Schur complement condition for positive definiteness of operators

Question

To verify if a symmetric block matrix is positive definite, one can check the definiteness of its diagonal blocks and the Schur complement of the respective blocks.

Is this also true in the infinite dimensional setting?

Precisely, being $A$, $B$ and $C$ be linear bounded operators defined on a Hilbert space $\mathcal{H}$, it is true that if $C$ is invertible and $C^{-1}$ is also a bounded linear operator, then the operator block matrix

$$ \begin{bmatrix} A & B^{*} \\ B & C \end{bmatrix} $$

on $\mathcal{H} \oplus \mathcal{H}$ is positive if and only if $C$ is positive and $A - B^{*} C^{-1} B$ is positive?

Answer 1

If $C$ is invertible, yes. The proof in the wikipedia article you quote does not use any finite-dimension-specific steps, so it works in the infinite dimension as well.

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Now why $u_0 = - A^{-1} B^* v$ is the minimizer of $f(u) = (u,Au) + 2(v,B u)$ (I omit the last term, which is independent of $u$; also there is the adjoint $B^*$, not $B$, in the formula for $u_0$). This is quite easy: just "complete a square", i.e. $$ f(u) = \big(u-u_0, A(u-u_0)\big) - (u_0,A u_0) +(u,Au_0) + (u_0,A u) + 2(v,Bu)\\ = \big(u-u_0, A(u-u_0)\big) - (u_0,A u_0) -2 (u, B^*v) + 2(v,Bu) \\ = \big(u-u_0, A(u-u_0)\big) - (u_0,A u_0). $$ In particular, $f(u_0) = - (u_0,A u_0)\le f(u)$ since $A$ is positive definite (note that in Boyd and Vanderberghe, the roles of $A$ and $C$ are opposite to your question).

I don't know what is there for Banach spaces, but you cannot define positive definiteness there, since the dual space is different.