Determining the dimension of span$\{AB-BA : A,B \in M_{n \times n}(\mathbb R) \}$ Let $S$ be the subspace , of  $M_{n \times n}(\mathbb R)$ (the vector space of all $n \times n$ real matrices ) , generated by matrices of the form $AB-BA$ , where $A,B \in M_{n \times n}(\mathbb R)$ , then how do we prove that $\dim S=n^2-1$ ? The only thing that I can determine is that the trace of all matrices of $S$ is $0$ . Please help  
 A: In fact, a matrix $M$ is of the form $AB - BA$ if and only if it has trace zero.  It suffices to verify that the set of matrices of trace $0$ is indeed an $n^2 - 1$ dimensional space.
For a proof that the two spaces are the same, we can use the line of reasoning given here.  I think there is also a theorem from Horn and Johnson that can be used here that states that all matrices are similar to a matrix with identical diagonal entries.

For the purposes of this problem, however, it is not necessary to show that every trace zero matrix can be expressed as $AB - BA$ (i.e. as a commutator). Instead, it suffices to show that we have a set of commutators than spans the trace-0 subspace. To that end, it suffices to observe the following.
Let $E_{ij}$ denote the matrix with a $1$ as the $i,j$ entry and $0$s elsewhere. We have
$$
E_{ij}E_{pq} - E_{pq}E_{ij} = \delta_{jp} E_{iq} - \delta_{ip} E_{qi},
$$
where $\delta_{ij}$ is a Kronecker delta. By judiciously selecting $i,j,p,q$ in the above, we see that the matrices $E_{ij}$ with $i \neq j$ and $E_{ii} - E_{jj}$ can all be expressed as a commutator
