# Lower bound on smallest eigenvalue of (symmetric positive-definite) matrix

If $M$ is a positive-definite symmetric matrix, is it possible to get a positive lower bound on the smallest eigenvalue of $M$ in terms of a matrix norm of $M$ or elements of $M$? Eg. I want $$\lambda_{\text{min}} \geq f(\lVert M \rVert)$$ or something like that.

$M$ is a Gram matrix if that helps. Thanks.

-
Hi @ampeo, have you found the answer? I have the same question and all the bounds I can get are negative, which doesn't make any sense for a PD matrix. –  D. Chen Nov 7 '13 at 7:07
There is an obvious bound in terms of the operator norm of $M^{-1}$, of course. –  user7530 Dec 21 '13 at 16:52
As user7530 wrote, $\lambda_{min}$ depends essentially on $M^{-1}$. Moreover to find an inequality in the form $\lambda_{min}\geq f(||M||)$ is beyond all hope. For instance let $A_{\epsilon}=diag(1,\epsilon)$. We should obtain, for every $\epsilon >0$, $\epsilon\geq f(1)$ (for $||.||_2$). –  loup blanc Jan 24 at 16:59
Lower bounds on the smallest eigenvalue of a positive definite matrix are related to estimates of the condition number, for which see my Answer to a SciComp.SE Question. –  hardmath Jan 27 at 2:36

There is one lower bound on minimum eigenvalue of symmetric p.d. matrix given in [Applied Math. Sc., vol. 4, no, 64] which is based on Forbenius norm (F) and Euclidean norm (E)

$$\lambda_{min} \gt \sqrt{\frac{||A||_F^2-n||A||_E^2}{n(1-||A||_E^2/|det(A)|^{2/n})}}$$

if it helps.

-
Rather than reposting (with the same "typos" as best I can tell), you could have edited this, your original post. See "edit" link just under the Answer. I took a guess at the mismatched parenthesis in the denominator; please check it is correct. –  hardmath Jan 27 at 2:40
It is all correct, Thanks! –  Saeed Manaffam Jan 27 at 6:16