Matrix diagonalization and operators Let $V=\Bbb F^{m\times n}$
$T: V\to V$ , by $T(B)=P^{-1}BP$ , for any $B$ in $V$ , where $P$ is an invertible matrix.
prove that if $A$ is an eigenvector of $T$, with eigenvalue $\lambda$
and $A$ is a diagonalizable matrix, then $\lambda=1$.

I know that $A$ and $\lambda A$ are similar, then they have the same eigenvalues.
then if $a$ is an eigenvalue of $A$, $(\lambda*a)$ is an eigenvalue of $(\lambda A)$,
I'm not sure that I can claim then $(\lambda*a)=a$, because we don't know they have same eigenvectors.

Thank you in advance.
Suppose that $A \neq 0$ is diagonalizable, then there exists an invertible (unitary) matrix $U$ such that $A = U\mathrm{diag(\lambda_{1}, ..., \lambda_{n})}U^{-1}$. 

Then $\lambda A = T(A) = P^{-1}U\mathrm{diag(\lambda_{1}, ..., \lambda_{n})}U^{-1}P$.

We get $\lambda U\mathrm{diag(\lambda_{1}, ..., \lambda_{n})}U^{-1} =  P^{-1}U\mathrm{diag(\lambda_{1}, ..., \lambda_{n})}U^{-1}P$ (*)

On taking the trace of the (*) and using the fact $\mathrm{Tr(AB)} = \mathrm{Tr(BA)}$, we should be able to force the condition we need on $\lambda$. Better yet, when are two diagonal matrices of the same size similar?
I'm sorry, I think I found a contradiction. Let

$$A= \left(\begin{array}{ccc}1 &amp; 0 &amp; 0\\0 &amp;-1&amp; 1\\0&amp; 0 &amp;0 \end{array}\right)$$

A is diagonalizable.

Then if

$$P= \left(\begin{array}{ccc}0 &amp; 1 &amp; -1\\1 &amp;0&amp; 1\\0&amp; 0 &amp;1 \end{array}\right)$$

We get $P=P^{-1}$, and $P^{-1}AP=-A$, therefore $T(A)=-A$, and $\lambda=-1$.