similar matrices have same eigenvalue… stuck

So $Ax=\lambda x$ where $A$ is the matrix, $\lambda$ and $x$ its eigenvalue and eigenvector resptectively.

Now I have another matrix similar to $A$ i.e. $B=TAT^{-1}$

I'll let a vector $y$ which is an eigenvector of $B$, i.e. $By=\mu y$ where $\mu$ is its ($B$'s) eigenvalue.

I have to now prove that $By=\mu x$.

$By = TAT^{-1}y$ but here I'm stuck. This answers wants me to assume $y=Tx$ which will essentially solve it but why/how is $y=Tx$?

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If you already have $\mathbf T\mathbf A\mathbf T^{-1}\mathbf y=\mu\mathbf y$, multiply both sides by $\mathbf T^{-1}$, remembering that $\mu$ is scalar, and note the correspondence with $\mathbf A\mathbf x=\lambda \mathbf x$... –  Ｊ. Ｍ. Feb 18 '12 at 12:50
You need to assume something more than what you have so far. A matrix may have several distinct eigenvalues and eigenvectors. So for example you may have $Ax = \lambda x$ and $Av = \eta v$, but there is no reason to suppose that $Ax = \eta v$; it isn't generally true. So you need to consider that in the context of your question. –  yasmar Feb 18 '12 at 12:50
You want to prove $By=\lambda y$, not $\lambda y$ (typo, fourth line). And you're there: $By=(TAT^{-1})$ $(Tx)$ $=TAx$ $=T\lambda x$ $=\lambda Tx=\lambda y$. Thus $y$ is an eigenvector of $B$ corresponding to the eigenvector $x$ of $A$. $T$ is a change of basis. –  bgins Feb 18 '12 at 13:01

It is telling you to assume $y= \mathbf Tx$ just momentarily, so you can verify that $\mathbf Tx$ is the corresponding $\lambda$-eigenvector to $\mathbf B$ when $x$ is the $\lambda$-eigenvector to $\mathbf A.$

I think this proof of this fact is very slick:

$$\mathbf B- \lambda \mathbf I = \mathbf T \mathbf A \mathbf T^{-1} - \lambda \mathbf I = \mathbf T \mathbf A \mathbf T^{-1} - \mathbf T \lambda \mathbf I \mathbf T^{-1} = \mathbf T (\mathbf A - \lambda \mathbf I)\mathbf T^{-1}$$

so $$\det(\mathbf B-\lambda \mathbf I) = \det \mathbf T \det (\mathbf A-\lambda \mathbf I) \det (\mathbf T^{-1})= \det(\mathbf A-\lambda \mathbf I) .$$

Thus $\mathbf A$ and $\mathbf B$ have the same characteristic polynomial, and thus the same eigenvalues (counting multiplicity).

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The characteristic polynomial of similar matrices is the same, they have then the same eigenvalues. If AX=tX then BTX=tTX.

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