How can we prove that the Schur complement S is positive definite?

[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?

 

[Nino MMP]

Let us assume that S is not positive definite. Let λ be an eigenvalue of S with corresponding eigenvector x such that Sx=λx. Then, we have (C−BA^{−1}B^{t})x=λx. By multiplying both sides by A^{−1}, we get A^{−1}(C−BA^{−1}B^{t})x=A^{−1}λx. Since A is positive definite, we can further write this equation as (A^{−1}C−B^{t})x=A^{−1}λx. Multiplying both sides by B, we get B(A^{−1}C−B^{t})x=B(A^{−1}λx). By substituting A^{−1}C=M−B^{t}, we have BMx=B(A^{−1}λx). Finally, by multiplying both sides by A, we obtain Mx=Aλx. This shows that λ is an eigenvalue of M which is a contradiction to our assumption that M is positive definite with all its eigenvalues being positive. Therefore, the only possible conclusion is that S must be positive definite.