# Unitary illustration

In mathematics, a unitary mapping or unitary transformation is a mapping between two complex scalar product spaces that contains the scalar product . Unitary mappings are always linear , injective , norm-preserving and distance- preserving , in some sources invertibility is also required. The bijective unitary images of a Skalarproduktraums in form with the sequential execution as linking a subset ofAutomorphism group of space. The eigenvalues of such a mapping all have the amount one. In finite-dimensional scalar product spaces, bijective unitary mappings can be represented by unitary matrices .

The corresponding counterparts for real scalar product spaces are orthogonal mappings . A bijective unitary mapping between two Hilbert spaces is also called a unitary operator .

## definition

A mapping between two complex scalar product spaces and is called unitary if for all vectors${\ displaystyle f \ colon V \ to W}$ ${\ displaystyle (V, \ langle \ cdot, \ cdot \ rangle _ {V})}$ ${\ displaystyle (W, \ langle \ cdot, \ cdot \ rangle _ {W})}$ ${\ displaystyle u, v \ in V}$ ${\ displaystyle \ langle f (u), f (v) \ rangle _ {W} = \ langle u, v \ rangle _ {V}}$ applies. A unitary mapping is therefore characterized by the fact that it receives the scalar product of vectors. In particular, a unitary mapping maps mutually orthogonal vectors and (that is, vectors whose scalar product is zero) onto mutually orthogonal vectors and . ${\ displaystyle v}$ ${\ displaystyle w}$ ${\ displaystyle f (v)}$ ${\ displaystyle f (w)}$ ## Examples

${\ displaystyle f \ colon V \ to V, \, x \ mapsto x}$ is trivially unitary. In coordinate space , unitary mappings are of straight form ${\ displaystyle \ mathbb {C} ^ {n}}$ ${\ displaystyle f \ colon \ mathbb {C} ^ {n} \ to \ mathbb {C} ^ {n}, \, x \ mapsto U \ cdot x}$ ,

where is a unitary matrix . In the space of square summable complex number sequences for example, the bilateral Shift${\ displaystyle U \ in \ mathbb {C} ^ {n \ times n}}$ ${\ displaystyle \ ell ^ {2}}$ ${\ displaystyle f \ colon \ ell ^ {2} ({\ mathbb {Z}}) \ rightarrow \ ell ^ {2} ({\ mathbb {Z}}), \, \, (a_ {n}) _ {n \ in {\ mathbb {Z}}} \ mapsto (a_ {n-1}) _ {n \ in {\ mathbb {Z}}}}$ represents a unitary mapping. Other important unitary mapping are integral transformations of the form

${\ displaystyle f \ colon L ^ {2} (\ mathbb {R}) \ to L ^ {2} (\ mathbb {R}), \, g \ mapsto \ int _ {\ mathbb {R}} K ( x, \ cdot) \, g (x) ~ dx}$ with a suitably chosen integral core . An important example of this is the Fourier transform , the unitarity of which follows from Plancherel's theorem. ${\ displaystyle K}$ ## properties

In the following, let the complex scalar product be linear in the first argument and semilinear in the second argument. The additions are omitted because the argument makes it clear which room is in question. ${\ displaystyle V, W}$ ### Linearity

A unitary mapping is linear , i.e. it holds for all vectors and scalars${\ displaystyle u, v \ in V}$ ${\ displaystyle a, b \ in \ mathbb {C}}$ ${\ displaystyle f (au + bv) = af (u) + bf (v)}$ .

