Stuck on finding rank of $T$ when $n=8$ of $T(A) = A - A^T$

Question

$T: M_{n\times n}(F) \to M_{n\times n}(F)$ is a linear transformation. I know from rank-nullity that $\text{rank}(T) + \text{nullity}(T) = \dim(M_{n\times n}(F))$.

I'm trying to find $N(T)$ and then the dimension of that. I found that $N(T)$ is the set of matrices where $A=A^T$, and therefore $A_{ij} = A_{ji}$.

After this, I'm stuck as to figure out what $\dim(N(T))$ is.

Answer 1

Clearly $A \in \ker T$ iff $A = A^T$.

So what is the dimension of symmetric matrices? It should be clear that ${\cal B} = \{ E_{ij} + E_{ji}\}_{i \le j} $ (where $E_{ij}$ is the matrix of zeros with a one in the $i,j$ position) is a basis for the symmetric matrices, and $|{\cal B}| = \frac{1}{2}n(n+1)$.

It follows that $\operatorname{rk} T = n^2-\frac{1}{2}n(n+1) = \frac{1}{2}n(n-1)$.

Addendum: To see why $|{\cal B}| = \frac{1}{2}n(n+1)$, note that $|{\cal B}| = |\{ (i,j) | i \le j \}|$.

We have $|\{(i,j)\}| = n^2$, $|\{(i,j) | i = j\}| = n$, and $|\{ (i,j) | i < j \}| = |\{ (i,j) | i > j \}| $.

Furthermore the set of indices can be written as the disjoint union $\{(i,j)\} = \{ (i,j) | i < j \} \cup \{ (i,j) | i = j \} \cup \{ (i,j) | i > j \}$, so $n^2 = |\{(i,j)\}| = |\{ (i,j) | i < j \}| + | \{ (i,j) | i = j \} | + | \{ (i,j) | i > j \} = 2 |\{ (i,j) | i < j \}| + n$, so $|\{ (i,j) | i < j \}| = \frac{1}{2}n (n-1)$.

Since $\{ (i,j) | i \le j \} = \{ (i,j) | i = j \} +\{ (i,j) | i < j \} $, we have $|\{ (i,j) | i \le j \} | = n+\frac{1}{2}n (n-1) = \frac{1}{2}n (n+1)$.

Answer 2

The simplest approach, IMO, is to find a basis for $N(T)$. It's not much more complicated than the standard basis for $M_{8\times8}(F)$. You could find a basis for the image of $T$ instead, if you like.