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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 $By=\mu x$.

Some maths leads me to

$By = TAT^{-1}y$ 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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    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.2012-02-18

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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.