It applies because of the sesquilinearity and the hermiticity of the scalar product

{\ displaystyle {\ begin {aligned} & \ langle f (u + v) -f (u) -f (v), f (u + v) -f (u) -f (v) \ rangle = \\ & = \ langle f (u + v), f (u + v) \ rangle -2 \ operatorname {Re} \ langle f (u + v), f (u) \ rangle -2 \ operatorname {Re} \ langle f (u + v), f (v) \ rangle + \ langle f (u), f (u) \ rangle +2 \ operatorname {Re} \ langle f (u), f (v) \ rangle + \ langle f (v), f (v) \ rangle = \\ & = \ langle u + v, u + v \ rangle -2 \ operatorname {Re} \ langle u + v, u \ rangle -2 \ operatorname {Re} \ langle u + v, v \ rangle + \ langle u, u \ rangle +2 \ operatorname {Re} \ langle u, v \ rangle + \ langle v, v \ rangle = \\ & = \ langle u + v, u + v \ rangle -2 \ langle u + v, u + v \ rangle + \ langle u + v, u + v \ rangle = 0 \ end {aligned}}} such as

{\ displaystyle {\ begin {aligned} & \ langle f (au) -af (u), f (au) -af (u) \ rangle = \ langle f (au), f (au) \ rangle -2 \ operatorname {Re} \ langle f (au), af (u) \ rangle + \ langle af (u), af (u) \ rangle = \\ & = \ langle f (au), f (au) \ rangle - 2 {\ bar {a}} \ operatorname {Re} \ langle f (au), f (u) \ rangle + | a | ^ {2} \ langle f (u), f (u) \ rangle = \ langle au, au \ rangle -2 \ langle au, au \ rangle + \ langle au, au \ rangle = 0. \ end {aligned}}} The additivity and the homogeneity of the mapping then follow from the positive definiteness of the scalar product.

### Injectivity

The kernel of a unitary mapping contains only the zero vector , because for holds ${\ displaystyle v \ in \ operatorname {ker} f}$ ${\ displaystyle \ langle v, v \ rangle = \ langle f (v), f (v) \ rangle = \ langle 0,0 \ rangle = 0}$ and it then follows from the positive definiteness of the scalar product . A unitary mapping is therefore always injective . If and are finite-dimensional with the same dimension, then, based on the ranking , applies${\ displaystyle v = 0}$ ${\ displaystyle V}$ ${\ displaystyle W}$ ${\ displaystyle \ dim V = \ dim \ mathrm {ker} (f) + \ dim \ mathrm {im} (f) = \ dim \ mathrm {im} (f)}$ and thus is also surjective and therefore bijective . Unitary mappings between infinite-dimensional spaces need not necessarily be surjective; an example of this is the right shift. ${\ displaystyle f}$ ### Standard maintenance

A unitary mapping receives the scalar product norm of a vector, that is

${\ displaystyle \ | f (v) \ | = \ | v \ |}$ ,

because it applies

${\ displaystyle \ | f (v) \ | ^ {2} = \ langle f (v), f (v) \ rangle = \ langle v, v \ rangle = \ | v \ | ^ {2}}$ .

Conversely, every linear mapping between two complex scalar product spaces that contains the scalar product norm is unitary. It applies on the one hand because of the sesquilinearity and the hermiticity of the scalar product

${\ displaystyle \ | f (u + v) \ | ^ {2} = \ | u + v \ | ^ {2} = \ langle u + v, u + v \ rangle = \ langle u, u \ rangle + 2 \ operatorname {Re} \ langle u, v \ rangle + \ langle v, v \ rangle = \ | u \ | ^ {2} +2 \ operatorname {Re} \ langle u, v \ rangle + \ | v \ | ^ {2}}$ and with the linearity of the mapping on the other hand

{\ displaystyle {\ begin {aligned} \ | f (u + v) \ | ^ {2} & = \ | f (u) + f (v) \ | ^ {2} = \ langle f (u) + f (v), f (u) + f (v) \ rangle = \\ & = \ | f (u) \ | ^ {2} +2 \ operatorname {Re} \ langle f (u), f (v ) \ rangle + \ | f (v) \ | ^ {2} = \ | u \ | ^ {2} +2 \ operatorname {Re} \ langle f (u), f (v) \ rangle + \ | v \ | ^ {2}. \ End {aligned}}} By equating the two equations, it then follows that the real parts agree. An analogous consideration of also results in the agreement of the imaginary parts and thus the unitarity of the mapping. ${\ displaystyle f (u + iv)}$ ### Isometry

Due to the maintenance of the standard and the linearity, a unitary mapping also contains the distance between two vectors, because the metric induced by the standard applies ${\ displaystyle d}$ ${\ displaystyle d (f (u), f (v)) = \ | f (u) -f (v) \ | = \ | f (uv) \ | = \ | uv \ | = d (u, v )}$ .

