Unilateral shift operator

Short description: Operator on a Hilbert space that shifts basis vectors

In operator theory, the unilateral shift is an operator on a Hilbert space. It is often studied in two main representations: as an operator on the sequence space ${\ell }^2$, or as a multiplication operator on a Hardy space. Its properties, particularly its invariant subspaces, are well-understood and serve as a model for more general theories.

Definition

Let ${\ell }^2$ be the Hilbert space of square-summable sequences of complex numbers, i.e., $${\ell }^2=\{(a_0,a_1,a_2,\dots ):a_n\in \mathbb{ℂ}\text{ and }\sum _{n=0}^{\mathrm{\infty }}|a_n{|}^2<\mathrm{\infty }\}$$The unilateral shift is the linear operator $S:{\ell }^2\to {\ell }^2$ defined by: $$S(a_0,a_1,a_2,\dots )=(0,a_0,a_1,a_2,\dots )$$This operator is also called the forward shift.

With respect to the standard orthonormal basis $(e_n{)}_{n=0}^{\mathrm{\infty }}$ for ${\ell }^2$, where $e_n$ is the sequence with a 1 in the n-th position and 0 elsewhere, the action of $S$ is $S{e}_n=e_{n+1}$. Its matrix representation is:$$S=\left[\begin{array}{ccccc}0& 0& 0& 0& \cdots \\ 1& 0& 0& 0& \cdots \\ 0& 1& 0& 0& \cdots \\ 0& 0& 1& 0& \cdots \\ ⋮& ⋮& ⋮& ⋮& \ddots \end{array}\right]$$This is a Toeplitz operator whose symbol is the function $f(z)=z$. It can be regarded as an infinite-dimensional lower shift matrix.

Properties

Adjoint operator

The adjoint of the unilateral shift, denoted $S^{*}$, is the backward shift. It acts on ${\ell }^2$ as: $$S^{*}(b_0,b_1,b_2,b_3,\dots )=(b_1,b_2,b_3,\dots )$$The matrix representation of $S^{*}$ is the conjugate transpose of the matrix for $S$: $$S^{*}=\left[\begin{array}{ccccc}0& 1& 0& 0& \cdots \\ 0& 0& 1& 0& \cdots \\ 0& 0& 0& 1& \cdots \\ 0& 0& 0& 0& \cdots \\ ⋮& ⋮& ⋮& ⋮& \ddots \end{array}\right]$$It can be regarded as an infinite-dimensional upper shift matrix.

Basic properties

• $S,S^{*}$ are both continuous but not compact.
• $S^{*}S=I$.
• $S,S^{*}$ make up a pair of unitary equivalence between ${\ell }^2$ and the set of ${\ell }^2$-sequences whose first element is zero.

The resolvent operator has matrix representation$$(zI-S{)}^{-1}=\left[\begin{array}{ccccc}{z}^{-1}& 0& 0& 0& \cdots \\ z^{-2}& z^{-1}& 0& 0& \cdots \\ z^{-3}& z^{-2}& z^{-1}& 0& \cdots \\ z^{-4}& z^{-3}& z^{-2}& z^{-1}& \cdots \\ ⋮& ⋮& ⋮& ⋮& \ddots \end{array}\right]$$which is bounded iff $|z|>1$. Similarly, $(zI-S^{*}{)}^{-1}=((z^{*}I-S{)}^{-1}{)}^{*}$.

For any $z\in \mathbb{ℂ},a\in {\ell }^2$ with $\Vert a\Vert =1$,$$\Vert (zI-S)a{\Vert }^2=1+|z{|}^2-2\mathrm{\Re }(⟨Sa,a⟩z),\Vert (zI-S^{*})a{\Vert }^2=1-|a_0{|}^2+|z{|}^2-2\mathrm{\Re }(⟨Sa,a⟩z^{*})$$where $\mathrm{\Re }$ is the real part.

Spectral theory

Spectrum of the forward shift — Let $\mathbb{𝔻}$ be the open unit disk, $\overline{\mathbb{𝔻}}$ the closed unit disk, and $\mathbb{𝕋}$ the unit circle.

• The spectrum of $S$ is $\sigma (S)=\overline{\mathbb{𝔻}}$.
• The point spectrum of $S$ is empty: ${\sigma }_p(S)=\mathrm{\varnothing }$.
• The approximate point spectrum of $S$ is the unit circle: ${\sigma }_{ap}(S)=\mathbb{𝕋}$.

Proof

To show $\sigma (S)=\overline{\mathbb{𝔻}}$, use the matrix representation of $(zI-S{)}^{-1}$, and note that it is bounded iff $|z|>1$. To show ${\sigma }_p(S)=\mathrm{\varnothing }$, directly show that $Sa=\lambda a$ implies $a=0$.

