# Let $M$ be a symmetric and positive definite block matrix. Prove that the matrix $D$, an element of $M$ is symmetric and positive definite

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.

• Any principal submatrix of a positive definite matrix must be positive definite. Nov 9, 2019 at 13:06
• Many thanks for your answer, but why is that true? Nov 9, 2019 at 13:53
• 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. Nov 9, 2019 at 13:59
• many thanks, no we've never looked towards block matrices in lectures....but that is so simple that I should have seen it myself! Nov 9, 2019 at 17:40

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.