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?
Best Answer
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.