To show ${\sigma }_{ap}(S)=\mathbb{𝕋}$, note that $\Vert (zI-S)a{\Vert }^2\ge 1+|z{|}^2-2|z|=(1-|z|{)}^2$ for any $z\in \mathbb{ℂ},a\in {\ell }^2$ with $\Vert a\Vert =1$, so ${\sigma }_{ap}(S)\subset \overline{\mathbb{𝔻}}\setminus \mathbb{𝔻}=\mathbb{𝕋}$. Conversely, for any $z\in \mathbb{𝕋}$, construct the following unit vector$$a=\frac{1}{\sqrt{N}}(1,z^{-1},z^{-2},\dots ,z^{-(N-1)},0,0,\dots )$$then $\Vert (zI-S)a{\Vert }^2=2/N$, which converges to 0 at $N\to \mathrm{\infty }$.

The spectral properties of $S^{*}$ differ significantly from those of $S$:

• $\sigma (S^{*})=\overline{\mathbb{𝔻}}$ (since $\sigma (A^{*})=\overline{\sigma (A)}$).
• The point spectrum ${\sigma }_p(S^{*})$ is the entire open unit disk $\mathbb{𝔻}$. For any $\lambda \in \mathbb{𝔻}$, the corresponding eigenvector is the geometric sequence $(1,\lambda ,{\lambda }^2,{\lambda }^3,\dots )$.
• The approximate point spectrum ${\sigma }_{ap}(S^{*})$ is the entire closed unit disk $\overline{\mathbb{𝔻}}$. To show this, it remains to show $\mathbb{𝕋}\subset {\sigma }_{ap}(S^{*})$, which can be proven by a similar construction as before, using $a=\frac{1}{\sqrt{N}}(1,z^1,z^2,\dots ,z^{(N-1)},0,0,\dots )$.

Hardy space model

The unilateral shift can be studied using complex analysis.

Define the Hardy space $H^2$ as the Hilbert space of analytic functions $f(z)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{z}^n$ on the open unit disk $\mathbb{𝔻}$ for which the sequence of coefficients $(a_n)$ is in ${\ell }^2$.

Define the multiplication operator $M_z$ on $H^2$: $$(M_{z}f)(z)=zf(z)$$then $S$ and $M_z$ are unitarily equivalent via the unitary map $U:{\ell }^2\to H^2$ defined by$$U(a_0,a_1,a_2,\dots )=\sum _{n=0}^{\mathrm{\infty }}{a}_n{z}^n$$which gives $U^{*}{M}_{z}U=S$. Using this unitary equivalence, it is common in the literature to use $S$ to denote $M_z$ and to treat $H^2$ as the primary setting for the unilateral shift.

Commutant

The commutant of an operator $A$, denoted $\{A{\}}^{\prime }$, is the algebra of all bounded operators that commute with $A$. The commutant of the unilateral shift is the algebra of multiplication operators on $H^2$ by bounded analytic functions.$$\{S{\}}^{\prime }=\{M_{\phi }:\phi \in H^{\mathrm{\infty }}\}$$Here, $H^{\mathrm{\infty }}$ is the space of bounded analytic functions on $\mathbb{𝔻}$, and $(M_{\phi }f)(z)=\phi (z)f(z)$.

Cyclic vectors

A vector $x$ is a cyclic vector for an operator $A$ if the linear span of its orbit $\{A^{n}x:n\ge 0\}$ is dense in the space. We have:

• For the unilateral shift $S$ on $H^2$, the cyclic vectors are the outer functions.
• A function $f\in H^2$ that has a zero in the open unit disk $\mathbb{𝔻}$ is not a cyclic vector. This is because every function in the span of its orbit will also be zero at that point, so the subspace cannot be dense.
• A function $f\in H^2$ that is bounded away from zero (i.e., $\underset{z\in \mathbb{𝔻}}{\inf }|f(z)|>0$) is a cyclic vector.
• A function $f\in H^2$, that is in the open unit disk $\mathbb{𝔻}$ is nonzero but $\underset{z\in \mathbb{𝔻}}{\inf }|f(z)|=0$, may or may not be cyclic. For example, $f(z)=1-z$ is a cyclic vector.

The cyclic vectors are precisely the outer functions.

Lattice of invariant subspaces

The $S$-invariant subspaces of $H^2$ are completely characterized analytically. Specifically, they are precisely $M_u(H^2)$ where $u$ is an inner function.

The $S$-invariant subspaces make up a lattice of subspaces. The two lattice operators, join and meet, correspond to operations on inner functions.

Given two invariant subspaces $M_u(H^2),M_v(H^2)$, we have $M_u(H^2)\subset M_v(H^2)$ iff $u/v\in H^2$.

See also

• Bilateral shift
• Hardy space
• H²
• Operator theory
• Beurling's theorem
• Toeplitz operator

References

• Garcia, Stephan Ramon; Mashreghi, Javad; Ross, William T. (2023). "5. The Unilateral Shift". Operator Theory by Example. Oxford University Press. pp. 109–132. doi:10.1093/oso/9780192863867.003.0005. ISBN 9780192863867. https://academic.oup.com/book/45766/chapter/399055174.