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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?

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  • $\begingroup$ The word 'invertible' takes on the context of 'being invertible in $B(H)$'. So that means the inverse has to be defined on all $H$ and must be bounded. Generally speaking--but not always--having an inverse and being invertible are terms to distinguish between these cases. $A \in B(H)$ may have an unbounded inverse, but that would not mean that $A$ is invertible. $\endgroup$ – DisintegratingByParts Jul 25 '15 at 21:11
  • $\begingroup$ @TrialAndError Thank you, but in a banach space, and hence a hilbert space, if it has an inverse, then the inverse is bounded? $\endgroup$ – user119615 Jul 25 '15 at 21:48
  • $\begingroup$ Even if $T \in B(H)$ is one-to-one, that does not mean that the range is all of $H$. The range may be a proper closed subspace, or it may be a non-closed subspace. If the range of $T$ is not closed, then the inverse is not bounded. This is true for a Banach space or a Hilbert space $H$. $\endgroup$ – DisintegratingByParts Jul 25 '15 at 22:03
  • $\begingroup$ @TrialAndError Thank you that is very interesting. Is it difficult to prove that?, that is: Let $T\in B(x)$,(or $T: X\rightarrow Y$, T linear and bounded), and assume T is 1-1, but also assume that $\{y_y=T(x), x \in X\}$ is not closed, then the inverse is not bounded. I tried proving it by using the fact that there must be a sequence in the image converging to a point that is not in the image, but I do not see how to end the proof. $\endgroup$ – user119615 Jul 25 '15 at 22:32
  • $\begingroup$ Suppose $T^{-1}$ is bounded, and let $y$ be in the closure of the range of $T$. Then there exists $\{ y_{n } \} \in TX$ such that $y_{n}\rightarrow y$. Hence, $\{ y_{n} \}$ is Cauchy and, because $T^{-1}$ is bounded, then $\{ T^{-1}y_{n} \}$ is Cauchy and, hence, converges to some $x$. Then, because $T$ is bounded, $\{TT^{-1}y_{n} =y_n \}$ converges to $Tx$, which means $y=Tx$ is in the range of $T$. Essentially, if $T^{-1}$ is bounded, then $X$ and $TX$ are topologically equivalent, which means that $TX$ is complete and, hence, closed. $\endgroup$ – DisintegratingByParts Jul 25 '15 at 22:56
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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.

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  • $\begingroup$ I think you only meant your final sentence in the case of $\lambda = 0$.... $\endgroup$ – Hurkyl Jul 26 '15 at 6:40
  • $\begingroup$ Isn't it non-surjective for all $\lambda$ though? Or is this related to domains and boundedness? $\endgroup$ – Henry Swanson Jul 26 '15 at 8:16
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    $\begingroup$ I don't see why it would be nonsurjective; to solve $(T-\lambda)f = g$, you can just solve for the leading term $f_0 = -\lambda^{-1} g_0$, and recursively solve for the remaining ones from $f_{n-1} - \lambda f_n = g_n$, and it's not "clear" that the result shouldn't be in the Hilbert space. On the space of all sequences, we can write the inverse as $S = -\lambda^{-1} \sum_{k=0}^{\infty} (T/\lambda)^{k}$. And I think for $|\lambda| > 1$ it maps square summable to square summable, since $$|(Sf)_n| \leq |\lambda|^{-1} \sum_{k=0}^n |f_{n-k} \lambda^{-k}| \leq \max_k(f_k) |\lambda - 1|^{-1}$$ $\endgroup$ – Hurkyl Jul 26 '15 at 14:22
  • $\begingroup$ Oh right, the leading term will be $-\lambda f_0$, not $-\lambda$... $\endgroup$ – Henry Swanson Jul 26 '15 at 21:03
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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.

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  • $\begingroup$ Not the OP, but I have a question anyways: why is the spectrum just the unit disk? Isn't $R - \lambda I$ non-invertible for all $\lambda$? $\endgroup$ – Henry Swanson Jul 25 '15 at 21:18
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    $\begingroup$ @HenrySwanson: No, the spectrum of a (continuous) operator $T$ is always bounded (in fact, compact). If $\lambda > \|T\|$, then $(T - \lambda)^{-1} = -\lambda^{-1} \left(1 + \lambda^{-1} T + \lambda^{-2} T^{-2} + \cdots\right)$. The full computation of $\sigma(L), \sigma(R)$ is done in detail in any introductory textbook on spectral theory (e.g., Barry Simon's book). $\endgroup$ – anomaly Jul 26 '15 at 3:48
  • $\begingroup$ @user119615: No thanks are necessary, but you're welcome. $\endgroup$ – anomaly Jul 26 '15 at 3:49
  • $\begingroup$ Ah, okay. The only things I know about spectra are from a QM course I took, and we never mentioned operator norms (in fact, we didn't distinguish eigenvalues and the spectrum at all). I'll go find that book then. :) $\endgroup$ – Henry Swanson Jul 26 '15 at 4:44
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$T-\lambda I$ 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.

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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 $\ell^\infty$.

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