TITLE: Finding a matrix from its product with its transpose
QUESTION [7 upvotes]: Suppose $A$ is a $3 \times 3$ matrix. If $A A^T = B$ and $A^T A = C$, where $B$ and $C$ are known and $B \neq C$, can I uniquely determine A?
$A$ has 9 independent elements. Since $B$ and $C$ are symmetric, they have 6 independent entries each. Thus I have an overdetermined nonlinear system of equations with 9 variables and 12 equations. Does a unique solution exist for this system? How can I prove that it does or doesn't?

REPLY [2 votes]: After requiring $B \neq C$ still no.
Consider $A =\begin{pmatrix}0 & a & 0 \\ b & 0 & 0 \\ 0 & 0 & c\end{pmatrix}$.
This satisfies your requirements, and yet the only non-zero elements of $AA^T$ and $A^T A$ are $a^2, b^2$ and $c^2$, so their sign cannot be determined.
Conjugation of $A \to O A O^T$ with any orthogonal will maintain this structure and generate a larger family of such cases.