Spectrum of Bounded Linear Operator on Finite-Dimensional Banach Space is equal to Point Spectrum

Theorem

Let $X$ be a finite-dimensional Banach space over $\C$.

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

Let $\map {\sigma_p} T$ be the point spectrum of $T$.

Let $\map \sigma T$ be the spectrum of $T$.

Then $\map \sigma T = \map {\sigma_p} T$.

Proof

We have that $\lambda \in \map \sigma T$ if and only if $T - \lambda I$ is not invertible as a bounded linear transformation.

So $T - \lambda I$ is not bijective or its inverse $\paren {T - \lambda I}^{-1}$ is not bounded.

From Linear Transformations between Finite-Dimensional Normed Vector Spaces are Continuous, every linear operator $S : X \to X$ is bounded.

So $T - \lambda I$ is not invertible as a bounded linear transformation if and only if $T - \lambda I$ is not bijective.

From Linear Transformation from Finite-Dimensional Vector Space is Injective iff Surjective, we have that $T - \lambda I$ is not bijective if and only if $T - \lambda I$ is not injective.

From Linear Transformation is Injective iff Kernel Contains Only Zero, we have $T - \lambda I$ is not injective if and only if $\ker T \ne \set {\mathbf 0_X}$.

That is, if and only if there exists $x \ne \mathbf 0_X$ such that:

$\paren {T - \lambda I} x = 0$

and hence $T x = \lambda x$.

So $\lambda \in \map {\sigma_p} T$.

$\blacksquare$

Sources

• 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $14.1$: The Resolvent and Spectrum