Shift operator

A shift operator translates the components of a function or sequence by a fixed displacement. Bilateral shifts act on doubly infinite sequences; the unilateral shift operator acts on a one-sided sequence space and has distinct forward and backward versions.

On ${\ell }^2({\mathbb{N}}_0)$, the forward unilateral shift operator is $S(x_0,x_1,\dots )=(0,x_0,x_1,\dots )$. Its adjoint operator is the backward shift $S^{\ast }(x_0,x_1,\dots )=(x_1,x_2,\dots )$. Thus $S^{\ast }S=I$ but $S{S}^{\ast }=I-P_0$, where $P_0$ projects onto the first coordinate.

The forward shift is an isometry but is not unitary, since it is not surjective. It has no eigenvalues: solving $Sx=\lambda x$ forces every coordinate to vanish. Nevertheless its spectrum is the closed unit disk. This separates the notions of spectrum and eigenvalue in infinite-dimensional operator theory.

On the absolutely summable sequence space, this weighted shift has $T^m{e}_j=e_{j+m}/[j(j+1)\cdots (j+m-1)]$, so $\Vert T^m\Vert =1/m!$. The spectral radius formula gives radius zero. Its spectrum is exactly $\{0\}$ because it is not onto, while it has no eigenvalues: zero is excluded by injectivity, and any nonzero eigenvalue equation forces all coordinates to vanish starting at the first. Thus it is a nonzero quasinilpotent element of the operator algebra.

On ${\ell }_1$, the unilateral shift operator $R$ and left shift operator $L$ satisfy $LR=I$ and $\Vert L\Vert =1$, while $L$ has nontrivial kernel. If $\Vert A\Vert <1$, the Neumann series makes $L(R+A)=I+LA$ invertible. Were $R+A$ invertible, this identity would express $L$ as a product of invertible operators, a contradiction. Thus the open unit ball about $R$ contains no invertible operator.

On the Hilbert space ${\ell }^2(\mathbb{N})$, the left shift deletes the first coordinate. It is the adjoint operator of the unilateral shift operator $S$. The identities $LS=I$ and $SLf=f-f_1{e}_1$ show that it is also the Moore–Penrose inverse of an operator for $S$; both maps have operator norm one.

An operator $T$ is a unilateral shift exactly when it is an isometry, $(\mathrm{Im}T{)}^{\perp }$ is one-dimensional, and ${\bigcap }_{n\ge 1}\mathrm{Im}(T^n)=\{0\}$. A unit vector $e_1$ in the first orthogonal complement is wandering, and $e_n=T^{n-1}{e}_1$ is a complete orthonormal basis.