Prove that if $M$ is positive definite, the Schur-complement is invertible.

If $$M=\begin{bmatrix} A & B \\ C & D \end{bmatrix}$$ is a positive definite $n\times n$ Matrix, prove that the Schur-complement $$S=D-CA^{-1}B$$ is invertible. Can anyone help? The hint in my textbook tells me to show that $A$ is invertible first.

Actually my argumentation of why $A$ has to have full rank was wrong, so I deleted this part.

• What is your definition of positive definite? Does positive definite imply symmetric for our purposes? – Omnomnomnom Feb 21 '17 at 18:42
• Good question. My textbook doesn't specify it more closely. I simply assumed that positive definite means that all Eigenvalues are strictly bigger than one. – bobo Feb 21 '17 at 19:05
• That is never what positive definite means! Do you mean bigger than $0$? Also, do you mean that the eigenvalues are necessarily real? – Omnomnomnom Feb 21 '17 at 20:09
• You are right, I wasn't aware of that. I assume the textbook meant positive definite in the sense that $x^tAx>0$ for all $x$, so that $A$ is invertible. – bobo Feb 21 '17 at 20:33
• Excuse me! "whether $A$ is necessarily symmetric" is what I mean. – Omnomnomnom Feb 21 '17 at 20:47

Now the Schur complement of $A$ shows up from multiplying $M$ by $$U = \begin{bmatrix} I & -A^{-1}B \\ 0 & I \end{bmatrix}$$ So that $$MU = \begin{bmatrix} A & 0\\ C & D -CA^{-1}B \end{bmatrix}$$ The determinant of $U$ is $1$, so this shows that $\det(M) = \det(A)*\det(S)$. The determinant of $M$ is positive since it is positive definite, so both $\det(A)$ and $\det(S)$ are non zero. In fact positive definiteness of $M$ implies $\det(A)$ is positive, so $\det(S)$ is positive too. In any case, nonzero determinant implies $S$ is ivertible.