Let $\boldsymbol{X}$ be an $n \times m$ matrix and $\boldsymbol{A}$ be an $n \times n$ invertible matrix with $n > m$. I'm trying to find a solution to $$\boldsymbol{X}^\top \boldsymbol{A} \boldsymbol{X} = \boldsymbol{I}_m$$ in a situation where I know that at least one (unknown to me) solution - call this $\boldsymbol{X}^*$ - must exist. I also know that $\boldsymbol{X}^*$ cannot be the only solution since for any $m \times m$ orthogonal matrix $\boldsymbol{Q}$, $\boldsymbol{Z} = \boldsymbol{X}^*\boldsymbol{Q}$ also solves this equation.
Edit:
- Just to be clear $\boldsymbol{A}$ may not be positive definite.
I might have something that works when $\text{eig}(A)>0$. Here we know that $$Av = \lambda v$$ As $v$ is a unitvector, this can be alternatively expressed as $$v^TAv = \lambda$$ which suggests that if $m=1$, then $X = \frac{1}{\sqrt{\lambda}}v$. For $m>1$, this can be easily expanded as follows: $$X = \begin{bmatrix}v_1 & v_2 & \ldots & v_m\end{bmatrix}\begin{bmatrix}\sqrt{\lambda_1} & & \\ & \ddots & \\ & & \sqrt{\lambda_m}\end{bmatrix}^{-1}$$
However, if the eigenvalues of $A$ are not strictly positive, it will not yield an identity, but some elements on the diagonal become -1 instead. Maybe an additional multiplication with the imaginary element $i$ can solve that for these eigenvalues.
Maybe this helps forming a formal solution