$AA^T = BB^T \rightarrow A =BU $

Question

Is it true that $AA^T = BB^T$ always implies $A =BU $ for some unitary matrix $U$. Or there may exist other scenarios also? All matrices are real and square.

Answer 1

This should work in general, although I think it is overkill if $A$ and $B$ are invertible:

The entry at position $(i,j)$ of $AA^T$ is the scalar product of the $i$th row of $A$ with the $j$th row. So the equality of these matrices shows that these scalar products are always the same, both for the rows of $A$ and the rows of $B$. As the scalar product is bilinear, we can extend this result to sums and coefficients.
This gives us a linear map $$\varphi : rs(A) \to rs(B),$$ where $rs$ denotes the row space such that $\varphi$ is compatible with the scalar product. Using that the zero vector is the only vector of length $0$, you can show that $\varphi$ is bijective, i.e. an isometry.

Now using a real version of Witt's theorem, you can extend $\varphi$ to an isometry $\Phi$ on the whole space. But every isometry $\Phi$ on the whole space is given by an orthogonal matrix (or a unitary matrix with real entries if you want).