How to get matrix $A$ from $A^\top A=B$ with given symmetric matrix $b$? Given a symmetric matrix $B \in \mathbb{C}^{n\times n}$.
How many coefficients of  $A \in \mathbb{C}^{n\times n}$ can you obtain from the following equation?
$$A^\top A=B$$
I think this problem is under determined? Isn't it? Sure $B$ must be symmetric. Thus  only $\frac{n(n+1)}{2}$ complex equations remain for $n\times n$ coefficients.
 A: Not completely sure about complex numbers, but... If $A^\top A=B$ then, $B$ is symmetric positive semi-definite. So you can take the eigenvalue/vector decomposition of $B$, such that $B=U\Lambda U^\top$.
Then $A=(U\Lambda^{1/2})^\top$.
A: From matrix analysis, if $B$  is complex-symmetric, there is a unitary matrix $U$ such that $B=UDU^{T}$ where the columns of $U$ are eigenvectors of $BB^{*} = B \bar{B}$ and $D$ is diagonal and the entries are the positive square roots of the corresponding eigenvalues. Now, if the columns of $U$  are real, then $U$ is orthogonal and so $B$ is orthogonally diagonalisable, and what I wrote then follows. So, if the eigenvectors of $B \bar{B}$  are real, then the eigenvalues of $B$ are real. The last part here may not be of any use but I think the former is.
A: The davcha and Ian posts are false when $\mathbb{C}$ is the underlying field. Of course, the OP gave a green chevron to davcha, although his solution is valid only on R....


*

*Any complex matrix is similar to a symmetric matrix. Then a symmetric matrix may be non-diagonalizable !

*The real case is treated in:    General Cholesky-like decomposition   . cf. my post (which has not been read) where all solutions are determined.

*If $A$ is a complex matrix , then the eigenvalues of $A^TA$ have no special property and $B$ is certainly not symmetric positive semi-definite.

*About the OP's question. Using the Autolatry's post, we can find particular solutions. Indeed there is a SVD of $B$ in the form $B=UDU^T$ (beware $U^T\not= U^{-1}$). Let $\Delta$ be one of the square roots of $D$. Then the $A=\Delta U^T$ are particular solutions. Yet, there are many other solutions. For example, for a generic symmetric $B$, if $n=3$ (resp $n=4$) then the degree of freedom of the solutions is $3$ (resp. $6$).
