This is my second attempt at calculating the spectrum of the left shift operator on a Hilbert space. I got stuck again and I would be grateful if someone could help. (You can find my previous (failed) attempt here).

Let $H$ be a Hilbert space with orthonormal base $e_i$ and let $L\in B(H)$ be the left shift operator, that is, $L(e_i) = e_{i-1}$.

In the following I will assume that if there exists a sequence $x_n\in H$ with $\|x_n\|=1$ and $Tx_n \to 0$ then $T$ is not invertible.

First note that $L$ is not invertible hence $0 \in \sigma (L)$.

Let $\lambda \in \mathbb C$ be non-zero.

I am stuck trying to find $x_n$ with $\|x_n\|=1$ and $(T-\lambda I)x_n \to 0$.


After doing some calculations I think that $L - \lambda I$ is never invertible for any $\lambda \in \mathbb C$ but I can't find $x \neq 0$ such that $(L-\lambda I)x = 0$. Could someone help me find such $x$ please? I'm sure that $\sigma(L) = \mathbb C$.

  • $\begingroup$ You can find examples among also the shifts gone through explicitely in Lax, Functional Analysis, page 226ff. $\endgroup$ Dec 21, 2014 at 22:42
  • $\begingroup$ Were you taught that the spectrum of a bounded operator on a complex Banach space is a non-empty closed and proper subset of $\mathbb{C}$? $\endgroup$ Dec 22, 2014 at 7:21

1 Answer 1


Since $\lVert L\rVert = 1$, it follows that $L-\lambda I$ is invertible for all $\lambda$ with $\lvert\lambda\rvert > 1$, we can see that using the Neumann series. For $\lvert \lambda\rvert > \lVert L\rVert$, the series

$$\sum_{n=0}^\infty \lambda^{-n}L^n$$

is absolutely convergent, and since $B(H)$ is a Banach space, it is convergent. One computes

$$(I-\lambda^{-1}L)\sum_{n=0}^\infty \lambda^{-n}L^n = \lim_{N\to\infty}(I-\lambda^{-1}L)\sum_{n=0}^N \lambda^{-n}L^n = \lim_{N\to \infty} I - \lambda^{-N-1}L^{N+1} = I,$$

and ditto for $\biggl(\sum\limits_{n=0}^\infty \lambda^{-n}L^n\biggr)(I-\lambda^{-1}L)$, and from that one sees that $L-\lambda I$ is invertible when $\lvert\lambda\rvert > \lVert L\rVert$, and the inverse is given by the series

$$(L-\lambda I)^{-1} = -\sum_{n=0}^\infty \lambda^{-n-1}L^n$$


This argument uses no specific property of $L$, it works for all bounded linear operators on Banach spaces. The spectrum of a bounded linear operator $T$ on a Banach space is always conatined in the closed disk with radius $\lVert T\rVert$ and centre $0$.

For the left shift operator $L$, we have - using the separable Hilbert space $\ell^2(\mathbb{N})$ to get a convenient notation -

$$(L-\lambda I)x = (x_1 -\lambda x_0, x_2 - \lambda x_1, x_3 - \lambda x_2,\dotsc).$$

Thus $(L-\lambda I)x = 0$ if and only if we have $x_1 = \lambda x_0$, $x_2 = \lambda x_1$, and so on, which becomes $x_n = \lambda^n x_0$. For $\lvert \lambda\rvert < 1$, that defines a one-dimensional subspace of $H$, spanned by

$$\nu_\lambda = \sum_{n=0}^\infty \lambda^n\cdot e_n.$$

Thus we have $\ker (L-\lambda I) \neq \{0\}$ for $\lvert\lambda\rvert < 1$, and hence $\sigma(L)$ contains the open unit disk.

Since the spectrum of a bounded linear operator is compact, and we saw above that $\sigma(L)$ is contained in the closed unit disk, it follows that

$$\sigma(L) = \overline{\mathbb{D}} = \{ \lambda\in\mathbb{C} : \lvert \lambda\rvert \leqslant 1\}.$$

  • $\begingroup$ Regarding the calculation of $\|L\|$: could you check if I'm doing it right? Here is how I do it: $$\\$$ (1) $\|L\|$ is $\le 1$ because $\|Lx\|\le\|x\|$ for all $x \in H$. $$\\$$ (2) $\|L\|$ is $\ge 1$ because $x=(0,1,0,\dots)$ has norm $1$ and so does $Lx$. $\endgroup$
    – user167889
    Dec 24, 2014 at 2:55
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    $\begingroup$ Confirmed, none of them is stupid. 1. Your calculation of $\lVert L\rVert$ is correct and pretty much the best way to do it (the "best way" is non-unique here). 2. Yes, even more general, all Hilbert spaces with orthonormal bases of the same cardinality are isometrically isomorphic. If you have an orthonormal basis indexed by $A$, your Hilbert space is isometrically isomorphic to $\ell^2(A)$ by Parseval's identity, and $\ell^2(A) \cong \ell^2(B)$ if and only if there is a bijection between $A$ and $B$. So every Hilbert space with a countable (infinite) orthonormal basis is isometrically ... $\endgroup$ Dec 25, 2014 at 0:26
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    $\begingroup$ ... isomorphic to $\ell^2(N)$. 3. Yes, the spectrum generally does not consist only of eigenvalues (but it can of course, consider the identity or $0$ for trivial examples). That is also the case here. But since we know that the spectrum is contained in the closed disk of radius $\lVert L\rVert$ and is closed, and here every point in the open disk with radius $\lVert L\rVert$ is an eigenvalue, we know that $\sigma(L)$ is a closed set with $$\{ z : \lvert z\rvert < \lVert L\rVert\} \subseteq \sigma(L)\subseteq \{ z : \lvert z\rvert\leqslant \lVert L\rVert\},$$ hence the latter is an equality. $\endgroup$ Dec 25, 2014 at 0:31
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    $\begingroup$ So in this case, to determine the spectrum, it is sufficient to find the eigenvalues. That is a nice property of the left-shift operator. In general, determining the spectrum of an operator can be much more difficult since one also must find the values $\lambda$ such that $\lambda I - T$ is not surjective, which tends to be harder than finding the eigenvalues. $\endgroup$ Dec 25, 2014 at 0:34
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    $\begingroup$ @student Right. You don't need to know that all separable Hilbert spaces are isometrically isomorphic to $\ell^2(\mathbb{N})$ to calculate the spectrum, I just find the notation nicer for $\ell^2$. If you take the eigenvector of $L-\lambda I$ as $x = (1,\lambda,\lambda^2,\lambda^3,\dotsc)$, you don't even need to special-case $\lambda = 0$. That tells us every $\lambda$ with $\lvert\lambda\rvert < 1$ is an eigenvalue, and hence every $\lambda$ with $\lvert\lambda\rvert\leqslant 1$ is a spectral value, since the resolvent set is open. $\endgroup$ Jan 6, 2015 at 17:07

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