Solution of vector equation in alpha

Question

Is there a closed solution for $\alpha$ in the following vector equation?

$$(\vec\alpha^T\Sigma\vec\alpha)\vec\mu - (\vec\alpha^T\vec\mu)\Sigma\vec\alpha = \vec0$$

$\Sigma$ is an $n$ by $n$ symmetric correlation matrix.

$\vec\alpha$ is an $n$ by $1$ variable column vector.

$\vec\mu$ is an $n$ by $1$ column vector.

Answer 1

I assume that $\Sigma$ is invertible, $\mu$ is non-zero, and the equation has a non-zero solution.

We begin with $$ (\alpha^T\Sigma \alpha) \mu = (\alpha^T \mu) \Sigma\alpha $$ We see that this equation can only have a solution if $\Sigma \alpha$ is a multiple of $\mu$. That is, there exists a scalar $k$ for which $$ k\mu = \Sigma \alpha \implies \alpha = k\Sigma^{-1}\mu. $$ It suffices to plug this into the original equation and solve for $k$. We have $$ (\alpha^T\Sigma \alpha) \mu = (\alpha^T \mu) \Sigma\alpha \implies\\ k^2(\mu^T\Sigma^{-1} \mu) \mu = k^2 (\mu^T\Sigma^{-1} \mu) \mu. $$ Both sides are the same. In other words, the equation holds for all $k$.

So, if $\Sigma$ is invertible and a non-zero solution exists, then $\alpha$ is a solution iff it has the form $$ \alpha = k \Sigma^{-1}\mu $$ for any $k \in \Bbb R$.