TITLE: Why isn't every element of the spectrum an eigenvalue? (Where is the error in my proof?)
QUESTION [13 upvotes]: 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?

REPLY [3 votes]: Left/right shift operators are the standard examples, but I personally think that the multiplication operator is the easiest way to see that there may be something else in the spectrum besides eigenvalues. Consider a multiplication operator $A_c$ on $\ell^\infty$ ($c\in\ell^\infty$)
$$
(A_c x)_n=c_n x_n.
$$
The inverse if exists is clearly a multiplication by $\frac{1}{c_n}$. So $\lambda I-A_c$ is invertible iff $\frac{1}{\lambda-c_n}$ is bounded. 
If $\lambda=c_k$ for some $k$ then $\lambda-c_k=0$ and this $\lambda$ is an eigenvalue (the kernel of $\lambda I-A_c$ contains the sequence $e_k$ with all zeros except identity at this particular $k$). So all $\lambda$ in the range of $c$ are eigenvalues. But it may happen that the image is not closed. For example, if we take $c_n=\arctan(n)$, we can observe that there are two special $\lambda=\pm\pi/2$ - endpoints, i.e. from the closure of the image of $c$ - such that the sequence $\lambda-c_n$ is never zero, but nevertheless does not have a bounded inverse, so that $\lambda I-A_c$ does not still  have a bounded inverse on $\ell^\infty$.