Why don't similar matrices have same eigenvectors and eigenvalues?

Question

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?

Answer 1

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.

Answer 2

It's wrong when you jump from $2S^{-1}v=TS^{-1}v$ to $2S^{-1}Sv=Tv$.

Answer 3

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 $\vec v \mapsto P^{-1} \vec v$.