MATH 20D Lecture 18

Case 5a. $A$ has repeated eigenvalue $\lambda >0$ and one independent eigenvector

• Origin is an improper source/degenerate source
• Unstable b/c only eigenvalue we have is

Case 5b. $A$ has repeated eigenvalue $\lambda <0$ and one independent eigenvector

• Origin is an improper source/degenerate source
• Stable b/c only eigenvalue we have is

Fundamental matrix

Systems with Repeated Eigenvalues

Let with repeated eigenvalue .
Then the general solution to the system is given by

Let with repeated eigenvalue and eigenvector . Find the general solution to the system

• We can still obtain one solution
• Next, we generate another solution based on our current eigenvalue/eigenvector
  WARNING: The second solution won’t be as nice as as we saw earlier

Generalized Eigenvector of $A$ with rank $k$

Let be an matrix and suppose that is an eigenvalue of . We say that a nonzero vector is a generalized eigenvector of with rank (corresponding to ) provided that for all integers .

Jordan Chain of Independent Vectors

If is a generalized eigenvector of with rank then there exists a Jordan Chain of independent vectors (corresponding to ) such that





Note: An eigenvector is a generalized eigenvector of rank 1

Fact

If is a Jordan chain of corresponding to then






are all solutions to

Basically, if is a with repeated eigenvalue and only one eigenvector , then the second solution is given by
where is the solution to

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Summary

Consider a system of two first-order ODEs


The equivalent matrix form (normal form) of this system is given by

There are three cases:

1. has two distinct, real eigenvalues
   Let be the eigenvalues, and let be the corresponding eigenvector. Then the general solution to is given by
   
2. has complex eigenvalues
   Here, the eigenvalues and eigenvectors must occur as conjugate pairs. So we let denote the eigenvalues and denote the corresponding eigenvectors. Then the general solution to is given by
   
3. A has repeated eigenvalue There are two sub-cases:
   3a. If has two linearly independent eigenvectors . Then the general solution to is given by
   
   3b. If has only one eigenvector then we need to solve the system of algebraic equations
   
   for any solution . Then the general solution to is