How I can calculate the signature of this inner product?
Let $\mathbb{R}$-vector space $\mathcal{M}_{n \times n} ( \mathbb{R}) \rightarrow \mathbb{R} $ defined by:
$ f: \space \mathcal{M}_{n \times n} ( \mathbb{R}) \rightarrow \mathbb{R}, \space \space f(A,B) := n\operatorname{tr}(AB) - \operatorname{tr}(A)\operatorname{tr}(B) $
I know that this inner product is bilinear and simmetric. So, because it's an inner product we can conclude that the matrix representation of $f$ is positive definite and it has real eigenvalues, but how I calculate the signature of $f$?
Thank you!
Let $\langle A,B \rangle$ denote the inner product $ \langle A,B \rangle = \operatorname{tr}(AB^T). $ We can express $f$ as $$ f(A,B) = n \langle A,B^T \rangle - \langle A,I \rangle \cdot \langle B^T,I \rangle \\ = \left\langle A,n\left[B^T - \frac{\langle B^T,I\rangle}{\langle I,I\rangle} I \right]\right\rangle = \langle A,\Psi(B)\rangle, $$ where $\Psi(B) = n\left[B^T - \frac{\langle B^T,I\rangle}{\langle I,I\rangle} I\right]$. Notably, we have $\Psi = n\cdot \tau \circ \pi$, where $$ \tau(B) = B^T, \quad \pi(B) = B - \frac{\langle B,I\rangle}{\langle I,I \rangle}I. $$ Because $\tau$ is self-adjoint relative to the inner product $\langle \cdot,\cdot\rangle$ and $I$ is an eigenvector of $\tau$ associated with the eigenvalue $1$ (equivalently, $\pi$ is the spectral projection associated with the eigenvector $I$), we can conclude that the eigenvalues of $\Psi$ are equal to those of $\tau$ except that one eigenvalue $n$ has been replaced by an eigenvalue $0$.
The eigenvalues of $\tau$ are $-1$ with multiplicity $n(n-1)/2$ and $+1$ with multiplicity $n(n+1)/2$. Thus, the eigenvalues of $\Psi$ are $-n$ with multiplicity $n(n-1)/2$, $+n$ with multiplicity $n(n+1)/2 - 1$, and $0$ with multiplicity $1$.
Thus, the signature of $f$ is $$ n_+ = \frac{n(n+1)}{2} - 1, \quad n_- = \frac{n(n-1)}{2}, \quad n_0 = 1. $$