$V \in \mathbb{C}^{n \times n}$. Does $\lVert I_n - V \rVert_\infty \leq \epsilon$ imply lower bounds on the smallest eigenvalue of $V$?

Question

Let $V \in \mathbb{C}^{n \times n}$ and $I_n$ is the $n \times n$ identity matrix. Suppose $\lVert I_n - V \rVert_\infty \leq \epsilon$, where the norm is the operator norm.

I would like to get a tight lower bound on $\lambda_{\mathrm{min}}$, the smallest eigenvalue of $V$. Intuitively, it seems the smallest eigenvalue should be at least $1 - \epsilon$, but I'm not sure how one would show that. Any ideas/suggestions?

Answer 1

Let $\lambda$ be any eigenvalue of $V$. Say $Vx = \lambda x$ where $x$ is a nonzero vector. We may assume $\|x\| = 1$. Then $$ (I-V)x = x-\lambda x = (1-\lambda) x $$ so $$ |1-\lambda| = |1-\lambda|\;\|x\| = \|(1-\lambda)x\| = \|(I-V)x\| \le \|I-V\|_\infty\;\|x\| = \|I-V\|_\infty \le \epsilon . $$ Every eigenvalue of $V$ lies in the disk $\{\lambda \in \mathbb C \mid |1-\lambda| \le \epsilon\}$ . I guess that whatever "smallest eigenvalue of $V$" means, it would at least be an eigenvalue of $V$.