Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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 '14 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 '14 at 2:36

2 Answers 2

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.

[reference]: K. H. Schindler, ``A New Lower Bound for the Minimal Singular Value for Real Non-Singular Matrices by a Matrix Norm and Determinant'', Journal of Applied Mathematical Sciences, Vol. 4, No. 64, 2010.

share|improve this answer
    
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 '14 at 2:40
    
It is all correct, Thanks! –  Saeed Manaffam Jan 27 '14 at 6:16
    
@Saeed Thanks for this! Is there a chance you could find the article on Google Scholar and post a link? I can't find it anywhere (not sure if Sc. is short for something...) –  Elements May 14 at 18:00
1  
1  
@Elements you can also search the title in google scholar: "A New Lower Bound for the Minimal Singular Value for Real Non-Singular Matrices by a Matrix Norm and Determinant" Applied mathematical Sciences Vol. 4, No. 64 –  Saeed Manaffam May 15 at 19:07

A quick comment: If you have diagonal dominance, then Gerhsgorin's circle theorem for eigenvalues will get you at least something. So for each row, subtract the diagonal term from the sum of the absolute values of the off-diagonal terms, and take the minimum over the rows. That is a bound on the eigenvalue that will be positive (again, if you have diagonal dominance, which may not hold for all Gram matrices).

share|improve this answer

Your Answer

 
discard

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.