- #1

randomafk

- 23

- 0

I've been trying to brush up on my linear algebra and ran into this bit of confusion.

I just went through a proof that an operator with distinct eigenvalues forms a basis of linearly independent eigenvectors.

But the proof relied on a one to one mapping of eigenvalues to eigenvectors. Is there any particular reason why for a distinct eigenvalue, there shouldn't be more than one (normalized) eigenvectors that satisfies the eigenvalue definition.

And if so, how do I prove it? I'm not mentally convinced, even if that is the case as the proofs seem to indicate!

Thanks!