Show that $ad(X)$ is also diagonalisable.

Question

Let $V$ be an dimensional vetor space over $\Bbb C$ and let $x \in gl(V)$. Suppose that $x$ is diagonalisable, with eigen values $\lambda_1, ..., \lambda_n$. Show that $ad(X)$ is also diagonalisable.

I was trying to prove it for $n=2$ first then suppose $X=\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}$

$ad(X)(E_{11})=XE_{11}-E_{11}X=\begin{pmatrix} 0 & -b \\ c & 0 \\ \end{pmatrix}$ where $E_{11}=\begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix}$

Similarly $ad(X)(E_{12})=\begin{pmatrix} -c & a-d \\ 0 & c \\ \end{pmatrix}$ where $E_{12}=\begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix}$

$ad(X)(E_{21})=\begin{pmatrix} b & 0 \\ d-a & -b \\ \end{pmatrix}$ where $E_{21}=\begin{pmatrix} 0 & 0 \\ 1 & 0 \\ \end{pmatrix}$ and

$ad(X)(E_{22})=\begin{pmatrix} 0 & b \\ -c & 0 \\ \end{pmatrix}$ where $E_{22}=\begin{pmatrix} 0 & 0 \\ 0 & 1 \\ \end{pmatrix}$

So, if we write the matrix of $ad(X)$ we get

$\begin{pmatrix} 0 & -c & b & 0 \\ -b & a-d & 0 & b \\ c & 0 & d-a & -c \\ 0 & c &-b & 0 \end{pmatrix}$

Now there is a way to calculate eigen values and then eigen vectors to show that the matrix is diagonalizable. But is there any other shorter way because we have to prove it for $n$.

I haven't understood from Here. If there is any other way. I am new in mathstack thats why I have posted as a question with my try.

Answer 1

The key point of the answer to the question "İf x is diagonalizable then ad(x) is also diagonalizable" is in the first line

Take $\mathcal{B}$ a basis wherein $x$ is diagonalizable and denote $A$ its matrix.

Since $x$ is diagonalizable with eigenvalues $\lambda_1,\dotsc,\lambda_n$, you can write $X=\begin{pmatrix} \lambda_1&0 \\ 0&\lambda_2 \end{pmatrix}$ with an appropriate basis $\mathcal{B}$, that is, $a=\lambda_1$, $d=\lambda_2$, and $b=c=0$ in your example for $n=2$.

Then as you wrote, $ad(X)$ becomes a diagonal matrix with diagonal entries $a-d$, $d-a$ and $a-a=d-d=0$.

Why don't you try to extend it for general $n$?