Can every symmetric regular complex matrix be decomposed into a product of a matrix times its transpose? Let $A\in GL(n,\mathbb{C})$ be symmetric. Is there a always a $B\in GL(n,\mathbb{C})$, such that $A=B^TB$?
 A: 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.
A: An special case is proved in [Gantmacher, Matrix theory, chapter II, §4, Cor.3]:
If $A$ has the rank $r$ and its first $r$ leading principal minors are non-zero, then $A=B^TB$ where $B$ is upper triangular.
A: The answer is yes and it can be proven by induction on $n$. First a few notes though:
Note that if $P^\mathsf{T}AP = B^\mathsf{T}B$ for any invertible matrix $P$, then $A = (BP^{-1})^\mathsf{T}(BP^{-1})$.
Secondly, note that if we prove the proposed decomposition for $A,B\in\mathbb{C}^{n\times n}$, equating the determinants shows that $B$ must be invertible whenever $A$ is and the proposed result follows.
Now we start the induction:
For $n=1$, the answer is clear, just let $b$ be a square root of $a$.
Suppose $A=B^\mathsf{T}B$ for all $A\in\mathbb{C}^{(n-1)\times(n-1)}$ and write $A\in\mathbb{C}^{n\times n}$ as $\begin{pmatrix}\alpha & \beta^\mathsf{T} \\ \beta & \Delta\end{pmatrix}$, where $\alpha\in\mathbb{C}$, $\beta\in\mathbb{C}^{n-1}$ and $\Delta\in\mathbb{C}^{(n-1)\times(n-1)}$.
The first case is $\alpha=0$, $\beta=\mathbf{0}$ and $\Delta=O$, which is trivially solved by $B=O$.
Secondly, suppose that $\alpha=0$, but $\Delta\neq O$ or $\beta\neq \mathbf{0}$, then consider $P=\begin{pmatrix} 1 & 0 \\ \mathbf{c} & I \end{pmatrix}$,
$$P^{\mathsf{T}}AP = \begin{pmatrix} 1 & \mathbf{c}^{\mathsf{T}} \\ 0 & I\end{pmatrix}\begin{pmatrix}0 & \beta^{\mathsf{T}} \\ \beta & \Delta\end{pmatrix}\begin{pmatrix} 1 & 0 \\ \mathbf{c} & I\end{pmatrix} = \begin{pmatrix} 2\mathbf{c}^{\mathsf{T}}\beta+\mathbf{c}^\mathsf{T}\Delta\mathbf{c} & \beta^{\mathsf{T}}+\mathbf{c}^\mathsf{T}\Delta \\ \beta+\Delta\mathbf{c} & \Delta\end{pmatrix}.$$
Since $\beta$ and $\Delta$ are not both zero there exists some $\mathbf{c}$ such that the upper left hand corner is non-zero.
Thus by the first note this case is reduced to the case $\alpha\neq 0$.
Thirdly suppose $\alpha\neq0$. By the induction hypothesis there is some $D\in\mathbb{C}^{(n-1)\times(n-1)}$ such that $D^\mathsf{T}D=\Delta-\frac{1}{\alpha}\beta\beta^\mathsf{T}$. Define $B=\begin{pmatrix}\sqrt{\alpha} & \frac{1}{\sqrt{\alpha}}\beta^\mathsf{T}  \\ 0 & D \end{pmatrix}$, then
$$B^\mathsf{T}B = \begin{pmatrix} \sqrt{\alpha} & 0 \\ \frac{1}{\sqrt{\alpha}}\beta & D^\mathsf{T} \end{pmatrix}\begin{pmatrix}\sqrt{\alpha} & \frac{1}{\sqrt{\alpha}}\beta^\mathsf{T}  \\ 0 & D \end{pmatrix} = \begin{pmatrix} \alpha & \beta \\ \beta^\mathsf{T} & \frac{1}{\alpha}\beta\beta^\mathsf{T}+D^\mathsf{T}D\end{pmatrix} = A.$$
This concludes the induction step and thereby the proof.
