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Prove that $D$ must also be positive definite matrix, given that $M$ is symmetric and positive definite matrix.

Consider the block matrix $$M=\begin{bmatrix} A & B\\ C & D \end{bmatrix}$$

where $A \in Mat_{n\times n}, \ B \in Mat_{n \times m} , \ C \in Mat_{m\times n}, \ D \in Mat_{m\times m}$.

D is invertible.

I have a notion that I have to use the Schur complement, but I have no Idea how. And I think I don't really get what the Schur complement really is.

I tried to write the Matrix as a multiplication of two matrices containing the Schur complement. but it did not get anywhere.

The other idea was that the I know that the diagonal entries of $D$ must be positive. but again this seems of no help.

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  • $\begingroup$ Any principal submatrix of a positive definite matrix must be positive definite. $\endgroup$ Nov 9, 2019 at 13:06
  • $\begingroup$ Many thanks for your answer, but why is that true? $\endgroup$
    – Lillys
    Nov 9, 2019 at 13:53
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    $\begingroup$ I was hoping that the term "principal submatrix" might jog your memory towards a result from you notes. In any case see my answer below. $\endgroup$ Nov 9, 2019 at 13:59
  • $\begingroup$ many thanks, no we've never looked towards block matrices in lectures....but that is so simple that I should have seen it myself! $\endgroup$
    – Lillys
    Nov 9, 2019 at 17:40

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Suppose that $M$ is positive semidefinite. The for any non-zero $y \in \Bbb R^{n+m}$ we have $y^TMy > 0$. So, for any non-zero $x \in \Bbb R^m$, we can set $y = (0,x)$ to find that $$ 0 < y^TMy = \pmatrix{0&x^T}\pmatrix{A&B\\C&D} \pmatrix{0\\x} = x^TDx. $$ So, $D$ is positive definite.

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