Why don't similar matrices have same eigenvectors and eigenvalues? What is wrong with this proof:
Suppose R and T are similar operators and R has eigenvalue 2 for some eigenvector $v$. 
By property of similar matrices:
$R=STS^{-1}$
Therefore:
$Rv = STS^{-1}v$
$2v = STS^{-1}v$
$2S^{-1}v = TS^{-1}v$
$2S^{-1}Sv = Tv$
$2Iv = Tv$
$Tv = 2v$
Thus $v$ is also an eigenvector of $T$ with eigenvalue 2.
Clearly this proof is incorrect, but where does it go wrong? 
 A: When you went from
$$
2S^{-1}v=TS^{-1}v
$$
to
$$
2S^{-1}Sv=Tv
$$
you completely ignored the rules of matrix multiplcation; you cannot arbitrarily commute matrices in a product! It should have read as follows:
$$
2S^{-1}vS=TS^{-1}vS
$$
Also not that it is not even clear that this is well defined as written, since $v$ is probably a column vector.
A: Since the error in our OP Justin Sanders' argument has been thoroughly vetted in other answers, here I will directly address the title question.
Similar matrices do have the same eigenvalues, to wit:
if 
$B = PAP^{-1}, \tag 1$
then
$B - \lambda I = PAP^{-1} - \lambda I = PAP^{-1} - \lambda PIP^{-1} = P(A - \lambda I)P^{-1}, \tag 2$
whence
$\det(B - \lambda I) = \det(P(A - \lambda I)P^{-1}) = \det(P) \det(A - \lambda I) \det (P^{-1}) = \det(A - \lambda I), \tag 3$
since $\det(P^{-1}) = (\det(P))^{-1}; \tag 4$
thus $A$ and $B$, having the same characteristic polynomials, also share the roots of these polynomials, i.e., their eigenvalues.
However, similar matrices do not in general share eigenvectors; if
$B \vec v = \lambda \vec v, \tag 5$
then 
$PAP^{-1} \vec v = \lambda \vec v, \tag 6$
or
$AP^{-1} \vec v = \lambda P^{-1} \vec v, \tag 7$
that is, $P^{-1} \vec v$ is an eigenvector of $A$ corresponding to $\lambda$; indeed, if $\lambda$ is a root of (3) of multiplicity one, then $P^{-1} \vec v$ is, up to a scale factor, the eigenvector of $A$ associated with $\lambda$.  Since we can in general choose $P$ so that $P^{-1} \vec v \ne \vec v$, the eigenvectors will not be shared 'twixt' $A$ and $B$.  
So while the eigenvalues are similarity invariant, the eigenvectors transform according to $\vec v \mapsto P^{-1} \vec v$.
A: It's wrong when you jump from $2S^{-1}v=TS^{-1}v$ to $2S^{-1}Sv=Tv$.
