This is the fourth part of a four-part problem in Charles W. Curtis's book entitled Linear Algebra, An Introductory Approach (p. 216). I've succeeded in proving the first three parts, but the most interesting part of the problem eludes me. Part (a) requires the reader to prove that $\operatorname{Tr}{(AB)} = \operatorname{Tr}{(BA)}$, which I was able to show by writing out each side of the equation using sigma notation. Part (b) asks the reader to use part (a) to show that similar matrices have the same trace. If $A$ and $B$ are similar, then
$\operatorname{Tr}{(A)} = \operatorname{Tr}{(S^{-1}BS)}$
$= \operatorname{Tr}(BSS^{-1})$
$= \operatorname{Tr}(B)$,
which completes part (b). Part (c) asks the reader to show that the vector subspace of matrices with trace equal to zero have dimension $n^2 – 1$. Curtis provides the hint that the map from $M_n(F)$ to $F$ is a linear transformation. From this, I used the theorem that $\dim T(V) + \dim n(T) = \dim V$ to obtain the dimension of the null space. Part (d), however, I'm stuck on. It asks the reader to show that subspace described in part (c) is generated by matrices of the form $AB – BA$, where $A$ and $B$ are arbitrary $n \times n$ matrices. I tried to form a basis for the subspace, but wasn't really sure what it would look like since an $n \times n$ matrix has $n^2$ entries in it, but the basis would need $n^2 – 1$ matrixes. I also tried to think of a linear transformation whose image would have the form of $AB – BA$, but this also didn't help me. I'm kind of stuck…
Many thanks in advance!
Best Answer
One way of proving this: Note that for all $A,B$ matrices, $AB−BA$ has trace equal to zero. Denote by $E_{ij}$ the matrix with entry $1$ in row $i$, column $j$ and $H_{ij}:=E_{ii}−E_{jj}$. Then $\{E_{ij}:i\ne j\}∪\{H_{i,i+1}:1\le i\le n-1\}$ form a basis for the space. Also, $H_{ij}=E_{ij}E_{ji}−E_{ji}E_{ij}$ and $2E_{ij}=H_{ij}E_{ij}−E_{ij}H_{ij}$. So, you have a basis formed by elements of the form $AB−BA$.