Find two non-singular matrices such that $AB=-BA$

Question

Question is precisely the title. I've been trying to come up with an example for an hour now and haven't found one. Do such matrices, $A$ and $B$ exist?

Answer 1

$A=$

\begin{bmatrix} 1& 1\\1&-1\end{bmatrix}

and $B=$

\begin{bmatrix} 0& -2\\ 2&0\end{bmatrix}

Answer 2

Here's how you might construct an example of this yourself, via brute force. I'm going to change the notation a bit to make it easier (for me at least). Let $$ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \\ X = \begin{bmatrix} x & y \\ z & w \end{bmatrix} $$ We require $AX+XA=O$, where $O$ is the $2\times 2$ matrix of all zeroes. To go further, I will require that $A,X,AX,XA \neq O$ (otherwise it is trivial). Computing the products gives us $$ AX = \begin{bmatrix} ax+bz & ay+bw \\ cx+dz & cy+dw \end{bmatrix} \\ XA = \begin{bmatrix} ax+cy & bx+dy \\ az+cw & bz+dw \end{bmatrix} $$ And so, the requirement that their sum is the zero matrix breaks down to 4 equations: $$ 2ax+bz+cy = 0 \\ 2dw + bz + cy = 0 \\ b(x+w) + y(a+d) = 0 \\ c(x+w) + z(a+d) = 0 $$ Subtract the first two equations and add the last two equations to get two new equations $$ ax=dw \\ (b+c)(x+w)+(y+z)(a+d) = 0 $$ One potential choice here is to just make $a=d=x=w=0$. That satisfies both conditions, and simplifies the remaining equations for us down to one: $$bz+cy = 0$$ If $bz$ and $cy$ are both 0, then $AX = XA = O$ which I wanted to avoid. So let's pick the next simplest one: $bz = -1, cy=1$. We can accomplish this by letting $b=c=y=1, z=-1$.

Thus, with this choice, we have $$ A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \\ X = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \\ AX = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} \\ XA = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} $$ and so $AX+XA = O$ as required.