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Suppose $M$ is a square matrix (with elements that are continuous functions which are bounded above and below) and $v$ is a vector. I want a lower bound like $$|Mv| \geq C|v|$$ for constant $C$.

Do I have any luck here?

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    I'd call that a lower bound. Do you mean $|Mv| \le C |v|$?2012-11-16
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    @RobertIsrael Doh, I actually do mean a lower bound.2012-11-16

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There can't be a better lower bound than $0$, because it is possible to have $Mv = 0$ with $v \ne 0$.

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    Does it help in some way if we know that $v^TM^TMv > 0$ for non-zero vectors $v$?2012-11-16
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    Well, that says there is some positive lower bound, but doesn't tell you what it is. In fact, the best lower bound is the square root of the lowest eigenvalue of $M^T M$ (the least singular value of $M$).2012-11-16
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    Thanks, do you know what I can search for to read more about your last sentence? "Least singular value" gives me just how to optimise the singular value.2012-11-17
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    You might look at http://en.wikipedia.org/wiki/Min-max_theorem2012-11-18