Unilateral shift ($S{e}_n=e_{n+1}$)

On the Hilbert space ${\ell }^2({\mathbb{N}}_0)$, the unilateral shift is the bounded linear operatorIt preserves the norm, so $S^{\ast }S=I$. Its adjoint operator shifts the other way: $S^{\ast }(x_0,x_1,\dots )=(x_1,x_2,\dots )$. Consequently $S{S}^{\ast }=I-P_{e_0}$, where $P_{e_0}$ is the orthogonal projection onto the first coordinate.

The range consists of sequences whose first coordinate is zero. Thus $S$ is an isometry but is not onto and is not unitary. Its adjoint has a one-dimensional kernel even though $S$ has zero kernel. These formulas give a concrete infinite-dimensional distinction: in finite dimensions an isometry from a space to itself is automatically onto, whereas the unilateral shift loses one output coordinate.