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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!


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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.

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Thanks so much for you answer! – Juan Carlos Castro Nov 28 '12 at 0:48

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!

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Nice factorisation. +1 – user1551 Jun 24 '13 at 8:54

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 $T$.

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@Ivan Does the spectral theorem really apply here? $A$ is a complex symmetric matrix, not (necessarily) a real symmetric matrix or a complex Hermitian matrix. – user1551 Nov 28 '12 at 1:12
Oh, that's right. Scuse me. – Ivan Nov 28 '12 at 8:37

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