I recently stumbled across this bilinear form: $\beta(A,B)=\operatorname{Tr}(AB)$ for $A,B \in \mathbb{R}^{n,n}$. I am searching for an orthogonal basis. With many difficulties I could find one for $\mathbb{R}^{2,2}$ namely:
$B_{2,2}=\left\{\left( \begin{array}{cc} \frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{2}} \end{array} \right),\left( \begin{array}{cc} 0 & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & 0 \end{array} \right),\left( \begin{array}{cc} \frac{1}{\sqrt{2}} & 0 \\ 0 & -\frac{1}{\sqrt{2}} \end{array} \right),\left( \begin{array}{cc} 0 & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & 0 \end{array} \right)\right\}$
This form was also made that the representation matrix of $\beta$ is very elegant:
$M_\beta=\left( \begin{array}{cccc} \beta \left(b_1,b_1\right) & \beta \left(b_1,b_2\right) & \beta \left(b_1,b_3\right) & \beta \left(b_1,b_4\right) \\ \beta \left(b_2,b_1\right) & \beta \left(b_2,b_2\right) & \beta \left(b_2,b_3\right) & \beta \left(b_2,b_4\right) \\ \beta \left(b_3,b_1\right) & \beta \left(b_3,b_2\right) & \beta \left(b_3,b_3\right) & \beta \left(b_3,b_4\right) \\ \beta \left(b_4,b_1\right) & \beta \left(b_4,b_2\right) & \beta \left(b_4,b_3\right) & \beta \left(b_4,b_4\right) \end{array} \right)=\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{array} \right)$
So I can read the positive index of inertia (basically the number of 1s on the diagonal) which is in this case $n_+=3$. I am looking for such bases for higher dimensions of $\mathbb{R}^{n,n}$ but could not succeed. Thank you in advance.