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 $\lambda \in \map \sigma T$ $T - \lambda I$ 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 $T - \lambda I$ is not bijective.

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

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

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


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

and hence $T x = \lambda x$.

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