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!
Best Answer
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.