1. ## [SOLVED] Eigenvalues

A nonzero vector can't correspond to two different eigenvalues of A.

True, can only correspond if the eigenvalue is of multiplicity > 1.

Correct?

2. Originally Posted by dwsmith
A nonzero vector can't correspond to two different eigenvalues of A.

True, can only correspond if the eigenvalue is of multiplicity > 1.

Correct?
What do you mean by saying that a "non-zero vector x corresponds to an eigenvalue $\lambda$ of A"? Du you mean that $Ax=\lambda x$ holds (i.e. that x is an eigenvector of A for the eigenvalue $\lambda$)?

And is what you wan't to show that, from x being non-zero and that $Ax = \lambda_1 x$ and $Ax=\lambda_2 x$, it follows that $\lambda_1=\lambda_2$?
If so, this has got nothing to do with multiplicities of those eigenvalues:
$Ax=\lambda_1 x=\lambda_2x\Leftrightarrow (\lambda_1-\lambda_2)x=0\Leftrightarrow \lambda_1-\lambda_2=0\Leftrightarrow \lambda_1=\lambda_2$.

3. Originally Posted by Failure
What do you mean by saying that a "non-zero vector x corresponds to an eigenvalue $\lambda$ of A"? Du you mean that $Ax=\lambda x$ holds (i.e. that x is an eigenvector of A for the eigenvalue $\lambda$)?

And is what you wan't to show that, from x being non-zero and that $Ax = \lambda_1 x$ and $Ax=\lambda_2 x$, it follows that $\lambda_1=\lambda_2$?
If so, this has got nothing to do with multiplicities of those eigenvalues:
$Ax=\lambda_1 x=\lambda_2x\Leftrightarrow (\lambda_1-\lambda_2)x=0\Leftrightarrow \lambda_1-\lambda_2=0\Leftrightarrow \lambda_1=\lambda_2$.

I mean exactly what it says. These questions from test I have taking this semester and I am just redoing them for the final. You getting all the information verbatim as it appears.

4. Did you understand what Failure (he really needs to change that!) is saying? He showed you exactly how to answer your question.