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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.

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    $\begingroup$ What is your definition of positive definite? Does positive definite imply symmetric for our purposes? $\endgroup$ – Omnomnomnom Feb 21 '17 at 18:42
  • $\begingroup$ 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. $\endgroup$ – bobo Feb 21 '17 at 19:05
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    $\begingroup$ That is never what positive definite means! Do you mean bigger than $0$? Also, do you mean that the eigenvalues are necessarily real? $\endgroup$ – Omnomnomnom Feb 21 '17 at 20:09
  • $\begingroup$ 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. $\endgroup$ – bobo Feb 21 '17 at 20:33
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    $\begingroup$ Excuse me! "whether $A$ is necessarily symmetric" is what I mean. $\endgroup$ – Omnomnomnom Feb 21 '17 at 20:47
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Here's one quick way to see it: I don't think positive definiteness is very necessary (it's overkill).

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.

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  • $\begingroup$ thanks for the quick answer! $\endgroup$ – bobo Feb 21 '17 at 19:07

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