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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 $|Mv| \le C |v|$? –  Robert Israel Nov 16 '12 at 22:11
    
@RobertIsrael Doh, I actually do mean a lower bound. –  soup Nov 16 '12 at 22:34

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up vote 2 down vote accepted

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 $v$? –  soup Nov 16 '12 at 22:39
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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 $M$). –  Robert Israel Nov 16 '12 at 22:42
    
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. –  soup Nov 17 '12 at 22:32
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You might look at en.wikipedia.org/wiki/Min-max_theorem –  Robert Israel Nov 18 '12 at 6:12

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