# Spectrum of Linear Operator

The spectrum of a linear operator on a finite dimensional space is pure point spectrum, that is, both continuous and residual spectrums are empty.

Or we can say that on a finite dimensional space, all spectral values of a linear operator are eigenvalues.

How can this be proved?

• Rank nullity theorem – user113529 Apr 26 '16 at 4:36
• In this situation injective, bijective & surjective amount to the same thing... This is not true in infinite dimensions. – copper.hat Apr 26 '16 at 4:38
• In finite dimensions, an operator $A-\lambda I$ is not invertible iff $A-\lambda I$ is not injective, hence $A$ can only have a point spectrum. – copper.hat Apr 26 '16 at 4:46

Proposition: Let $T:H \to H$ be a linear operator on a finite dimensional complex Hilbert space. If $(T-\lambda I)\mathbf{x} \neq 0$ for all nonzero $\mathbf{x}$, then $R_{\lambda}:= (T-\lambda I)^{-1}$ exists, is continuous, and densely defined.
On the other hand, we know that the kernel of $T-\lambda I$ is trivial by assumption, so our function is injective. Of course, this in turn implies that our function is surjective, since $H/\ker T \cong \mathrm{Im} T$ (or you can argue by rank nullity theorem.) So, $$\sigma (T) \subseteq \{\textrm{eigenvalues}\}$$
invertible $\iff$ bijective $\iff$ injective