# If a matrix $Q$ is symmetric and positie definite, is it possible to show that the matrix $Q-A^T(AQ^{-1}A^T)^{-1}A$ is also positive definite?

If I have a symmetric and positive definite $$n\times n$$ matrix $$Q$$ and a full row-rank totally unimodular $$m\times n$$, where $$m, matrix $$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.

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.

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