Trace zero matrix that can't be written as $AB - BA$?

Question

According to this paper, every trace zero matrix over a field can be written in the form $AB - BA$. However, here's a basic counterexample:

Let $A = diag(a, -a)$ for some nonzero number a. Then $A = BC - CB$ if and only if a system of equations involving entries of $B,C,A$ is true. Then the entries of $A$ are:

$$\begin{align} a = b_{1,1}c_{1,1} + b_{1,2}c_{2,1} - b_{1,1}c_{1,1} - b_{2,1}c_{1,2} \\ 0 = b_{1,1}c_{1,2} + b_{1,2}c_{2,2} - b_{1,1}c_{2,1} - b_{2,1}c_{2,2} \\ -a = b_{2,1}c_{1,1} + b_{2,2}c_{2,1} - b_{1,2}c_{1,1} - b_{2,2}c_{1,2} \\ 0 = b_{2,1}c_{1,2} + b_{2,2}c_{2,2} - b_{1,2}c_{2,1} - b_{2,2}c_{2,2} \end{align} $$

For $a$ nonzero the first and the last equations can't both be true. So where did I mess up?

Answer 1

Consider $$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 0 \\ a & 0 \end{bmatrix} -\begin{bmatrix} 0 & 0 \\ a & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$

What made you conclude that the first and last of those equations can't be true?

Answer 2

$$B=\left(\begin{array}{rr}0&1\\0&0\end{array}\right).$$ $$C=a\left(\begin{array}{rr}0&0\\1&0\end{array}\right).$$ $$BC-CB=\left(\begin{array}{rr}a&0\\0&-a\end{array}\right).$$