Take the 2-minute tour ×

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.

share|improve this question

migrated from mathematica.stackexchange.com Nov 5 '12 at 11:30

This question came from our site for users of Mathematica.

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

1 Answer 1

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).

share|improve this answer
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

Your Answer


