Is $\vert \lambda \vert \leq \Vert A \Vert$ for eigenvalue $\lambda$ of bounded operator $A$ on $\infty$-dimensional space?

Question

The following question came to my mind while reading $\S 90.$ Minimax principle from Paul R. Halmos's Finite Dimensional Vector Spaces (second edition), and the exercises therefrom.

If $\lambda$ is an eigenvalue of a bounded linear operator $A$ on a not-necessarily finite-dimensional inner product space, does it follow that $\vert \lambda \vert \leq \Vert A \Vert$?

The norm $\Vert \cdot \Vert$ of a bounded operator on a non-trivial inner product space is defined in $\S 88.$ Expressions for the norm as follows.

$\Vert A \Vert = \sup \left\{ \frac{\Vert Ax \Vert}{\Vert x \Vert}: \Vert x \Vert \neq 0 \right\}.$

I believe that the answer to the above question is "yes". Proof: If $y$ is an eigenvector for eigenvalue $\lambda$, then $Ay = \lambda y$. Thus $\Vert Ay \Vert = \Vert \lambda y \Vert = \vert \lambda \vert \cdot \Vert y \Vert$. It follows that $\frac{\Vert Ay \Vert}{\Vert y \Vert} = \vert \lambda \vert \leq \Vert A \Vert$.

However, I am little unsure about this conclusion due to my general lack of comfort in case of infinite-dimensional spaces, and would appreciate help. Thanks.

Answer 1

More generally, if $T: V \to V$ is a bounded operator on a normed space $V$ and if $\lambda$ is an eigenvalue of $T$ then $\|T\|\ge |\lambda|$.

Proof: Choose a unit vector $v \in V$ such that $Tv = \lambda v$ (just normalise an eigenvector associated to $\lambda$). Then $$|\lambda| = \|\lambda v\|_V = \|Tv\|_V \le \|T\|$$ and we are done!