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?


1 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.

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.

  • $\begingroup$ Dear @zhoraster, thanks for the reply and sorry for my delay in accepting the answer. The proof of Boyd's book for the finite dimensional case uses the fact that $u = -A^{-1}Bv$ is the minimizer of $$u^{T}Au + 2v^{T}B^{T}u + v^{T}Cv.$$ Could you give more details of how this is derived (in the infinite dimensional case) or indicate to me a reference that deals with optimization problems in Hilbert (or even Banach) spaces? $\endgroup$
    – shamisen
    Commented Oct 6, 2015 at 1:48
  • $\begingroup$ @shamisen, here you are. $\endgroup$
    – zhoraster
    Commented Oct 6, 2015 at 15:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .