TITLE: Can every symmetric regular complex matrix be decomposed into a product of a matrix times its transpose?
QUESTION [5 upvotes]: Let $A\in GL(n,\mathbb{C})$ be symmetric. Is there a always a $B\in GL(n,\mathbb{C})$, such that $A=B^TB$?

REPLY [4 votes]: Yes. On any finite-dimensional vector space over a field of characteristic $\ne2$, every symmetric bilinear form is diagonalisable by congruence. In particular, this is true when the underlying field is $\mathbb C$. So, if $A=VDV^T$ is such a diagonalisation, we may take $B=\sqrt{D}V$ where $\sqrt{D}$ is any entrywise square root of $D$. Since $A$ is assumed to be invertible, $B$ must be invertible too.
Alternatively, every complex symmetric matrix $A$ has a Takagi factorisation $A=VDV^T$, where $V$ is (complex) unitary and $D$ is nonnegative and diagonal. (Tagaki factorisation is a special form of singular value decomposition. Note that $V$ is unitary but not necessarily orthogonal; we have $VV^\ast=I$ but not necessarily $VV^T=I$.) Hence $A=B^TB$ where $B=\sqrt{D}V^T$ and $\sqrt{D}$ is the usual entrywise square root of $D$. Again, since $A$ is invertible, $B$ must be invertible too.