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?

  • $\begingroup$ Rank nullity theorem $\endgroup$ – user113529 Apr 26 '16 at 4:36
  • $\begingroup$ In this situation injective, bijective & surjective amount to the same thing... This is not true in infinite dimensions. $\endgroup$ – copper.hat Apr 26 '16 at 4:38
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    $\begingroup$ 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. $\endgroup$ – 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.

Every linear operator is continuous in finite dimension, so it is clear that the continuous spectrum is empty.

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, $$\sigma (T) \subseteq \{\textrm{eigenvalues}\}$$

So, it's bijective, so it has an inverse that is defined everywhere . The basic point is that

invertible $\iff$ bijective $\iff$ injective

doesn't hold in infinite dimension, and the issue is complicated by continuous and densely defined.


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