Eigenvalues of the Left-Shift Plus Right-Shift Operator

The phrase “left shift plus right shift” usually means the sum of the unilateral shift on sequences and its adjoint. On the Hilbert space of square-summable sequences indexed by $0,1,2,\ldots$, this operator has a simple spectrum: the spectrum is $[-2,2]$, but there are no eigenvalues.

The distinction between spectrum and eigenvalues is essential here. Every point of the interval belongs to the spectrum, yet the eigenvalue equation has no nonzero solution in the Hilbert space.

Define the shifts

Let $H=\ell^2(\{0,1,2,\ldots\})$. For $x=(x_0,x_1,\ldots)$, define the right shift $S$ and left shift $S^*$ by

$$(Sx)_0=0,\qquad (Sx)_n=x_{n-1}\quad(n\geq 1),$$

and

$$(S^*x)_n=x_{n+1}\quad(n\geq 0).$$

These are adjoints of one another. The operator in question is

$$A=S+S^*,\qquad (Ax)_n=x_{n-1}+x_{n+1},$$

where the boundary convention is $x_{-1}=0$. In particular, $A$ is bounded and self-adjoint.

Solve the eigenvalue equation

Suppose $Ax=\lambda x$. Then the coordinates satisfy

$$x_{n-1}+x_{n+1}=\lambda x_n\quad(n\geq 0),\qquad x_{-1}=0.$$

For $-2<\lambda<2$, write $\lambda=2\cos\theta$ with $0<\theta<\pi$. The recurrence has solutions built from $e^{in\theta}$ and $e^{-in\theta}$. Applying the boundary condition gives

$$x_n=C\sin((n+1)\theta)$$

for a constant $C$. If $C\ne 0$, this sequence does not tend to zero and is not square-summable. Thus it cannot be an eigenvector in $H$.

At the endpoints, the recurrence gives $x_n=C(n+1)$ for $\lambda=2$ and $x_n=C(-1)^n(n+1)$ for $\lambda=-2$. These are not square-summable unless $C=0$. For $|\lambda|>2$, the characteristic equation $r^2-\lambda r+1=0$ has roots with different magnitudes; the boundary condition leaves a solution that grows, rather than a nonzero square-summable sequence. Hence there are no eigenvalues outside the interval either.

So the point spectrum—the set of eigenvalues—is empty.

Why the spectrum is still $[-2,2]$

Because $A$ is self-adjoint and $\|A\|\leq \|S\|+\|S^*\|=2$, its spectrum must be contained in $[-2,2]$. To see that every point of this interval is in the spectrum, take any $\lambda=2\cos\theta$ and use a long, finite block of the sequence $e^{in\theta}$, supported far from the boundary. In the interior of the block, the recurrence is satisfied exactly. The error is confined to the two ends of the block, while the norm of the block grows like the square root of its length. After normalization, the error therefore tends to zero as the block gets longer.

This produces unit vectors $v$ with $\|(A-\lambda I)v\|$ arbitrarily small. If $A-\lambda I$ had a bounded inverse, such vectors could not exist. Thus $\lambda$ is in the spectrum. Since this works for each $\lambda\in[-2,2]$, the spectrum is exactly that interval.

A concrete check: $\lambda=0$

For $\lambda=0$, the recurrence says $x_{n+1}=-x_{n-1}$. Starting from $x_0=c$, the boundary equation at $n=0$ gives $x_1=0$, and then

$$x=(c,0,-c,0,c,0,-c,0,\ldots).$$

If $c\ne 0$, the squared coordinates do not have a finite sum. So $0$ is in the spectrum, but it is not an eigenvalue. Finite truncations of this alternating pattern, placed farther and farther from the boundary, give approximate eigenvectors instead.

Common mistakes

• Calling every spectral value an eigenvalue. A spectral value need not have an eigenvector. Here the entire interval is spectrum, but the point spectrum is empty.
• Ignoring the boundary condition. The equation at index $0$ is $x_1=\lambda x_0$, not an interior recurrence with a freely chosen $x_{-1}$.
• Using a bounded oscillating sequence as an eigenvector. Boundedness is not enough: an eigenvector must belong to $\ell^2$, so its squared coordinates must have a finite sum.
• Confusing $\ell^2$ with $L^2([0,\infty))$. The notation in the title is commonly used informally for sequences on the nonnegative integers. If instead the space is $L^2$ of the continuous half-line and the shifts are translations by a fixed distance $h$, the analogous sum of right translation and its adjoint has the same spectrum and no eigenvalues; the coordinates can be grouped into sequences along each arithmetic progression modulo $h$.