Suppose $x,y$ are vectors and $A$ is a symmetric invertible matrix. Show that $(x'Ay)x=(xx'A)y$.
How can one prove the above? I am aware that matrix multiplication is not necessarily associative. But it seems associativity has been used here.
Is it true that $\dfrac{(A^{1/2}xx'A^{1/2})x}{x'Ax}=x$? Here $x$ is a vector, and $A$ is a symmetric pd matrix.
I think this is true and "somehow" follows from the previous result. But how does it follow?
$$(x'Ay)x = x(x'Ay)$$
the result is obvious.
O/w,
$x'Ax$ is an eigenvalue of $A^{1/2}xx'A^{1/2}$.
This can be seen by defining $a := (x'A^{1/2})'$ s.t.
$$A^{1/2}xx'A^{1/2} = (x'A^{1/2})'x'A^{1/2} = aa'$$
$$x'Ax = x'A^{1/2}A^{1/2}x = (A^{1/2}x)'(A^{1/2}x) = aa'$$
Is $a'a$ an eigenvalue of $aa'$? I believe so. Corresponding eigenvector is $x = a$ as can be seen with:
$$(aa')a = (a'a)a$$
which follows the same idea as in #1.