Find $AB=-BA$ with not null product.

Question

Given $A,B,\text{ both }2\times2$ matrices, find an example such that $AB=-BA$ with $AB\neq{0}$. I just can't find an example where that happens, it's easy to see where $AB=BA$ but no idea how to get the $(-1)$ scalar.

Answer 1

Pick $$ A = \left(\begin{matrix} -1 & 0 \\ 0 & 1 \end{matrix}\right) $$ and

$$ B = \left(\begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix}\right) $$

Then

$$ AB = \left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right) $$ and

$$ BA = \left(\begin{matrix} 0 & -1 \\ -1 & 0 \end{matrix}\right) $$

The intuition behind this answer is to exploit the property of anti-commutativity of reflection about the y-axis and rotation by 90 degrees anti-clockwise.

Answer 2

A simple example would be $A = \begin{bmatrix}1 &0 \\ 0 & -1\end{bmatrix}$ and $B = \begin{bmatrix}0 &0 \\1 & 0\end{bmatrix}$.

Answer 3

Try $A=\pmatrix{0&1\\-1&0}$ and $B=\pmatrix{0&1\\1&0}$.