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

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.

• The result holds. In general the norm on a subspace must be $\le$ the norm on the ambient space. Oct 13 '21 at 3:47
• Your argument is correct - and basically the way I would have shown it. Oct 13 '21 at 4:07

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!