The lower bound of the smallest eigenvalue of a symmetric positive definite matrix 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!
 A: 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.
A: 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
$$\lambda_{\min} (\mathrm A) \geq 1 - \| \mathrm A - \mathrm I_n \|_{\infty} \geq 1 - \sqrt n \, \| \mathrm A - \mathrm I_n \|_{F}$$
