2
$\begingroup$

I encounter a symmetric positive definite matrix whose features are

  • all diagonal entries are $1$.
  • all the other entries are in $[0, 1)$, but the matrix is not diagonally dominant.

Now I am looking for a positive lower bound of the smallest eigenvalue, expressed by trace and Frobenius norm.

I have seen a lot of papers related to this topic. Especially, the result in this paper is very close to my goal. But that expression still involves the maximum eigenvalue and determinant. I have seen the answer of Lower bound on the smallest eigenvalue. I'm happy if the answer posted in that is correct. But I think it's wrong. Does that kind of lower bound exist?

Anyone could give me any tips? Thanks so much!

$\endgroup$
  • $\begingroup$ Are your matrices correlation matrices? What is the context? $\endgroup$ – Rodrigo de Azevedo Jul 27 '16 at 0:31
2
$\begingroup$

If it's an $n \times n$ matrix with all diagonal entries $1$, the trace is $n$, so that won't help. The Frobenius norm is bounded by $n$. Since the smallest eigenvalue can be arbitrarily close to $0$, I don't see how you could possibly get a nonzero lower bound in terms of the Frobenius norm.

$\endgroup$
  • $\begingroup$ Thanks so much for your reply. I think I have to detour this question. $\endgroup$ – jack Jul 27 '16 at 16:12
0
$\begingroup$

Let $\mathrm A \in [0,1)^{n \times n}$ be a symmetric, positive definite, nonnegative matrix whose diagonal entries are equal to $1$. Using the Gershgorin circle theorem, the minimum eigenvalue of $\mathrm A$ is bounded by

$$\lambda_{\min} (\mathrm A) \geq 1 - \max_{1 \leq i \leq n} \, \sum_{j \neq i} a_{ij} = 1 - \| \mathrm A - \mathrm I_n\|_{\infty}$$

Unfortunately, unless $a_{ij} \ll 1$, this bound is likely too loose. I suspect that this bound is only potentially useful for small perturbations of the identity matrix.

If the bound is not too loose, then, using the inequality $\| \cdot \|_{\infty} \leq \sqrt n \, \| \cdot \|_{F}$, we obtain

$$\lambda_{\min} (\mathrm A) \geq 1 - \| \mathrm A - \mathrm I_n \|_{\infty} \geq 1 - \sqrt n \, \| \mathrm A - \mathrm I_n \|_{F}$$

$\endgroup$
  • $\begingroup$ Thanks so much for your reply. But your bound cannot be guaranteed to be positive, right? $\endgroup$ – jack Jul 27 '16 at 16:34
  • $\begingroup$ @jack That is correct. $\endgroup$ – Rodrigo de Azevedo Jul 27 '16 at 16:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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