Eigenvalues of Left Shift + Right Shift in $l_2([0,\infty))$

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Solution 1

There is a general method for finding sequences which satisfy linear recurrent relations with constant coefficients. In this case, we have the relation $$ x_{n+1}+x_{n-1} = \mu x_n \text{ for } n\geq 1. \text{ (*)} $$ Here is how we solve it: consider all the sequences (not nesessarily from $l^2$) which satisfy (*). They form a linear space (it's easy to check the axioms); let's denote it $M$.

What is $\dim M$? One can see that

$\forall u,v\in \mathbb{C}~ \exists y^{u,v}\in M: y^{u,v}_0 = u, y^{u,v}_1 = v$;

if $x,y\in M$, $x_0 = y_0, x_1 = y_1$, then $x = y$.

From this we can see that any $x\in M$ can be represented as a linear combination of two linearly independent sequences $y^{1,0}$ and $y^{0,1}$: $$ x = x_0\cdot y^{1,0} + x_1\cdot y^{0,1}, $$ so $\dim M = 2$.

How can we find a nice basis for $M$? For $\lambda\in \mathbb{C}$, let $p^\lambda$ denote the sequence of its powers, $(p^\lambda)_n = \lambda^n$. It's easy to see that $p^\lambda\in M$ if and only if $\lambda^2 + 1 = \mu\lambda$. For $|\mu|<2$, this equation has two roots: $$ \lambda_1 = \frac{\mu +\sqrt{4-\mu^2}i}{2}; \lambda_2 = \lambda_1^* = \lambda_1^{-1}. $$ So $p^{\lambda_1}$ and $p^{\lambda_2}$ form a basis for $M$. Now, for $x$ to satisfy $(S_l+S_r)x = \mu x$, one more condition is needed in addition to (*): $x_1 = \mu x_0$. So for $x = c_1 p^{\lambda_1} + c_2 p^{\lambda_2}$ we have $$ \begin{cases} c_1+c_2 = x_0\\ c_1\lambda_1 + c_2\lambda_2 = \mu x_0 \end{cases} $$ from which we get $$ c_1 = \frac c2x_0, c_2 = \frac{c^*}2x_0, \text{ where }c = 1 - \frac{\mu}{\sqrt{4-\mu^2}}i, $$ so we can write $$ x_n = c_1\lambda_1^n + c_2\lambda_2^n = \Re(c \lambda_1^n) x_0. $$ Since $|\lambda_1| = 1$, the sequence $z_n = \Re(c \lambda_1^n)$ would converge to $0$ only if it was constant $0$, but $z_0 = 1$, so $z\not\in l^2[0,\infty)$. This means that $x \in l^2[0,\infty)$ only if $x_0= 0$, i.e., the values from $(-2, 2)$ are not eigenvalues.

Note, though, that the sequence $z$ is bounded. So when you say that if $|\mu|\geq 1$, and $x$ satisfies $(S_l+S_r)x = \mu x$, $x_0\neq 0$, then $|x_n|\to \infty$, it's not true: if $1<|\mu|<2$, then $x$ is bounded. The problem is that the term which you said is $O(\mu^4+1)$ is a sum of many terms with coefficients which depend on $n$ and have different signs, so you cannot say so easily how it behaves when $n\to \infty$.

So you need a new way to show that when $|\mu|\geq 2$, and $x$ satisfies $(S_l+S_r)x = \mu x$, then either all $x_n$ are equal to $0$ or $|x_n|\to \infty$ as $n \to \infty$.

If $\mu = \pm 2$, then $x = x_0(1,\pm 2, 3, \pm 4, \dots)$;

And if $|\mu|>2$, then you can apply the same method as before and see that $|x_n| = |x_0|O(\left(\frac{|\mu|+\sqrt{\mu^2-4}}{2}\right)^n)$.

(You could also use the same method to find the solutions for $\mu = \pm 2$, but with a little modification: in these cases the equation on $\lambda$ has a single root of multiplicity $2$. And one can show that the equation has a root $\lambda$ of multiplicity $m$, then the sequences $n^k\lambda^n$, where $k = 0, \dots, m-1$, lie in $M$.)

Solution 2

Even though this question is almost three years old: for anyone that stumbles across this problem like me, I want to give a simple argument on why the operator given by the sum of the right shift $S_r$ and left shift $S_l$ on $\ell_2(\mathbb N)$ does not have any eigenvalues. I will need a few well known results which can for example be looked up in Chapter VII.§1 of the book "Introduction to Hilbert Space" by Berberian (1976).

  • All eigenvalues of a bounded operator $T$ are bound by $\Vert T\Vert$.
  • All eigenvalues of a self-adjoint operator are real.
  • $\mu\in\mathbb C$ is eigenvalue of $S_l$ if and only if $|\mu|<1$.

