Let $B\in \mathbb{R}^{n\times n}$ be a symmetric and positive definite matrix and $D\in\mathbb{R}^{n\times q}$ a matrix with full (column) rank. Then the matrix $D^TBD$ is positive definite and we can define the symmetric matrix
$C:=D(D^TBD)^{-1}D^T$.
Can one prove that the matrix $M:= B^{-1} – C\ $ is positive semidefinite?
The case where $q=n$ is of course trivially true, since then $C=B^{-1}$ and $M=0$. However, if $q<n$ I do not see how the eigenvalues/eigenvectors of $C$ and $B^{-1}$ are related. In fact, I do not even see how the eigenvalues/eigenvectors of $B$ and $D^TBD$ are related. And that $M\succeq 0$ is equivalent to
$B-BCB\succeq 0$
does not seem to help either.
Any help would be appreciated.
Best Answer
Yes. The statement $B^{-1}\succeq C$ is equivalent to $I\succeq (B^{1/2}D) (D^TBD)^{-1} (D^TB^{1/2})$, or $I\succeq A(A^TA)^{-1} A^T$ where $A=B^{1/2}D$. Now the latter statement can be easily proved by using performing singular value decomposition on $A$ or simply by noting that $\rho\left(A(A^TA)^{-1} A^T\right)=\rho\left(A^TA(A^TA)^{-1}\right)=1$.