Matrix diagonalization and operators

Question

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.

Answer 1

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?

Answer 2

I'm sorry, I think I found a contradiction. Let

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

A is diagonalizable.

Then if

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

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