Assume we have the matrix equation:
$(U^{\top} A) (Q \Lambda Q^{\top}) (A^{\top} U) = \Sigma$
such that all matrices in this equation have real values and (d > n):
$U$ is $d \times d$ and is orthonormal
$A$ is $d \times n$
$Q$ is $n \times n$ and is orthonormal
$\Lambda$ is $n \times n$ and is diagonal (with only non-negative values)
$\Sigma$ is $d \times d$ and is diagonal (with only non-negative values)
Is there a way to extract the matrix $U^{\top} A$ if we know: $\Sigma$, $\Lambda$ and $Q$ but don't know $A$, $U$? (I think this can be phrased more "mathematically" as a question about a uniqueness of a solution, but not sure how.)
I wouldn't mind getting $U^{\top} A$ up to an invertible linear transformation, as long as this transformation is a function of $U$ only. What if $A$ contains $n$ orthonormal vectors of length $d$? I think this has an easier solution.
Here is a SOLUTION which I think is correct:
Here is my conclusion, I am not sure if it is completely correct.
Set $F = U^{\top} A Q$. Then we have an equation of the form:
$F \Lambda F^{\top} = \Sigma$ where $\Lambda$ and $\Sigma$ are diagonal. To solve for $F$, we note that a solution would be a diagonal matrix that equals:
$F = \Sigma^{1/2} \Lambda^{-1/2}$. (I am assuming $d = n$ for simplicity, this can be easily generalized to any d > n).
So we have:
$U^{\top} A Q = \Sigma^{1/2} \Lambda^{-1/2}$ and therefore,
$U^{\top} A = \Sigma^{1/2} \Lambda^{-1/2} Q^{\top}$.
The solution, I am guessing is very much not unique (for $U^{\top} A$).
I just found a specific $F$ for which it works, there could be others.