Let us consider $V=M_n(\mathbb{R})$ the space of the matrices $n\times n$. Consider $\phi:V\times V \rightarrow \mathbb{R}$ the scalar product definited by $\phi(A,B)=tr(AB)$ where $tr$ indicates the trace of the matrix.
It's asked to find the metric signature of this product.
Thoughts: I have already determined that the product is not degenerate. Now I am a little bit in trouble. I can actually say that the signature is $\sigma(\phi)=(k,n-k,0)$ but I have to find such $k$. I already know that if the product was with the transpose, it would have been simple (just noticing that $tr(AB^T)$ can be considered as the canonical product in $\mathbb{R}^{n\times n}$). But what about this?
I tried to approach this problem by considering this decomposition of the space: $$V=S_n \oplus A_n$$ where $S_n$ and $A_n$ indicate the symmetrical matrices and the skew symmetric ones respectively. What do I do now? I think that I should consider an orthonormal basis but I really find it difficult. Any help appreciated.
As you note, the canonical inner product on $\Bbb R^{n \times n}$ is given by $\langle A,B \rangle = \operatorname{tr}(AB^T)$. As such, every bilinear form $\phi$ can be uniquely expressed as $$ \phi(A,B) = \langle \Phi(A), B \rangle $$ for some linear transformation $\Phi: \Bbb R^{n \times n} \to \Bbb R^{n \times n}$. If the bilinear form is symmetric, then $\Phi$ must be self-adjoint. The signature of a symmetric bilinear form corresponds to the signs of the eigenvalues of the corresponding map.
In our case, we find that $\Phi(A) = A^T$. That is, our bilinear form can be written as $$ \phi(A,B) = \langle A^T,B \rangle $$ The map $\Phi$ has eigenspaces $\ker(\Phi - \operatorname{id}) = S_n$ of dimension $n(n+1)/2$ and $\ker(\Phi + \operatorname{id}) = A_n$ of dimension $n(n-1)/2$. Thus, we find $$ \sigma(\phi) = \left(\frac{n(n+1)}{2},\frac{n(n-1)}{2},0 \right) $$