Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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 $v \ne 0$.

share|cite|improve this answer
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
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
You might look at – Robert Israel Nov 18 '12 at 6:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.