$x,y$ are vectors and $\Lambda$ is a symmetric square matrix, $y = \frac{x}{\sqrt{x^T \Lambda x}}$. $y$ is known, solve for vector $x$
This would be trivial for scalar cases, but somehow I struggle to wrap my head around for vectors and matrixes. We can get $y^Ty = \frac{x^Tx}{x^T \Lambda x}$, but how do we isolate $x$ out?
On
$$\begin{aligned} y = \frac{x}{\sqrt{x^T \Lambda x}} &\implies x = \lambda y \implies \big(1-\frac{\lambda}{|\lambda|\sqrt{y^T\Lambda y} }\big)y = 0\implies\operatorname{sign}(\lambda) = \sqrt{y^T\Lambda y} \end{aligned}$$
But then (within the reals) a solution exists if and only if $y^T\Lambda y=+1$. In this case $x=cy$ is a solution for any $c>0$.
One approach: Since $\Lambda$ is symmetric, it has real eigenvalues and an orthonormal eigenbasis. Let the eigenvectors be $\{\vec{v}_1,\dots,\vec{v}_n\}$ and the cooresponding eigenvalues $\{\lambda_1,\dots,\lambda_n\}$.
Now, we write $$ x=\sum a_i\vec{v}_i\qquad\text{and}\qquad y=\sum b_i\vec{v}_i. $$ Here, we think of the $a_i$'s as variables and the $b_i$'s as known constants. Now, we observe that $$ \Lambda x=\sum a_i\lambda_i\vec{v}_i. $$ Since the eigenvectors are orthonormal, $$ x^T\Lambda x=\sum a_i^2\lambda_i. $$
Putting this all together, we have that $$ \sum b_i\vec{v}_i=\sum\frac{a_i}{\sqrt{\sum a_i^2\lambda_i}}\vec{v}_i. $$ In other words, we must simultaneously solve the system $$ b_i=\frac{a_i}{\sqrt{\sum a_i^2\lambda_i}} $$ for all $i$.
Since the denominator is the same across all $i$'s, it's really just some constant scaling factor. This implies that $a_i=\mu b_i$ for all $i$. Plugging this in, we get $$ b_i=\frac{\mu b_i}{|\mu|\sqrt{\sum b_i^2\lambda_i}}=\operatorname{sign}(\mu)\frac{b_i}{\sqrt{\sum b_i^2\lambda_i}}. $$ Therefore, we conclude that if this equation has a solution, it must be that $x=y$ is a solution (and that $y^T\Lambda y=1$). In fact, since the formula is homogeneous of degree zero, $x$ could be any positive multiple of $y$.