Factorization of an invertible symmetric matrix Given any invertible symmetric matrix:
$A=\begin{bmatrix}a&b&c\\ b&d&e\\ c&e&f\end{bmatrix}$
over the complex number,
Can be it factored as $A=T^\top T$?
where $T^\top$ is the transpose matrix of $T$, for some invertible matrix $T$.
Any suggestions are welcome!
Thanks!
 A: Yes. This is a direct consequence of Takagi's factorisation, which is a special form of singular value decomposition. If $A$ is a complex symmetric matrix, then by Takagi's factorisation, there exists a unitary matrix $U$ such that $A=U\Sigma U^\top$, where $\Sigma$ is a diagonal matrix containing the singular values of $A$. It follows that $A=TT^\top$, where $T=U\Sigma^{1/2}$. Since your $A$ is invertible, $T$ is obviously invertible.
A: Consider the example of
$$ T = \begin{bmatrix} 
\sqrt{\left(a-\frac{c^2}{f}\right) + \frac{(c e-b f)^2}{f ( e^2-d f)} } & 0 & 0 \\ 
\frac{b f - c e}{\sqrt{f ( d f-e^2)}} & \frac{\sqrt{d f-e^2}}{\sqrt{f}} & 0 \\
\frac{c}{\sqrt{f}} & \frac{e}{\sqrt{f}} & \sqrt{f} \end{bmatrix}$$
where if you multiply it out as $T^\top T=A$. Try it!
A: For the Spectral theorem we have a matrix $P$ such that $PAP^t=PAP^{-1}=D$, where $D$ is diagonal and $P \in GL(n)$.
Now $\exists S: D=S^2$, and that's obvious.
So  $PAP^{-1}=S^2 \rightarrow  A = (P^{-1}S)(SP)$.
We've just finished because $(SP)^t=P^tS^t=P^{-1}S$.
So $SP$ is your $T$.
