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:
- $\cmod \lambda \le \norm T$
$\blacksquare$
Sources
- 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $14.1$: The Resolvent and Spectrum