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Repeated Eigenvalues in Math - Lecture 24

Explore the concept of repeated eigenvalues in matrix A and its implications in solving X' = AX. Understand how to find the second solution in the case of one eigenvector.

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Repeated Eigenvalues in Math - Lecture 24

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  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. 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)!

  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

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