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For any matrix A, does there exist a matrix decomposition such that:

$$ A = Z^T Z $$

A is not necessarily positive definite, so the Cholesky decomposition does not apply.

My motivation for asking this question is because I believe it will allow me to solve the system below for both A and b: $$ A^TA = X^TX+Y^TY\\ A^Tb = X^TP+Y^TQ\\ $$

where all symbols are matrices, except b is a vector.

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migrated from Nov 5 '12 at 11:30

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you ask for "any matrix A" - but if A is not symmetric, then it has no cholesky-factorization. So you seem to assume, A is not "any" but is symmetric. Second, symmetric but non-semi-positive definite matrices A can have a cholesky decomposition if you use complex numbers. This might be helpful when your analytical formulae involve more steps and intermediate steps cannot be expressed otherwise, even if there will be a real over-all result... – Gottfried Helms Nov 5 '12 at 12:05
up vote 6 down vote accepted

If $A=Z^TZ$ then $A^T=A$ and $x^TAx=\|Zx\|^2\geqslant0$ for every $x$, hence $A$ is positive semi-definite. In the other direction, Cholesky factorizations for positive semi-definite matrices always exist (only, they may not be unique).

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Is a corollary of this: if A is not positive semi-definite, then no such decomposition exists? – carl Nov 5 '12 at 11:42
Yes. You have such a factorization if and only if it is positive semi-definite. – anonymous Nov 5 '12 at 11:47

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