Assume we have a positive definite real matrix $P$ and we define an inner product on a finite dimensional Hilbert space
$\langle x, y \rangle = x^\top P y\,$
and clearly the induced norm is $\| x\| = \sqrt{\langle x , x \rangle}$. Is there a way to compute the induced operator norm of a real linear symmetric operator represented by the matrix $A = A^\top$, i.e. $\| A\|$?
I know that for the special case when $P$ is the identity matrix we get that $\|A\|$ is the maximum singular value of $A$ but I couldn't figure out if there is something similar for general matrices $P \succ 0$.
Thanks very much for every hint.
In general, it will still be the maximum singular value of some matrix, although this matrix is a more complicated one: let $P=U_PSU_P^T, A=U_A\Sigma{U_A}^T$ be the SVD of $P$ and $A$ respectively. Then, substituting it into the scalar product, we can find the norm of A by definition: $$ ||A||=\sup_{||x||=1,||y||=1}(x, Ay)=\sup_{||x||=1,||y||=1} x^TPAy=\sup_{||x||=1,||y||=1}(U_Px)^TS\underbrace{U_PU_A^T}_R\Sigma(V_Ay)^T=\sup_{||x||=1,||y||=1}x^TSR\Sigma y=||SR\Sigma||_2, $$ where $||.||_2$ is the usual maximum singular value operator norm.
However, is you are so lucky that the singular vectors of $P$ and $A$ coincide, then $R=U_AU_A^T=I$ and $||A||=||S\Sigma||_2 = \max_i\sigma_i(P)\sigma_i(A)$, i.e. the maximum product of corresponding singular values.
That being said, I suspect that it may be a bit late for the author by now...