MATH 374 Lecture 24 Repeated Eigenvalues

8.6: Repeated Eigenvalues For the problem X’ = AX (1) what happens if some of the eigenvalues of A are repeated?

Repeated Eigenvalues

X’ = AX (1) Theorem 8.9

Example 1

Example 1

Example 1

Example 1

Example 1

Example 1

X’ = AX (1) One Eigenvector Case 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 X1(t) = Cemt and Theorem 8.9 says a second linearly independent solution of the form X2(t) = K21temt + K22emt (5) can be found (here, K21 and K22 are vectors).

X’ = AX (1) One Eigenvector Case X2(t) = K21temt + K22emt (5) We will now figure out what K21 and K22 can be to make (5) a solution of (1). Substituting (5) into (1): (K21temt + K22emt)’ = A(K21temt + K22emt) ) K21emt + K21mtemt + K22memt = A K21temt + AK22emt ) (A-mI)K21temt + [(A-mI)K22 – K21]emt = 0 (6)

One Eigenvector Case (A-mI)K21temt + [(A-mI)K22 – K21]emt = 0 (6) X’ = AX (1) One Eigenvector Case X2(t) = K21temt + K22emt (5) (A-mI)K21temt + [(A-mI)K22 – K21]emt = 0 (6) Note that (6) will hold if: (A-mI)K21 = 0 (7) and (A-mI)K22 = K21 (8) Thus, if we take K21 to be an eigenvector of A corresponding to eigenvalue m, (7) will hold! To find K22, all we need to do is to solve (8)!

(A-mI)K21 = 0 (7) (A-mI)K22 = K21 (8) Example 2

(A-mI)K21 = 0 (7) (A-mI)K22 = K21 (8) Example 2

(A-mI)K21 = 0 (7) (A-mI)K22 = K21 (8) Example 2