Example of two matrices such that $LM+ML=0$ I am looking for some examples of $n\times n$ complex matrices $L$ and $M$ satisfying $LM=aML$, for some fixed $a\in \mathbb{C}$. In particular if we take $n>2$ and $a=-1$ then we can see that matrices of the form $AB=0=-BA$ gives the required result. But I need some non trivial examples. Please give some hints so that I can proceed.

Should I look some different rings other than the matrix rings?

 A: Take $M = \begin{pmatrix} 0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 &\cdots & 1\\
0 & 0 & 0 &\cdots & 0\end{pmatrix}$
Then for any $a \neq 0 \in \mathbb{C}$, the matrix $aM$ is nilpotent of order $n$, so it is similar to $M$. So there exists $L \in \mathcal{GL}_n({\mathbb{C}})$ such that $aM = LML^{-1}$, so $LM=aML$.
A: We can do the $a = -1$ case in your title using Clifford algebras. Namely, the Clifford algebra $\text{Cl}(\mathbb{R}^2)$ can be given a presentation with two generators $i, j$ subject to the relations
$$i^2 = j^2 = -1, ij + ji = 0.$$
This is exactly the algebra of quaternions $\mathbb{H}$ and in particular is $4$-dimensional, with basis $\{ 1, i, j, ij \}$. The quaternions act on themselves by left multiplication and this gives a $4$-dimensional faithful representation of $\text{Cl}(\mathbb{R}^2)$ which is also its unique simple module. In this representation $i$ and $j$ act via the $4 \times 4$ matrices
$$L_i = \left[ \begin{array}{cccc} 0 & -1 & 0 & 0 \\
 1 & 0 & 0 & 0 \\
 0 & 0 & 0 & 1 \\
 0 & 0 & -1 & 0 \end{array} \right]$$
$$L_j = \left[ \begin{array}{cccc} 0 & 0 & -1 & 0 \\
 0 & 0 & 0 & -1 \\
 1 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 \end{array} \right].$$
