Why isn't every element of the spectrum an eigenvalue? (Where is the error in my proof?)

Question

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?

Answer 1

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.

Answer 2

For finite-dimensional vector spaces, injectivity and surjectivity are equivalent. That's not the case for an arbitrary Hilbert space. The classic examples are the left- and right-shift operators $L, R:\ell^2 \to \ell^2$, given by \begin{align*} L(x_1, x_2, \dots) &= (x_2, \dots) \\ R(x_1, x_2, \dots) &= (0, x_1, x_2, \dots). \end{align*} The map $L$ is clearly surjective but not injective, and $R$ is clearly injective but not surjective. It's easy to see that $R$ has no eigenvalues, but its spectrum is certainly not empty; in fact, it's the closed unit disk.

Answer 3

$T-\lambda I$ being non-invertible does not imply there is a non-zero $x$ with $(T-\lambda I)x=0$. That is true when $H$ is finite-dimensional, but not necessarily when $H$ is infinite-dimensional. The classic counterexample is the right-shift operator $R:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N})$. Take a look at the Wikipedia article on the notion of spectrum.

Answer 4

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 $\ell^\infty$.