As stated in the original post, $S_l+S_r$ is self-adjoint and $\Vert S_l+S_r\Vert\leq \Vert S_l\Vert+\Vert S_r\Vert=2$ so if that operator has eigenvalues, those have to be elements of the real interval $[-2,2]$. Let $\lambda\in[-2,2]$ be given and assume there exists $x\in\ell_2(\mathbb N)$ such that $(S_l+S_r)x=\lambda x$ (our goal will be to show that $x$ has to be the zero sequence). Applying $S_l$ to this equation from the left gives

$$ 0=(S_l^2+S_lS_r-\lambda S_l)x=(S_l^2-\lambda S_l+\operatorname{id})x $$

since $S_lS_r$ obviously is the identity $\operatorname{id}$ on $\ell_2(\mathbb N)$. Looking at the related equation $t^2-\lambda t+1=0$ yields the solutions

$$ t_\pm=\frac{\lambda}2\pm i\sqrt{1-\frac{\lambda^2}4} $$

and thus we have

$$ ( S_l-t_+\operatorname{id} )( S_l-t_-\operatorname{id} ) x=0. $$

Note that $|t_\pm|=1$ since $\lambda\in[-2,2]$ so $t_\pm$ can not be an eigenvalue of $S_l$ as stated at the beginning. This implies $( S_l-t_-\operatorname{id} ) x=0$ and further $ x=0$ so $\lambda$ is not an eigenvalue of $S_l+S_r$. In total we showed that $S_l+S_r$ does not have any eigenvalues.

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Updated on October 24, 2020

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  • JMoravitz
    JMoravitz about 2 years

    This question appeared on an old final exam and I am having difficulty completing it for practice.

    Let $S_r$ and $S_l$ be defined on the hilbert space $l_2[0,\infty)\to l_2[0,\infty)$ as the following: $S_r (x_0, x_1, x_2, \dots) \mapsto (0,x_0, x_1, \dots)$ and $S_l (x_0, x_1, x_2,\dots) \mapsto (x_1, x_2, x_3,\dots)$. That is to say, $S_r$ is the right shift operator and $S_l$ is the left shift operator.

    I fully understand how to describe the spectrum of each individually, however things become clouded once they are combined.

    The question asks to:

    Describe the spectrum of $S_r + S_l$

    That is, $(S_r + S_l)(x_0, x_1, x_2,\dots) \mapsto (x_1, x_0+x_2, x_1+x_3,\dots)$

    To begin I noted first that $S_r + S_l$ is self-adjoint which implies two things. First, the spectrum will necessarily be a subset of $\mathbb{R}$. Second, it implies that $R_\sigma(S_r+S_l)$ is empty.

    Suppose that $\mu\in P_\sigma(S_r+S_l)$, then $\mu x = (S_r+S_l)x$ for some $x=(x_0,x_1,x_2,\dots)$.

    Then $\mu(x_0,x_1,\dots) = (x_1,x_0+x_2, x_1+x_3, \dots, x_{n-1}+x_{n+1},\dots)$ and $x_1 = \mu x_0$, $x_{n+1} = \mu x_n - x_{n-1}$

    So, for $x$ to be an eigenvector of $\mu$, we must have $x=x_0(1, \mu, \mu^2-1, \mu^3-2\mu, \mu^4-3\mu^2+1,\dots,\mu^n - (n-1)\mu^{n-2} + O(\mu^{n-4}+1),\dots) ~~~(\star)$

    It follows that as $n\to\infty$ if $|\mu|>1$, then $|((S_l+S_r)x)_n|\to\infty$ and so no eigenvalues of $S_l+S_r$ are greater than 1.

    For $\mu=1$ we must have $x_1 = x_0$, $x_2 = 0$, $x_3 = -x_0$, $x_{n+1} = x_n - x_{n-1}$

    And we get that $x = x_0(1,1,0,-1,0,1,0,-1,\dots)$ which is only in $l_2[0,\infty)$ if $x_0 = 0$, so $1$ is not an eigenvalue of $S_l+S_r$.

    In the case that $\mu=0$, by $(\star)$

    $x = x_0(1,0,-1,0,1,\dots)$, and since we are in $l_2[0,\infty)$, $x_0=0$ and $x=0$, so $0$ is not an eigenvalue of $S_l+S_r$.


    With this work so far, I was able to show as well that the operator norm is $2$ and the operator is definitely not compact (else $\pm 2$ would have been an eigenvalue)

    It is at this point that I lose track of where to continue from here. With what argument can I show that for $|\mu|<1$ that they are or aren't eigenvalues? I expect but cannot find the right wording to show that there are in fact no eigenvalues.