Matrix identites, Derivative, Determinant, and a kind of Duality involving Traces

Question

I am reading the blog entry Matrix identities as derivatives of determinant identities by Terence Tao, everything is quite clear, up to ~1/3 of the text, where he goes

[...] we conclude that $$ \operatorname{tr}(\begin{pmatrix} A & B \\ C & D \end{pmatrix}^{-1}\begin{pmatrix} 0 & 0 \\ 0 & H \end{pmatrix} ) = \operatorname{tr}((D - CA^{-1} B)^{-1} H). $$ As $H$ was an arbitrary $m\times m$ matrix, we conclude from duality that the lower right $m\times m$ block of $\begin{pmatrix} A & B \\ C & D \end{pmatrix}^{-1}$ is given by the inverse $(D - CA^{-1}B)^{-1}$ of the Schur complement.

In most steps before this he is quite clear about what he is using, but here I do not understand. What kind of duality is he referring to, and in what sense does the result follows? If I have an identity between traces, what does this imply about the inverse matrix?

Answer 1

The duality he refers to is the non-degenerateness of the bilinear map $(A,B)\mapsto tr(AB)$. A bit more holds: The bilinear map $(A,B)\mapsto tr(A^T B)$ is positive definite. In fact: It is the scalar product associated to the Frobenius matrix norm.

Anyway: Whenever one has a non-degenerate bilinear form and a statement of the form $\forall H:\langle x,h\rangle = \langle y,h\rangle$, then one can conclude $x=y$ which is exactly what is happening here.

Answer 2

It is the duality associated to the lower right block. The idea is to fix the three other blocks -there is a hidden derivative-

In fact we can directly reason as follows: let $\begin{pmatrix}A&B\\C&D\end{pmatrix}^{-1}=\begin{pmatrix}U&V\\W&X\end{pmatrix}$; then, for every $H$, $tr(\begin{pmatrix}A&B\\C&D\end{pmatrix}^{-1}\begin{pmatrix}0&0\\0&H\end{pmatrix})=tr(XH)= tr((D - CA^{-1} B)^{-1} H)$.

EDIT. To obtain $X=(D - CA^{-1} B)^{-1}$, it suffices to choose $H=(X-(D - CA^{-1} B)^{-1})^T$.