1. ## Schur Complement problem

Let M be a real symmetric and positive definite matrix with blocks A, Bt, B and C.

$M= [[A,B^{t}] ; [B,C]]$

where $A$ is a $p\times p$ matrix; $B$ is $q\times p$; and $C$ is $q\times q$.

Let $S=C-BA^{-1}B^{t}$ be the Schur complement. We prove that S is symmetric positive definite.

I can prove that S is symmetric but I am having trouble proving that it is positive definite.
I know that for S a symmetric matrix, S positive definite is equivalent to say that all eigen values of S are positive.

I guess my question is how do we prove that all eigen values of S are positive?

2. Originally Posted by math8
Let M be a real symmetric and positive definite matrix with blocks A, Bt, B and C.

$M= [[A,B^{t}] ; [B,C]]$

where $A$ is a $p\times p$ matrix; $B$ is $q\times p$; and $C$ is $q\times q$.

Let $S=C-BA^{-1}B^{t}$ be the Schur complement. We prove that S is symmetric positive definite.

I can prove that S is symmetric but I am having trouble proving that it is positive definite.

I know that for S a symmetric matrix, S positive definite is equivalent to say that all eigen values of S are positive.

I guess my question is how do we prove that all eigen values of S are positive?
I think it's easier here to describe positive definiteness in terms of inner products rather than eigenvalues. If $\langle x,y\rangle$ denotes the inner product in $\mathbb{R}^p$, and $A$ is a $p\times p$ matrix, then A is positive definite if (and only if) $\langle Ax,x\rangle\geqslant0$ for all x in $\mathbb{R}^p$.

The condition for M to be positive definite is thus $\textstyle \left\langle \bigl[ \begin{smallmatrix} A&B^{\textsc t}\\B&C \end{smallmatrix}\bigr]\bigl[{x\atop y}\bigr],\bigl[{x\atop y}\bigr]\right\rangle = \langle Ax,x\rangle + \langle B^{\textsc t}y,x\rangle + \langle Bx,y\rangle + \langle Cy,y\rangle \geqslant0$ (for all x in $\mathbb{R}^p$ and all y in $\mathbb{R}^q$). Now take $x = -A^{-1}B^{\textsc t}y$ in that inequality and simplify the result. (You will need the property of an inner product, that $\langle Bx,y\rangle = \langle x,B^{\textsc t}y\rangle$.) You should end up with the inequality $\langle Sy,y\rangle \geqslant0$, which shows that S is positive definite.

3. ## Schur Complement problem

Thanks a lot, this was very helpful.

When I used that specific value of x, it was very straightforward to get to

I actually didn't need to use the property that