Why is $ A^{-1} - P \left( P^T A P \right)^{-1}P^T$ positive definite here?

Question

Assume $A \in \Bbb R^{n \times n}$ is a symmetric and positive definite matrix. For matrix $P \in \Bbb R^{n \times m}$ with $m \le n$ and $\text{rank}(P)=m$, is the following matrix positive definite?

$$ A^{-1} - P \left( P^T A P \right)^{-1}P^T$$

If it is not positive definite, can we say that this matrix is positive semidefinite? Any idea is appreciated.

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This expression seems something like projection, but how can we prove it?

Answer 1

The matrix is indeed positive semidefinite. Note that $$ A^{-1} - P \left( P^T A P \right)^{-1}P^T = \\ A^{-1/2}\left[I - [A^{1/2}P] \left( [A^{1/2}P]^T [A^{1/2}P] \right)^{-1}[A^{1/2}P]^T\right]A^{-1/2}. \tag{*} $$ For the matrix $B = A^{1/2}$, the matrix $M = I - B(B^TB)^{-1}B^T$ is positive semidefinite since it satisfies $M = MM^T$ (in fact, $M$ is the orthogonal projection onto the column space of $B$). As equation $(*)$ above indicates, your matrix is equal to $[A^{-1/2}]^T M[A^{-1/2}]$ and is therefore positive semidefinite.

(In general, if $M$ is positive semidefinite, then $C^TMC$ must be positive semidefinite).

Answer 2

The matrix is not positive definite. We have $$ P^TA\cdot ( A^{-1} - P(P^TAP)^{-1}P^T) =0. $$ If the matrix would be positive definite, then it would be invertible, then the above equation would imply $P=0$.