What is the difference in the definition of a spectral value and an eigenvalue. My notes from functional analysis says
$\lambda$ is an eigenvalue of an operator $A$ if $\,\exists \, x \in \mathbb{C^n}$ such that $$Ax = \lambda x$$ This implies $(A – \lambda I)x = 0 \Rightarrow \ker(A – \lambda I) \neq {0}$. This is equivalent
of saying that $A$ is not injective.
On the other hand, the definition of a spectral value is
$\lambda$ is called a spectral value of $A$ if $A – \lambda I$ is not
invertible.
What is the difference here? How is it that some operators can have spectral values and not eigenvalues (eigenvalues $\subset$ spectrum(A)) and lastly, how do they conincide when the space is finite dimensional ?
Best Answer
Consider the unilateral shift on $\ell^2(\mathbb N)$, i.e. $$ S(a_1,a_2,\ldots)=(0,a_1,a_2,\ldots). $$ It is easy to verify that $S$ is injective, so $0$ is not an eigenvalue of $S$. But $S$ is not invertible (because it is not surjective). So we have $0$ as an element of the spectrum of $S$, but not an eigenvalue.
In finite dimension, two things happen: since the domain and codomain are of equal finite-dimension, an operator is injective if and only if it is surjective. And all linear operators are continuous. So if $T-\lambda I$ is not invertible, this means that it is not bijective; then it is not injective, so it has a nonzero kernel, i.e. there exists nonzero $x$ with $Tx=\lambda x$. Thus, in finite dimension the spectrum consists only of eigenvalues.