Question

When a 2x2 matrix has a single, repeated eigenvalue, how is it possible to get 2 linearly independent eigenvectors from only that one eigenvalue? Can someone give me an example of a matrix that has a repeated eigenvalues with linearly independent eigenvectors? This question arose because in the section on solving systems of 3 differential equations, this phenomenon was discussed, but there are no examples of it in the notes.

Answer 1

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Answer 2

Okay: When you have repeat e-values you (typically) can only get two linearly DEPENDENT vectors. However, there is a trick to get around it. You take the first e-vector. And using the equation.
\[Au=\lambda u\]
But you alter it a little.
You take your first e-vector (call it v) and set it equal to A (your constant matrix) multiplied with the vector you are trying to find (call it w). So you have:
\[Aw=v\]
Then you can solve your new, linearly independent, e-vector.

Answer 3

Try example one on this if that doesn't help:
http://tutorial.math.lamar.edu/Classes/DE/RepeatedEigenvalues.aspx

Answer 4

Sorry, my question wasn't clear. I already understand how to solve systems of ODEs with repeated eigenvalues; what I am looking for is the atypical case in which you do end up getting two eigenvectors from one repeated eigenvalue. You say that "When you have repeat e-values you (typically) can only get two linearly DEPENDENT vectors." Which would imply that SOMETIMES you would get two linearly INDEPENDENT vectors. When would this happen?

Answer 5

The identity matrix for example. A diagonalized matrix. Something with coefficients on the diagonal and zeros everywhere else. That would produce a single eigenvalue, but two (or however many) L.I. e-vectors.

Answer 6

Maybe this is a better link: http://www.jirka.org/diffyqs/htmlver/diffyqsse22.html

Answer 7

Thanks, that was helpful.