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

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?

• qaphla, how did you do that so fast? – BananaCats Category Theory App Aug 12 '13 at 18:07
• Lots of practice with LaTeX and I type about 85-90wpm. – qaphla Aug 12 '13 at 18:08
• I believe you want to interchange -a and 0 in the last two equations. – user84413 Aug 12 '13 at 18:08
• @user84413. And there it is! Thanks! – BananaCats Category Theory App Aug 12 '13 at 18:09
• For reference: I think this open access 1957 paper of Albert and Muckenhoupt was the first published proof of this fact for a general field of any characteristic. – user85486 Aug 14 '13 at 2:11

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