Given a square matrix $Q$ with full rank, it is always possible to obtain from it a matrix $R$ such that $R^T R = I$, i.e. $R$ is orthonormal, for example by the Gram-Schmidt process. Moreover, we can also get from it a matrix $R$ with $R^T P R = I$, whenever $P$ is Hermitian and positive semidefinite. Namely we can run the Gram-Schmidt process using the inner product defined by $\langle u, v \rangle = u^T P v$ instead of the usual dot product.
Is it possible to find $R$ such that $R^T P R = I$ when $P$ is Hermitian but not positive semidefinite? In particular, is it possible to do it when $P$ is a permutation matrix?
As an aside, is there a name for the condition $R^T P R = I$? If $P$ were the identity, we would call $R$ orthonormal. Perhaps $P$-orthonormal?
Any reference or suggestion will be appreciated. Thanks!
EDIT: In a nutshell, I need a basis $R$ of the column space of $Q$ such that $R^T P R = I$. Does it always exist?
I’ll assume we are talking about real matrices only (if we used adjoints instead of transposes, this carries over to complex matrices as well.)
Since $R$ is full rank, the decomposition $R^T P R =I$ is possible if and only if $P$ is positive definite.
You just gave one direction, namely that is $P$ is positive definite you can find a (full rank) $R$ such that $R^TPR=I$.
On the other hand, if $R^T P R =I$ holds for an invertible matrix $R$, then we can multiply by $R^{-T}$ on the left and $R^{-1}$ on the right to see that $P=R^{-T}R^{-1}$ is both full rank and positive definite.