nLab linear isometry

Idea

A linear isometry is a linear map

$$\phi \,\colon\, \mathscr{H}_1 \longrightarrow \mathscr{H}_2$$

between normed vector spaces which preserves the norm:

$$\big\Vert \psi(-) \big\Vert_2 \;=\; \big\Vert - \big\Vert_1 \,.$$

In particular, when the norm is induced by (Hermitian) inner products ${\langle \cdot \vert \cdot \rangle}_{i}$ (as on Hilbert spaces)

$${\big\Vert \psi \big\Vert}_i \,\equiv\, \sqrt{ \left\langle \psi \vert \psi \right\rangle_i }$$

then a linear isometry is an isometry in that it preserves these inner products:

$$\big\langle \phi(-) \vert \phi(-) \big\rangle_2 \;=\; \left\langle - \vert - \right\rangle_1$$

or equivalently, in terms of adjoint operators:

(1)$$\phi^\dagger \cdot \phi \,=\, Id_{\mathscr{H}_1} \,.$$

Remark

(relation to unitary operators)
If in addition to (1) also the reverse condition $\phi \cdot \phi^\dagger \,=\, Id_{\mathscr{H}_2}$ holds, then $\phi$ is called a unitary operator, which is the case iff $\phi$ is surjective map.

Properties

Proposition

Linear isometries are injective maps.

Proof

For $\left\vert \psi \right\rangle, \left\vert \psi' \right\rangle \,\colon\, \mathscr{H}$ we need to show that

$$\phi\left\vert \psi \right\rangle \;=\; \phi\left\vert \psi' \right\rangle \;\;\;\; \Rightarrow \;\;\;\; \left\vert \psi \right\rangle \;=\; \left\vert \psi' \right\rangle \mathrlap{\,.}$$

But by non-degeneracy of the norm this is equivalent to showing

$${ \Big\Vert \phi \big( \left\vert \psi \right\rangle - \left\vert \psi' \right\rangle \big) \Big\Vert } \,=\, 0 \;\;\;\;\;\;\; \Rightarrow \;\;\;\;\;\;\; { \Big\Vert \left\vert \psi \right\rangle - \left\vert \psi' \right\rangle \Big\Vert } \,=\, 0 \mathrlap{\,.}$$

But since $\phi$ is an isometry, here the left hand side already coincides with the right hand side and hence certainly implies it.

Related concepts

• isometry

• linear isomorphism

• unitary operator

References

See also:

• ProofWiki, Linear Isometry

• Wikipedia, Linear isometry