Unilateral shift operator

The unilateral shifts on sequences move every component one place left or right. Their spectra illustrate behavior absent in finite dimensions.

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.