Extract matrix from this form of linear algebra term? $$(Mu)\cdot(Mv)$$
In this form, I don't like the fact that matrix $M$ is on both sides of the dot product operator. Is there any way to extract matrix $M$ so I can get something nice along the line of $...(u\cdot v)...$?
 A: This can be done essentially through using the spectral theorem and the fact that when you compute the Euclidean dot product we essentially have
$$
(Mx)\cdot (My) \;\; =\;\; x^TM^TMy.
$$
We can extract the matrix $M^TM$ and note that this is a symmetric positive semi-definite matrix, hence we can use the spectral theorem to find that
$$
A \;\; =\;\; M^TM \;\; =\;\; VDV^T
$$
where $D$ is diagonal with the eigenvalues of $A$ and $V$ is orthogonal.  Note that since  $A$ is semi-definite, then the matrix $D$ has eigenvalues that are either $0$ or positive.  It therefore makes sense to decompose $D = \sqrt{D}\sqrt{D}$ where we take the square root of the entries.  We therefore find that we can write
$$
A \;\; =\;\; V\sqrt{D}\sqrt{D}V^T \;\; =\;\; \left (\sqrt{D}V^T\right )^T\left (\sqrt{D}V^T\right ).
$$
What we find here as a result is that $M = \sqrt{D} V^T$, however this is only unique up to the ordering of the eigenvalues and eigenvectors.  Therefore there isn't a single matrix $M$ that accomplishes this, but there is a family of matrices generated by permutations.
