Showing if $f_{A}(X)=AX-XA$ is diagonalizable, then $A$ is diagonalizable.

Question

Let $A=\begin{pmatrix} a_{1} & a_{2} \\ a_{3} & -a_{1}\end{pmatrix}\in\mathfrak{sl}_{2}(\mathbb{C})$. Let $f_{A}:\mathfrak{sl}_{2}(\mathbb{C})\rightarrow\mathfrak{sl}_{2}(\mathbb{C})$ by $f_{A}(X)=AX-XA$.

Let $\mathcal{B}=\left\{\begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix},\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix},\begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}\right\}$.

Show that if $f_{A}$ is diagonalizable, then $A$ is diagonalizable.

My attempt:

Let $\mathcal{B}_{0}=\{B_{1},B_{2},B_{3}\}$ be a basis of $\mathfrak{sl}_{2}(\mathbb{C})$ be such that $[f_{A}]_{\mathcal{B}_0}$ is a diagonal matrix. This means that $f_{A}(B_{i})=\lambda_{i}B_{i}$ for $i=1,2,3$.

I am unable to find an approach that would get me to a direction where I can express $P^{-1}AP=D$ where $D$ is a diagonal matrix. Any hints or answers would be greatly appreciated.

Answer 1

This is true for a general $n\times n$ matrix $A$ that is not necessarily traceless, but we will consider only your question's setting here.

Suppose $A$ is non-diagonalisable. Since $A$ is $2\times2$ and traceless, its minimal polynomial must be $x^2$. Therefore $A^2=0\ne A$. Pick two vectors $u$ and $v$ such that $v^TA$ and $Au$ are nonzero and let $Y=uv^T$. Then $f_A^2(Y)=A^2Y-2AYA+YA^2=-2AYA\ne0$ but $f_A^3(X)=A^3X-3A^2XA+3AXA^2-XA^3=0$ for every $X$. Hence $f_A^3=0\ne f_A^2$ and the minimal polynomial of $f_A$ is $x^3$. Now we arrive at a contradiction because $f_A$ is supposed to be diagonalisable. Therefore $A$ must be diagonalisable.

Answer 2

Suppose that $A$ is not diagonalizable. Then it has only one eigenvalue $\lambda$, with multiplicity $2$. But then $2\lambda=\operatorname{tr}(A)=0$, and therefore $\lambda=0$. So, the characteristic polynomial of $A$ is $\lambda^2$. But then, since the characteristic polynomial of $A$ is $\lambda^2-a_1^{\,2}-a_2a_3$, then $a_1$ is a square root of $-a_2a_3$. So, for two numbers $\alpha,\beta\in\Bbb C$, if $r$ is a square root of $-\alpha\beta$, then$$A=\begin{bmatrix}\pm r&\alpha\\\beta&\mp r\end{bmatrix}.$$And we have

• $f_A\left(\begin{bmatrix}1&0\\0&-1\end{bmatrix}\right)=\begin{bmatrix}0&-2\alpha\\2\beta&0\end{bmatrix}=-2\alpha\begin{bmatrix}0&1\\0&0\end{bmatrix}+2\beta\begin{bmatrix}0&0\\1&0\end{bmatrix}$;
• $f_A\left(\begin{bmatrix}0&1\\0&0\end{bmatrix}\right)=\begin{bmatrix}-\beta&\pm2r\\0&\beta\end{bmatrix}=-\beta\begin{bmatrix}1&0\\0&-1\end{bmatrix}+2r\begin{bmatrix}0&1\\0&0\end{bmatrix}$;
• $f_A\left(\begin{bmatrix}0&0\\1&0\end{bmatrix}\right)=\begin{bmatrix}\alpha&0\\\mp2r&-\alpha\end{bmatrix}=\alpha\begin{bmatrix}1&0\\0&-1\end{bmatrix}\mp2r\begin{bmatrix}0&0\\1&0\end{bmatrix}$.

So, the matrix of $f_A$ with respect to the basis $\mathcal B$ is$$\begin{bmatrix}0&-\beta&\alpha\\-2\alpha&\pm2r&0\\2\beta&0&\mp2r\end{bmatrix},\tag1$$whose characteristic polynomial is $-\lambda^3$. But $(1)$ is not the null matrix, which is the only diagonalizable matrix whose characteristic polynomial is $-\lambda^3$. So, $(1)$ is not diagonalizable, and therefore neither is $f_A$.

Answer 3

I'll do the general case.

For $n \ge 2$, and arbitrary $\lambda$, consider the $n \times n$ matrix $A = \pmatrix{\lambda & 1 & 0 & \dots & 0\\ 0&\lambda &1 &&0\\ 0&0&\lambda &&0\\ &&&\ddots&1\\ 0& &&&\lambda}$.

Quick calculation shows that for $x := \pmatrix{0 & 0 & \dots & 0\\ 1&0 &&0\\ &&\ddots&\\ 0& &&0}$, we have $f_A(x) = \pmatrix{1 & 0 & \dots & 0\\ 0&-1 &&0\\ &&\ddots&\\ 0& &&0}$ which is not the zero matrix, but another quick calculation or thought shows that $f_A^n(x)$ is the zero matrix, as indeed $f_A^{n-1}$ is identically zero on all upper triangular (including diagonal) matrices.

That is, the operator $f_A$ is nonzero but operates nilpotently on a certain vector $x$. Thus, it cannot be diagonalisable.

Now if $A$ is a general non-diagonalisable matrix, we decompose into Jordan blocks (over an algebraic closure, say), and use the above.

The other two answers treat your special case $n=2$ and, because you assume traceless matrices, $\lambda =0$.