1. MATH 374 Lecture 24
Repeated Eigenvalues

2. 8.6: Repeated Eigenvalues
For the problem
X’ = AX (1)
what happens if some of the eigenvalues of A are repeated?

3. Repeated Eigenvalues

4. X’ = AX (1)
Theorem 8.9

5. Example 1

6. Example 1

7. Example 1

8. Example 1

9. Example 1

10. Example 1

11. 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).

12. 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)

13. 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 = (7)
and
(A-mI)K22 = K (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)!

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

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

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