= Unilateral shift
{title2=$Se_n=e_{n+1}$}
On the <Hilbert space> $\ell^2(\mathbb N_0)$, the unilateral shift is the <bounded linear operator>
$$
S(x_0,x_1,x_2,\ldots)=(0,x_0,x_1,\ldots),\qquad Se_n=e_{n+1}.
$$
It preserves the norm, so $S^*S=I$. Its <adjoint operator> shifts the other way: $S^*(x_0,x_1,\ldots)=(x_1,x_2,\ldots)$. Consequently $SS^*=I-P_{e_0}$, 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.
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