My book defines the spectrum like this:
Let $H$ be a complex Hilbert space, let $I \in B(H)$ be the identity
operator and let $T \in B(H)$. The spectrum of $T$, denoted $\sigma(T)$,
is defined to be:$$\sigma(T)=\{\lambda \in \mathbb{C}: T-\lambda I\text{ is not invertible}\}$$
Later there is a lemma that says that all eigenvalues are in the spectrum:
Let $H$ be a complex Hilbert space and let $T \in B(H)$. If $\lambda$ is
an eigenvalue of T then $\lambda \in \sigma(T)$.
But why does the converse not hold? I mean, if $\lambda \in \sigma(T)$, why is not $\lambda$ and eigenvalue? What is wrong with this proof?:
Let $\lambda \in \sigma(T)$, then $T-\lambda I$ is not invertible. Then there is an $x \in H, x \ne0$ such that $T(x)-\lambda x=0$, if not, only 0 will be sent to 0, and then the operator is invertible. But then we have that $T(x)=\lambda x$, and hence $\lambda$ is an eigenvalue.
Do you see where the error is?
Best Answer
If something is non-invertible, there's two (non-disjoint) possibilities: it fails to be injective, or it fails to be surjective. In finite dimension, these are the same, but in infinite-dimensional spaces, weird things can happen.
If it fails to be injective, there's $x \ne y$ such that $(T - \lambda I)(x) = (T - \lambda I)(y)$. So $(T - \lambda I)(x - y) = 0$, implying $T(x - y) = \lambda (x - y)$, showing that $\lambda$ is an eigenvalue.
But if it fails to be surjective, then we can't do that kind of thing. For example, let $T$ be the "right shift" operator, where sequences are padded on the left with zeroes. Clearly $T - 0I$ has no inverse, but that doesn't mean $0$ is an eigenvalue.
EDIT: After getting some sleep, fixed stupid mistake.