Dimensionality of null space when Trace is Zero
Solution 1:
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 $AB−BA$.