I am reading a book and the author provides the following relationship without much explanation:
$$E(x'Axx'Bx) = E(\operatorname{tr}(x'Axx'Bx)) = E[\operatorname{tr}(\left(x' \otimes x'\right)\left(A \otimes B\right)\left(x \otimes x)\right)]$$
I understand the introduction of the trace since $x'Axx'Bx$ is $1\times1$, but I do not understand the third part.
I guess the underlying question is this:
How do I show that $x'Axx'Bx = (\left(x' \otimes x'\right)\left(A \otimes B\right)\left(x \otimes x)\right)?$
$$ (x' \otimes x')(A \otimes B) (x \otimes x) = (x'Ax) \otimes (x'Bx). $$ The expression on the right is a $1\times1$ matrix whose only entry is the scalar $(x'Ax)(x'Bx)$.