Let \(W\) be the space of \(n \times n\) matrices over the field \(F\), and let \(W_0\) be the subspace spanned by the matrices \(C\) of the form \(C = AB - BA\). Prove that \(W_0\) is exactly the subspace of matrices which have trace zero. (Hint: What is the dimension of the space of matrices of trace zero? Use the matrix ‘units,’ i.e., matrices with exactly one non-zero entry, to construct enough linearly independent matrices of the form \(AB - BA\).)
Solution
Every commutator has trace zero, since \[ \operatorname{tr}(AB-BA)=0. \] Hence \(W_0\) is contained in the trace-zero subspace \[ Z=\{A\in W:\operatorname{tr}A=0\}. \]
Let \(E_{ij}\) denote the matrix units. If \(i\ne j\), then \[ [E_{ii},E_{ij}]=E_{ij}, \] so every off-diagonal matrix unit lies in \(W_0\). Also, for \(i\ne j\), \[ [E_{ij},E_{ji}]=E_{ii}-E_{jj}. \] Thus \(W_0\) contains all off-diagonal \(E_{ij}\) and the \(n-1\) diagonal differences \[ E_{11}-E_{nn},\ldots,E_{n-1,n-1}-E_{nn}. \] These \(n(n-1)+(n-1)=n^2-1\) matrices are linearly independent. Therefore \[ \dim W_0\ge n^2-1. \] The trace map \(W\to F\) is a nonzero linear functional, so its kernel \(Z\) has dimension \(n^2-1\). Since \(W_0\subseteq Z\), equality of dimensions gives \[ W_0=Z. \] Thus the span of the commutators is exactly the subspace of trace-zero matrices.