Confusion on symmetric bilinear form exercise I'm trying to solve an exercise on the symmetric bilinear form $f(A,B)=\operatorname{tr}A\operatorname{tr}B-n\operatorname{tr}AB$. I have already found that $V^\perp, \!^\perp V$ are the scalar matrices. Now I'm supposed to calculate $\operatorname{rank}f$. I thought about calculating the representing matrix $(f(E_{k\ell},E_{\mu\nu}))$.
Now


*

*$\operatorname{tr}E_{k\ell}E_{\mu\nu}=\sum_j\sum_i\delta_{\ell
    k}(ij)\delta_{\nu\mu}(ji)\neq 0\iff\mu=\ell$ and $\nu=k$;

*$\operatorname{tr}E_{k\ell}\operatorname{tr}E_{\mu\nu}\neq 0\iff
    k=\ell,\mu=\nu$.


I think we want the XOR of these conditions, but I don't know how to proceed...
Also, I need to find the rank of the restriction of $f$ to matrices with zero trace, in which case I only need $\operatorname{tr}E_{k\ell}E_{\mu\nu}=\sum_j\sum_i\delta_{\ell
    k}(ij)\delta_{\nu\mu}(ji)\neq 0$ which has, I think, $n^2$ choices.
The answer to both questions is $n^2-1$ I'm told, so I'd appreciate some help!
 A: Consider first the symmetric bilinear form $g(A,B) = \operatorname{tr}(AB)$ on the subspace 
$$W := \{ A \in M_n(\mathbb{C}) \, | \operatorname{tr}(A) = 0 \}. $$
If $\operatorname{tr}(A) = 0$ then $\operatorname{tr}(A^{*}) = \sum_{i=1}^n \overline{a_{ii}} = \overline{\sum_{i=1}^n a_{ii}} = \overline{\operatorname{tr}(A)} = 0$ and so $A^{*} \in W$ and we have
$$ g(A^{*},A) = \operatorname{tr}(A^{*}A) = \sum_{i,j=1}^n |a_{ij}|^2. $$
Thus, if $A \neq 0$ then $g(A^{*},A) \neq 0$ and $g$ is non-degenerate and so has rank $\dim W = n^2 - 1$.
Returning back to $f$, we have $f|_{W} = -n g$ and so $f|_{W}$ is also non-degenerate and has rank $n^2 - 1$. Finally, if we choose a basis $A_1,\dots,A_{n^2 - 1}$ for $W$ and set $A_{n^2} = I$ we obtain a basis for $M_n(\mathbb{C})$ with respect to which the form $f$ is represented by a matrix of the form
$$ \begin{pmatrix} C_{(n-1) \times (n-1)} & 0_{(n-1) \times 1} \\ 
0_{1 \times (n-1)} & 0 \end{pmatrix} $$
where $C$ is the matrix representing $f|_{W}$ with respect to the basis $(A_1, \dots, A_{n^2-1})$. Thus, $f$ also has rank $n^2 - 1$.
