Notation: If $\Bbb F=\Bbb {R}$ or $\Bbb C$, denote by $M_n(F)$ the $n\times n$ matrices with entries in $\Bbb F$.
Let $V=M_3(C)$ be a $9$-dimension vector space over $\Bbb C$ and let $$A=\begin{pmatrix} 0 & 0 & 2 \\ 1 & 0 & 1 \\ 0 & 1 & -2 \\ \end{pmatrix}.$$ Define the linear transformation $T:V\to V$ by $T(B)=ABA^{-1}.$ Show that $T$ is also diagonalizable.
I have found that $\begin{array}{c}T(A)=A\\T(A^2)=A^2\\T(A^3)=A^3\\\vdots\end{array}$
I also try to discuss the eigenvalue of $T$, if $m$ is the eigenvalue of $T$, $T(B)=ABA^{-1}=mB$, so $AB=mBA$, $\cdots$
But then, I can't get more information, thanks for help!
Your $A$ has eigenvalues $-2,-1,1$, so it is diagonalizable with eigenvectors $x_1,x_2,x_3$, each of which is a $3 \times 1$ column vector. Note also that $A^T$ has eigenvectors $y_1,y_2,y_3$.
Show that each vector $x_iy_j^T$ is an eigenvector of $T$, and that the set $\{x_iy_j^T \mid i,j = 1,2,3\}$ is linearly independent. Then, since $M_3(\Bbb C)$ has a basis consisting of eigenvectors of $T$, we can conclude that $T$ is diagonalizable.