Re: Are eigenvectors independent

Luke Wu wrote:

I know about (and have seen proof of) the fact that eigenvectors
corresponding to distinct eigenvalues form a lin. independent seet.
(**)

But what about this type of situation:

Say a matrix A has 5 eigenvalues (where one of them as multiplicity 2
as a root of characteristic equation of A).

We are able to get 2 independent eigenvectors for this multiplicity 2
eigenvalue.

Not necessarily. For instance, take A equal to

1 1

0 1

Then 1 is a root of the characteristic polynomial with multiplicity 2,
but you can't find two linearly independent eigenvectors with eigenvalue
1.

Is it true that these 2 eigenvectors along with the other 4 (each
related to a multiplicity 1 root) make an independent set?

*If* these two eigenvectors exist, then, yes, it is true, as Jannick
Asmus has already explained.

Best regards,

Jose Carlos Santos
.

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