# Show that the linear transformation $T:V\to V$ defined by $T(B)=ABA^{-1}$ is diagonalizable

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 $AB=mBA$, $\cdots$

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.
• For a matrix $M$, $M^T$ denotes its transpose. – Ben Grossmann Feb 27 '16 at 15:56
• How do you show that the $x_iy_j^T$ are linearly independent (I mean without explicitly computing them)? – Augustin Feb 27 '16 at 17:40
• @Augustin all you need to know is that the $x_i$ are independent of each other, as are the $y_i$. – Ben Grossmann Feb 27 '16 at 18:09
• I think I figured it out. We assume that $\sum_i\sum_j\alpha_{ij}x_iy_j^t=0$. First we use the independence of the $x_i$ to prove that for all $i$, $\sum_j\alpha_{ij}y_{ji}=0$, where $y_j=(y_{j1},y_{j2},y_{j3})^t$. Then we use, for each $i$, the independence of the $y_j$ to prove that $\alpha_{ij}=0$ for all $j$. – Augustin Feb 27 '16 at 20:06