Matrix Multiplication, Trace and Integration

Question

Let $\omega(x)$ be a $p\times 1$ vector-valued function defined on a random variable $X$ with CDF $F$. Now define $$V:=\int \omega(x)[\omega(x)]^T dF(x).$$ Then define $\gamma$ as follows. $$ \gamma := \sup_x\sqrt{[\omega(x)]^TV^{-1}\omega(x)}. $$ It is clear that $$ \gamma^2 \geq [\omega(x)]^TV^{-1}\omega(x). $$ Then it is claimed that $$ \int [\omega(x)]^TV^{-1}\omega(x) dF(x) = \int \mathrm{trace} \left\{ V^{-1}\omega(x) [\omega(x)]^T \right\} dF(x) = p. (*) $$ I do not get how the two equal signs in $(*)$ was derived. Could anyone help me, please? Thank you!

Answer 1

This uses the cyclical invariance of the trace: $\mathrm{tr}\,ABC=\mathrm{tr}\,CAB$, which you can derive by writing it out with indices and noting that the trace "completes the cycle" of pairwise index summations.

That, together with the fact that you can insert a trace on the left-hand side because it's a scalar, explains the first equals sign, and then the second one readily follows, since the trace of the identity matrix is its dimension $p$.