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


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.


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

  • $\begingroup$ 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

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