3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $\rho\left(A(A^TA)^{-1} A^T\right)=\rho\left(A^TA(A^TA)^{-1}\right)=1$.

$\endgroup$
  • $\begingroup$ Thanks for this great (and simple) solution! Using the SVD one obtains that $A(A^TA)^{-1}A^T$ only has the eigenvalues 0 and 1 which proves the inequality. However, I prefere your argument via the spectral radius. The fact that the matrices $A(A^TA)^{-1}A^T$ and $A^TA(A^TA)^{-1}$ have the same nonzero eigenvalues (and therefore the same spectral radius) makes this prove quite short and elegant. $\endgroup$ – Robert Jun 20 '16 at 8:48
  • $\begingroup$ @Robert Thanks. By the way, $(A^TA)^{-1}A^T$ is actually the Moore-Penrose pseudo inverse of $A$. $\endgroup$ – user1551 Jun 20 '16 at 9:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.