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If I have a symmetric and positive definite $n\times n$ matrix $Q$ and a full row-rank totally unimodular $m\times n$, where $m<n$, matrix $A$, is it posible to show that the matrix $$Q-A^T(AQ^{-1}A^T)^{-1}A$$ is also positive definite?

I have show that it is true when the matrix $Q$ has dimention $2\times 2$, by doing all the posible situations. Also, If $m=n$, I have that the matrix $Q-A^T(AQ^{-1}A^T)^{-1}A$ can be zero.

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The matrix $Q - A^T(AQ^{-1}A^T)^{-1}A$ will generally be positive semidefinite. Let

$$M/Q : = Q - A^T(AQ^{-1}A^T)^{-1}A$$

and note that $M/Q$ is the lower Schur complement of the block matrix

$$ M= \begin{bmatrix} AQ^{-1}A^T & A \\ A^T & Q\\ \end{bmatrix}. $$

Since $Q$ is positive definite, it follows that $Q^{-1}$ is positive definite and that $Q^{1/2}$ exists. Furthermore $(Q^{-1})^{1/2}$ exists, hence let $Q^{-1/2} := (Q^{-1})^{1/2}.$ Note that

$$ \begin{bmatrix} AQ^{-1}A^T & A \\ A^T & Q\\ \end{bmatrix} = \begin{bmatrix} AQ^{-1/2} \\ Q^{1/2}\\ \end{bmatrix} \begin{bmatrix} Q^{-1/2}A^T & Q^{1/2} \\ \end{bmatrix}. $$

Therefore $M$ is positive semidefinite. Since $M$ is positive semidefinite, so too is its lower Schur complement.

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  • $\begingroup$ Why does $Q^{-1}$ being positive definite imply that $M$ is positive definite? $\endgroup$ – Omnomnomnom Nov 26 '18 at 20:03
  • $\begingroup$ @Omnomnomnom you're right; it should be positive SEMIdefinite. $\endgroup$ – SZN Nov 26 '18 at 20:05
  • $\begingroup$ then why does $Q^{-1}$ being positive definite imply that $M$ is positive semi-definite? Still don't see it. $\endgroup$ – Omnomnomnom Nov 26 '18 at 20:07
  • $\begingroup$ @Omnomnomnom thank you for your comment. I've tried to add more detail. $\endgroup$ – SZN Nov 26 '18 at 20:27
  • $\begingroup$ for non-symmetric matrices, non-negative eigenvalues alone do not imply that the matrix is positive definite. $\endgroup$ – Omnomnomnom Nov 26 '18 at 20:44

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