Originally Posted by
Failure What do you mean by saying that a "non-zero vector x corresponds to an eigenvalue $\displaystyle \lambda$ of A"? Du you mean that $\displaystyle Ax=\lambda x$ holds (i.e. that x is an eigenvector of A for the eigenvalue $\displaystyle \lambda$)?
And is what you wan't to show that, from x being non-zero and that $\displaystyle Ax = \lambda_1 x$ and $\displaystyle Ax=\lambda_2 x$, it follows that $\displaystyle \lambda_1=\lambda_2$?
If so, this has got nothing to do with multiplicities of those eigenvalues:
$\displaystyle Ax=\lambda_1 x=\lambda_2x\Leftrightarrow (\lambda_1-\lambda_2)x=0\Leftrightarrow \lambda_1-\lambda_2=0\Leftrightarrow \lambda_1=\lambda_2$.