Is there an analytic solution to this matrix equation?
$$K(\vec{\mu}-\vec{x})=||\vec{\mu}-\vec{x}||^2\textbf{C}(\vec{\mu}-\vec{x})$$
$\textbf{C}$ is a symmetric positive semi-definite matrix of dimension $n\times n$, $\vec{x}$ and $\vec\mu$ are vectors or dimension $n$ and $K$ is a constant. I am looking for a solution for $\vec{x}$, $\vec\mu$ is simply a constant.
Without the norm term $||\vec{\mu}-\vec{x}||^2$ the problem would be very straight forward, but in this particular problem I am not sure what to do.
Is this a case of non linearity and there isn't an analytical approach to solve this? Or am I just missing something?
Let $v = \mu - x$. We can express $v$ as $v = rw$ with $||w||=1$ and $r\geq 0$. Then we can write the equation as
$$Kv=r^2Cv$$
Obviously $v=0$ (so $\mu = x$) is a solution, so let us assume $v \neq0$ i.e. $r>0$. Then we can rewrite the equation as
$$Cv = \frac{K}{r^2} v$$
or after canceling the $r$s
$$Cw = \frac{K}{r^2} w$$
This means $v$ (and $w$) is an eigenvector of $C$, let us say with eigenvalue $\lambda$. If $K=0$ then we are looking for eigenpairs $(w,\lambda)$ with $\lambda=0$, otherwise we need $\lambda \neq 0$. In the case $\lambda=0$ we can choose $r$ arbitrarily. If $\lambda \neq 0$ then $r$ is determined by $r = \sqrt{K/\lambda}$, with that $v$ is determined (up to sign) and so is $$x = \mu -v.$$