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

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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:

\(\ds \norm {\map T {\frac x {\norm x} } }\) \(=\) \(\ds \norm {\frac 1 {\norm x} \map T x}\) Definition of Linear Transformation on Vector Space
\(\ds \) \(=\) \(\ds \frac {\norm {\map T x} } {\norm x}\) using Norm Axiom $\text N 2$: Positive Homogeneity
\(\ds \) \(=\) \(\ds \frac {\cmod {\lambda x} } {\norm x}\)
\(\ds \) \(=\) \(\ds \frac {\cmod \lambda \norm x} {\norm x}\) using Norm Axiom $\text N 2$: Positive Homogeneity
\(\ds \) \(=\) \(\ds \cmod \lambda\)

while:

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

So we have:

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

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

$\cmod \lambda \le \norm T$

$\blacksquare$


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