Decomposition of positive semi-definite matrices I know that any positive semi-definite matrix $A$ can be written as $A = B^T B$. My question is that if we know that $A = B^TB = C^TC$, then is there any relation with $B$ and $C$? I know that it is possible that $B \neq C$ for example:
\begin{equation}
\begin{bmatrix}
1 &0 \\0 &1
\end{bmatrix}\begin{bmatrix}
1 &0 \\0 &1
\end{bmatrix} = \begin{bmatrix}
-1 &0 \\0 &-1
\end{bmatrix}\begin{bmatrix}
-1 &0 \\0 &-1
\end{bmatrix}.
\end{equation}
Thanks,
 A: If $A=B^TB$ take any orthogonal matrix $P$ such that $P^TP=PP^T=I$ then if $C=PB$ one has
$$C^TC=B^TP^TPB=B^TB=A$$
Now assume we have two matrices such that $A=B^TB=C^TC$. If A is definite positive $B$ and $C$ are invertible and we have $B=\left(B^{-1}\right)^TC^TC$. Write $P=\left(B^{-1}\right)^TC^T$ and compute
$$PP^T=\left(B^{-1}\right)^TC^TCB^{-1}$$
If $B$ and $C$ are not invertible use density property of invertible matrices to get to the same result.
A: Let $B,C$ be $m\times n$ matrices. If $B^*B=C^*C$ $(=A)$ then the SVD of $B$ and $C$ are related as
$$
B=U_1\Sigma V^*,\qquad C=U_2\Sigma V^*
$$
that is, they share the matrices $V$ (with the columns being the normalized eigenvectors of $A$) and $\Sigma$ (with $\Sigma^*\Sigma$ being the diagonal form of $A$). It gives immediately that
$$
C=U_2\Sigma V^*=U_2U_1^*U_1\Sigma V^*=U_2U_1^*B=WB
$$
where $W$ is unitary.
P.S. It is straightforward to generalize the relation to the case when $C$ has more rows than $B$, but $W$ will be a rectangular matrix with $W^*W=I$.
