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Is this always true?

"If the rows of $M$ are a set of non-zero linearly independent vectors, then $M M^T$ is invertible."

CONTEXT: The context is that I am a musician working on a algorithmic composition system. I am using Lagrange multipliers to find a "closest point" on polytope to a given point. This is a computer program, not automated proof system, so I have to carefully refactor the algorithm so that it will reliably and efficiently find the vector I need (especially considering floating point errors). In the course of a long refactoring process, I realized that I had assumed the statement in one step of the refactor. I thought I would just take a second to make sure that this "obvious" fact,was true, because in my experience it is very easy for a non-expert to overlook some consideration that it isn't obvious at first glance, and this is my first time using LaGrange multipliers for optimization, and it is still not obvious to my how they work. But for my purposes, it is sufficient that I know that they work. I think also at the time I had a bug, and was carefully checking every step of the proof, to try to find the bug.

Charles Gillingham
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  • Over the real numbers, yes. Over the complex numbers, not in general. – Angina Seng Aug 11 '17 at 06:05
  • The term is "full row rank". Is that always true? Only if $\rm M$ has full row rank. – Rodrigo de Azevedo Aug 11 '17 at 06:12
  • I don't understand the down votes ... what's wrong with the question? – Charles Gillingham Aug 11 '17 at 08:39
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    Questions without showing efforts gets downvoted regularly on this site. Those who answer your question might get downvoted too. – Siong Thye Goh Aug 11 '17 at 23:29
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    This is a reasonable problem, but in posting a Question that states the problem without any context, you invite Readers to suspect that you have not digested the problem and are merely passing it along for them to analyze. Context can be added in many ways: What motivates the problem? What special cases can you already answer? What difficulty did you encounter in trying a particular approach? – hardmath Aug 12 '17 at 17:17
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    Charles, I see you asked about this question on the meta.math site, blaming it's downvotes on the presumption that you received them because the question is "elementary". None of the comments in existence at the time of your meta post mentioned anything with respect to being elementary. They rather speak of your failure to provide any context wrt your question. Please visit this link: https://math.stackexchange.com/help/asking. It will help answer on meta, and your question about "why downvotes" and also about how to ask a good question, etc. – amWhy Aug 12 '17 at 20:16
  • You can find probably a few related posts here on the site, have a look, for example, here: Gram matrix invertible iff set of vectors linearly independent. Wikipedia article Gramian matrix might also be interesting. – Martin Sleziak Aug 13 '17 at 13:34
  • Probably it would be helpful to clearly state whether $M$ is a square matrix ($n\times n$) or whether it is a $m\times n$ matrix where $m$ and $n$ can be different. – Martin Sleziak Aug 13 '17 at 13:37
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    This post (about non-square case) is rather similar, too: Under what conditions is $AA^T$ invertible? – Martin Sleziak Aug 13 '17 at 13:43
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    That's exactly the same question. You could close this as a duplicate. – Charles Gillingham Aug 14 '17 at 02:31
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    Per the OPs suggestion (see the previous comment) I voted to close as a duplicate of Under what conditions is $AA^T$ invertible?. – Martin Sleziak Aug 14 '17 at 04:33
  • I have upvoted now, since you've provided context; now a tip: you might be interested in looking up Cholesky decomposition, since trying that algorithm on a symmetric matrix is a perfectly usable method for determining invertibility. – J. M. ain't a mathematician Aug 14 '17 at 05:12
  • @J.M.isnotamathematician: A minor quibble: Cholesky decomposition is guaranteed to prove invertibility for a positive definite symmetric matrix, but failure is possible for a symmetric matrix that happens to be invertible but not positive definite. – hardmath Aug 19 '17 at 17:04
  • @hardmath, right, some pivoting is necessary in general. In this specific case tho, the matrices being generated are positive (semi)definite, so Cholesky is a usable test. – J. M. ain't a mathematician Aug 19 '17 at 17:23

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