I am interested in verifying that my understanding of the basis for $3 \times 3$ (or even $n \times n$ matrices) that have a trace $= 0$ is along the right path.
$$ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{pmatrix} $$ With the above $3 \times 3$ matrix with trace $0$ know that
$a + e + i = 0$
$a = -e - i$
$e = -i -a$
$i = -a -e$
Outside of the elementary bases making up everything except the diagonal, I thought my last basis was
$$ \begin{pmatrix} -e-i & 0 & 0 \\ 0 & -i-a & 0 \\ 0 & 0 & -a-e \\ \end{pmatrix} $$
But by inspection, this does not seem to account for the instance where one of the diagonals is 0 and the other two are not.
My last consideration that I think is more accurate would be splitting the above matrix into the following combinations: $$ \begin{pmatrix} a & 0 & 0 \\ 0 & -a & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} $$ $$ \begin{pmatrix} 0 & 0 & 0 \\ 0 & -a & 0 \\ 0 & 0 & a \\ \end{pmatrix} $$
But this also doesn't feel quite correct. I feel I am close but missing one additional consideration.
I have also found other posts concerning the dimensions like here, but there was never a concrete basis provided to go off of.
Thanks!