) (A-m. I)K 21 temt + [(A-m. I)K 22 – K 21]emt = 0 (6) l 12

X’ = AX (1) X 2(t) = K 21 temt + K 22 emt (5) One Eigenvector Case l l l 13 (A-m. I)K 21 temt + [(A-m. I)K 22 – K 21]emt = 0 (6) Note that (6) will hold if: (A-m. I)K 21 = 0 (7) and (A-m. I)K 22 = K 21 (8) Thus, if we take K 21 to be an eigenvector of A corresponding to eigenvalue m, (7) will hold! To find K 22, all we need to do is to solve (8)!

(A-m. I)K 21 = 0 (A-m. I)K 22 = K 21 Example 2 14 (7) (8)

(A-m. I)K 21 = 0 (A-m. I)K 22 = K 21 Example 2 15 (7) (8)

(A-m. I)K 21 = 0 (A-m. I)K 22 = K 21 Example 2 16 (7) (8)