Finding the rank and signature of $\phi = tr (A B) - tr(A)tr(B)$ Here is a question I would like to answer:

Thoughts and attempts: First I tried using the hint. Through some play, I found that the linear map representing the bilinear form is $D:$
$$
D(A) = A^T - tr(A)I_n. 
$$
Then the symmetric traceless matrices have an eigenvalue of $1$. However, I can't figure out the complementary space.
Without the hint, I tried using a basis $E_{ij}$ to find a matrix representing the bilinear form explicitly, but this got too convoluted in the sense that I couldn't find the rank and signature from this form.
Can anybody help with this question?
 A: This is closely related to Exercise about signature of a scalar product
In general for fields not of characteristic 2 you may write $V$ as a direct sum of symmetric and skew-symmetric matrices.  In this case the hint prompts a refinement-- split the space of symmetric matrices into $E=\big\{\lambda I\big\}$ and $W$ the subspace of traceless symmetric matrices.  Note: in the subspace of symmetric matrices, $W$ is the orthogonal complement of $E$, $\dim E=1$ and $I$ isn't self-orthogonal under the bilinear form hence the subspace of symmetric matrices is given by $E\oplus W$. We may thus write
$V=E\oplus W\oplus S$
(where $S$ is the subspace of skew-symmetric matrices)
now evaluate the bilinear form
$\langle a,b\rangle :=\text{trace}\big(ab\big)-\text{trace}\big(a\big)\text{trace}\big(b\big)$
on each subspace
(a) the quadratic form $=\lambda^2\cdot \big(n-n^2\big)\lt 0$ for non-zero $e\in E$, (b) is equivalent to $\text{trace}\big(ab\big)= \text{trace}\big(a^Tb\big)$ the standard Frobenius inner product for $a,b \in W$ (c) is equivalent to $\text{trace}\big(ab\big)= -\text{trace}\big(a^Tb\big)$ the negative of the standard Frobenius inner product for $a,b \in S$.
If these subspaces are orthogonal, this implies a signature of
$\big(\frac{n(n+1)}{2}-1,\frac{n(n-1)}{2}+1\big)$
(hence the form is non-degenerate)
Finally confirm that these subspaces are in fact orthogonal under the bilinear form.  For $e\in E$ and $a\in S\oplus W$ we have $\langle a,e\rangle =\text{trace}\big(ae\big)-\text{trace}\big(a\big)\text{trace}\big(e\big)=\lambda \cdot \text{trace}\big(a\big)-0=0$
and for $s\in S$ and $w\in W$ we have
$\langle s,w\rangle =\text{trace}\big(sw\big)-\text{trace}\big(s\big)\text{trace}\big(w\big)=\text{trace}\big(sw\big)=\text{trace}\big((sw)^T\big)=-\text{trace}\big(sw\big)$
$\implies \langle s,w\rangle=0$
