I'm having a dickens of a time solving a particular matrix problem:
$$M^TB^T \Sigma B - 2 \Sigma BMB = 0$$
$M$ is a 1 by k matrix, with feature means.
$B$ is a k by 1 matrix, with feature coefficients.
$\Sigma$ is a k by k co-variance (symmetric) matrix.
I'm trying like the dickens to solve for $B$, but given the arrangement of the matrices, it's looking impossible, that could be just my limited knowledge of matrices though.
This isn't a complete answer to get $B$, but I think it's relevant for the discussion.
Let $m$ and $b$ be the (column) vectors associated with $M$ and $B$. Note that the quantities $B^T \Sigma B$ and $MB$ are scalars, given by $b \cdot \Sigma b$ and $m \cdot b$, respectively. Moving these scalars to the left of the vectors they multiply, your equation is $$ (b \cdot \Sigma b) m = 2(m \cdot b) \Sigma b, $$ which is an equality of vectors. In particular, if this equality holds, it should hold when we dot both sides with $b$, which gives $$ (b \cdot \Sigma b) (m \cdot b) = 2(m \cdot b) (\Sigma b \cdot b). $$ If the quantities in parentheses are both nonzero, then they cancel, leaving us with the nonsensical equation $1 = 2$. So at least one of them must be $0$.
