Modulus of Eigenvalue of Bounded Linear Operator is Bounded Above by Operator Norm

Theorem
Let $\struct {X, \norm \cdot}$ be a normed vector space.

Let $T : X \to X$ be a bounded linear operator.

Let $\lambda$ be a eigenvalue of $T$.

Then:


 * $\cmod \lambda \le \norm T$

where $\norm T$ denotes the norm of $T$.

Proof
Since $\lambda$ is an eigenvalue of $T$, there exists $x \ne 0$ such that:


 * $T x = \lambda x$

Then we have:

while:


 * $\ds \norm {\frac x {\norm x} } = 1$

So we have:


 * $\cmod \lambda \in \set {\cmod {T x} : \norm x = 1}$

So, from the definition of supremum and of the norm of a bounded linear transformation, we have:


 * $\cmod \lambda \le \norm T$