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Let $M$ be an $n\times n$ symmetric positive definite matrix $$ M = \left[\begin{array}{cc} M_{1, 1} & M_{1, 2}\\ M_{2, 1} & M_{2, 2} \end{array}\right] $$ where $M$ is separated into blocks.

Consider the matrix $M_{2,2} - M_{2,1} M_{1,1}^{-1} M_{1,2}$. Prove it is symmetric and positive definite.

We are given that $M_{1,1}$ and $M_{2,2}$ are also symmetric.

Definition of positive definite matrix --

A matrix $n\times n$ matrix $A$ is positive definite if:

1.) $A$ is symmetric, and

2.) $x^TAx>0$ for all $x\in\mathbb R^n$ and $x\neq 0$.

I'm really struggling with this problem but here's what I got so far which is not much and possibly worthless.

So then there exists some $0<r<n$ such that we know:

$M_{1,1}$ is a $r\times r$ matrix,

$M_{2,1}$ is a $(n-r)\times r$ matrix,

$M_{1,2}$ is a $r\times (n-r)$ matrix, and

$M_{2,2}$ is a $(n-r)\times (n-r)$ matrix.

And the final matrix of $M_{2,2} - M_{2,1} M_{1,1}^{-1} M_{1,2}$ is of course $(n-r)\times (n-r)$ in size.

Since we know $M$ is positive definite we know that for all $x\in\mathbb R^n$ and $x\neq 0$ that:

$$x^TMx=\left[\begin{array}{cc} x_{1} & x_{2} \end{array}\right] \left[\begin{array}{cc} M_{1, 1} & M_{1, 2}\\ M_{2, 1} & M_{2, 2} \end{array}\right] \left[\begin{array}{c} x_{1} \\ x_{2} \end{array}\right] =x_1^T\cdot \left[\begin{array}{c} M_{1,1}x_{1}+M_{1,2}x_2 \\ \end{array}\right]+x_2^T\cdot \left[\begin{array}{c} M_{2,1}x_{1}+M_{2,2}x_2 \\ \end{array}\right]>0$$

Where the first dot product is one $r$ terms and the second of $n-r$ terms for a total of $n$ terms.

Since we know that $M_{1,1}$ and $M_{2,2}$ are symmetric, we know their inverses are symmetric as well but past that I have no idea how to tackle this.

It's been a long time thinking about this so the rest of this might have some errors I didn't catch yet, especially since I don't have much experience with subblocked matrices, but I'm looking at doing this:

$$x_1^T\cdot \left[\begin{array}{c} M_{1,1}x_{1}+M_{1,2}x_2 \\ \end{array}\right]+x_2^T\cdot \left[\begin{array}{c} M_{2,1}x_{1}+M_{2,2}x_2 \\ \end{array}\right]=\left[\begin{array}{c} x_1^TM_{1,1}x_{1}+x_1^TM_{1,2}x_2 \\ \end{array}\right]+ \left[\begin{array}{c} x_2^TM_{2,1}x_{1}+x_2^TM_{2,2}x_2 \\ \end{array}\right]>0 $$

and I believe there is something interesting regarding $x_1^TM_{1,1}x_{1}$ and $x_2^TM_{2,2}x_2$.

DoubleV
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    Your are not "given that $M_{1,1}$ and $M_{2,2}$ are also symmetric": it is a consequence of the fact that $M$ is symmetric. Another consequence is that $M_{1,2}$ and $M_{2,1}$ are transpose of eachother. It should now be obvious to you that $M_{2,2} - M_{2,1} M_{1,1}^{-1} M_{1,2}$ is symmetric. – Anne Bauval Jan 30 '23 at 07:25

1 Answers1

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First note that the matrix $N:=M_{2,2} - M_{2,1} M_{1,1}^{-1} M_{1,2}$ is well-defined, i.e. that $M_{1,1}$ is invertible. This is due to the fact that $M_{1,1}$ is positive definite.

Now, proving that $N$ is symmetric is easy (see comment).

To prove that it is positive definite, apply the result of your calculation to the particular case $$x_1=-M_{1,1}^{-1}M_{1,2}x_2.$$

Anne Bauval
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