If we have two matrices $M,N$ that relate two non-zero vectors $u,v$ according to $$\begin{align} u &= Mv \\ u &= Nv \end{align}$$ what can be concluded about $M$ and $N$?
According to this question:
If two matrices both multiplied by the same vector are equal are the matrices equal?
we unfortunately cannot conclude they are equal.
Can we conclude that their determinants are the same?
I do not think so, because maybe $\det(M)\ne 0,\det(N)=0$ with $v$ not in the null space of $N$.
But maybe we can say this weaker statement? The matrices are equal or at least one of the matrices has a determinant of zero.
Also, because $(M-N)v = 0$, it looks like we can at least conclude $v$ is in the null space of $(M-N)$.
Looking at the norm of $u$, we have: $$u^T u = v^T M^T M v = v^T N^T N v = v^T M^T N v = v^T N^T M v.$$ Can we somehow use this to conclude something stronger about $M^T M, N^T N, M^T N, N^T M$ ?
It feels really surprising to me that we can conclude so little about $M$, $N$, from what appears to be a strong relationship. So I'm hoping there is something useful for linear equation solving that I'm missing.
A simple counterexample: $$M = \begin{bmatrix}0&1\\1&0\end{bmatrix}, N=\begin{bmatrix}0&-1\\1&0\end{bmatrix} \\ M\begin{bmatrix}1\\0\end{bmatrix} = N\begin{bmatrix}1\\0\end{bmatrix} = \begin{bmatrix}0\\1\end{bmatrix}.$$ You only have one pair of vectors providing a constraint, which doesn’t make for a particularly “strong” relationship, even in a two-dimensional space. In the above example, for instance, the second columns of both $M$ and $N$ can be anything, and you will still have $Mu=Nu=v$. With larger matrices, you will have even more degrees of freedom. The same goes for the various inner products that you’re examining: equality for only one specific pair of vectors doesn’t provide as much information as one might think.