Find the spectrum of a given operator.

Question

Let $A$ be an operator on a Hilbert space $H$ defined by $$ A(x) = \sum_{n=2}^{\infty} \langle x, x_n \rangle x_{n-1} $$ for all $x \in H$, where $(x_n)$ is an orthonormal basis of $E$.

Find the spectrum of $A$.

I know all the definitions, but I have never done such exercises, so I don't even know where to start. Would appreciate any tips.

Answer 1

Notice that $A$ is a shift operator since $A(x_1)=0$ while $$A(x_j)=x_{j-1}\,\,\,\, \forall\, j\geq 2$$ Rather obviously we have $||A||=1$ so that all numbers $\lambda$ with $|\lambda|>1$ can not be in the spectrum.

We show that if $|\lambda|<1$ then $\lambda$ is an eigenvalue (hence it is in the spectrum). Fix $\lambda$ such that $|\lambda|<1$. We look for an $x\not=0$ such that $A(x)=\lambda x$. Write $$x=\sum_{n=1}^{\infty} a_nx_n$$ where the $a_n$ are scalars. Then $$A(x)=\sum_{n=2}^{\infty} a_n x_{n-1}=\sum_{n=1}^{\infty} a_{n+1}x_n$$ The equality $A(x)=\lambda x$ is equivalent to $$a_{n+1}=\lambda a_n\,\,\,\forall n\geq 1$$ hence $a_n=C\lambda^n$. Since $|\lambda|<1$ the series $\sum_{n=1}^{\infty}\big(\lambda^n)^2$ converges hence $$x=\sum_{n\geq 1} \lambda^nx_n$$ is a well-defined element of $H$, which satisfies $A(x)=\lambda x$.

Now the spectrum is always a closed subset of $\mathbb{C}$. In our example, it contains the open unit disc but contains no $\lambda$ with $|\lambda|>1$. It follows that the spectrum of $A$ is the closed unit disc.

Answer 2

Shift operators are nicely represented in terms of division and multiplication by $z$. It's not a bad idea to learn about this formalism because you'll probably see this pop up again. If you associate $\{ a_1,a_2,a_3,\cdots \}$ with a power series on the unit disk given by $$ a(z)=a_1 + a_2 z + a_3 z^2 + \cdots, $$ then $$ Aa=\frac{a(z)-a(0)}{z} = a_2 + a_3 z + a_4 z^2 + \cdots. $$ This operation corresponds with the action of your operator. Solving $$ (A-\lambda I)a =b $$ is reduced to an algebraic problem involving power series. For example, if $|\lambda| < 1$ and $b=0$, then there is a non-trivial solution $$ \frac{a(z)-a(0)}{z}-\lambda a(z)=0 \\ % a(z)-a(0)-\lambda z a(z) = 0 \\ a(z) = \frac{a(0)}{1-\lambda z}=a(0)\sum_{n=0}^{\infty}\lambda^nz^n. $$ So $(1,\lambda,\lambda^2,\lambda^3,\cdots)$ is an eigenvector of $A$ with eigenvalue $\lambda$. You don't get a solution for $b=0$ in the case that $|\lambda| > 1$ because the resulting function does not have a power series in $|z| < 1$ due to a singularity of $1/(1-\lambda z)$ at $z=1/\lambda$. So every $|\lambda| < 1$ is an eiegenvalue of $A$.

For $|\lambda| > 1$, you can solve for a unique $a$ such that $(A-\lambda I)a=b$: $$ a(z)-a(0)-\lambda z a(z) = zb(z) \\ a(z) = \frac{a(0)+zb(z)}{1-\lambda z} $$ Here $a(0)$ must be uniquely chosen so that the numerator vanishes at $z=1/\lambda$, thereby cancelling the singularity of the denominator. That is, $$ a(0)+\frac{1}{\lambda}b(1/\lambda)=0 \\ a(z) = \frac{-\frac{1}{\lambda}b(1/\lambda)+zb(z)}{1-\lambda z} = -\frac{1}{\lambda}\frac{b(1/\lambda)-\lambda z b(z)}{1-\lambda z}. $$ So $|\lambda| > 1$ is in the resolvent set of $A$.