# What is the Cholesky Decomposition used for? [closed]

It seems like a very niche case, where a matrix must be Hermitian positive semi-definite. In the case of reals, it simply must be symmetric.

How often does one have a positive semi-definite matrix in which taking it's Cholesky Decomposition has a significant usage?

## closed as unclear what you're asking by Axoren, Claude Leibovici, haqnatural, Daniel W. Farlow, user98602 Sep 29 '16 at 17:05

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• That is not at all a niche case. Hermitian positive semidefinite matrices are everywhere and are so important. For example, if $A$ is any real matrix, then $A^T A$ is positive semidefinite. Try reading some of Gilbert Strang's books that emphasize the ubiquity of $A^T CA$. – littleO Sep 28 '16 at 1:27
• @littleO Could you give a real world practical example in which the Cholesky decomposition of $A^TA$ has a practical application? I've had someone try to convince me with this line of thought but I feel like that alone is incomplete. Generally, if I'm working with $A$ as a linear transformation, $A^T A$ will have a different meaning. – Axoren Sep 28 '16 at 1:30
• The focus seems to be on whether positive definite symmetric matrices are commonly needed (they are). The Cholesky factorization will likely fail for semi-definite but not definite forms. – hardmath Sep 28 '16 at 1:40
• I will have to think on both given answers for a while. These are definitely cases in which we can be certain that we have a matrix matching the criteria. They don't immediately answer why the Cholesky decomposition benefits them, however. The answers seemed to focus on where $A^TA$ will appear, rather than what the Cholesky factorization brings to the table in those cases. "Why would you do Cholesky factorization in this case?" has not been expressly addressed by either question. – Axoren Sep 28 '16 at 2:01
• I understood your wording of the Question to focus on whether applications permitting a Cholesky decomposition are really common in practice. I considered posting some remarks about the advantages of this factorization, but concluded it was not of concern to you. – hardmath Sep 28 '16 at 22:27

Positive semidefinite matrices are everywhere, largely due to the ubiquity of the pattern $A^T C A$ (which is emphasized by Gilbert Strang). For example, in least squares problems we find $x$ to minimize $(1/2)\|Ax - b\|_2^2$. To solve this problem, set the gradient equal to zero: $$A^T(Ax- b) = 0.$$ This is just one of many examples.