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Let $W$ be the space $$\operatorname{span}\{AB-BA\} ,$$ where A and B are square matrices, and let $H$ be the space of all square matrices of trace $0$. Then prove that $W=H$.

The fact that $W$ is a subset of $H$ is fairly easy to prove. I could prove the converse.

Travis Willse
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tori
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1 Answers1

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Let $E_{i,j}$ be the matrix with $1$ at position $(i,j)$ and $0$ otherwise. For $i\ne j$ let $A=E_{i,j}$ and $B=E_{j,j}$. Then we have $AB=E_{i,j}$ and $BA=0$, hence $E_{i,j}\in W$. It is clear that the $E_{i,j}$ with $i\ne j$ span $H$, hence $H\subseteq W$.

Hagen von Eitzen
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    Perhaps I'm missing something, but (for $n > 1$ anyway) $E_{11} - E_{22}$ is in $H$ but is not in the span of ${E_{ij} : i \neq j}$. – Travis Willse Oct 13 '14 at 11:12
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    (One can remedy this by using that $E_{ii} - E_{jj} = E_{ij} E_{ji} - E_{ji} E_{ij} \in H$, and with the matrices $E_{ij}$, $i \neq j$, these matrices span $H$ as desired.) – Travis Willse Oct 13 '14 at 11:58
  • The basis that you've made for $H$ is not correct! As @Travis has mentioned, the correct one is: $${ E_{ij} ; \mid ; i \ne j } \cup { E_{ii} - E_{i+1, i+1} ; \mid ; 1 \le i < n }.$$ – F.K Apr 01 '16 at 16:16