A unitary mapping thus represents an isometry . Conversely, every linear mapping between two scalar product spaces is unitary if it contains distances. It follows from the polarization formula

{\ displaystyle {\ begin {aligned} 4 \ langle f (u), f (v) \ rangle & = \ | f (u) + f (v) \ | ^ {2} - \ | f (u) - f (v) \ | ^ {2} + i \ | f (u) + if (v) \ | ^ {2} -i \ | f (u) -if (v) \ | ^ {2} = \ \ & = \ | f (u) -f (-v) \ | ^ {2} - \ | f (u) -f (v) \ | ^ {2} + i \ | f (u) -f ( -iv) \ | ^ {2} -i \ | f (u) -f (iv) \ | ^ {2} = \\ & = \ | u + v \ | ^ {2} - \ | uv \ | ^ {2} + i \ | u + iv \ | ^ {2} -i \ | u-iv \ | ^ {2} = 4 \ langle u, v \ rangle. \ End {aligned}}} If there is a bijective unitary mapping between two scalar product spaces, then the two spaces are isometrically isomorphic .

## Unitary endomorphisms

### Group properties

A unitary mapping represents an endomorphism . The sequential execution of two unitary endomorphisms is again unitary, because it holds ${\ displaystyle f \ colon V \ to V}$ ${\ displaystyle f \ circ g}$ ${\ displaystyle \ langle (f \ circ g) (u), (f \ circ g) (v) \ rangle = \ langle f (g (u)), f (g (v)) \ rangle = \ langle g (u), g (v) \ rangle = \ langle u, v \ rangle}$ .

If a unitary endomorphism is bijective, then its inverse is due to ${\ displaystyle f ^ {- 1}}$ ${\ displaystyle \ langle f ^ {- 1} (u), f ^ {- 1} (v) \ rangle = \ langle f (f ^ {- 1} (u)), f (f ^ {- 1} (v)) \ rangle = \ langle u, v \ rangle}$ also unitary. The bijective unitary endomorphisms of form a subgroup of the automorphism group . If the space is finite dimensional with the dimension , then this group is isomorphic to the unitary group . ${\ displaystyle V}$ ${\ displaystyle \ mathrm {Aut} (V)}$ ${\ displaystyle n}$ ${\ displaystyle \ mathrm {U} (n)}$ ### Eigenvalues

If an eigenvalue is a unitary mapping with an associated eigenvector , then ${\ displaystyle \ lambda \ in \ mathbb {C}}$ ${\ displaystyle f \ colon V \ to V}$ ${\ displaystyle v}$ ${\ displaystyle \ | v \ | = \ | f (v) \ | = \ | \ lambda v \ | = | \ lambda | \, \ | v \ |}$ and with it . The eigenvalues ​​of a unitary mapping therefore all have the amount one and are therefore of the form ${\ displaystyle | \ lambda | = 1}$ ${\ displaystyle \ lambda = e ^ {it}}$ with . ${\ displaystyle t \ in \ mathbb {R}}$ ### Mapping matrix

The mapping matrix of a unitary mapping with respect to an orthonormal basis of is always unitary , that is to say ${\ displaystyle A_ {f}}$ ${\ displaystyle f \ colon V \ to V}$ ${\ displaystyle \ {e_ {1}, \ ldots, e_ {n} \}}$ ${\ displaystyle V}$ ${\ displaystyle A_ {f} ^ {H} A_ {f} = I}$ ,

because it applies

${\ displaystyle \ langle f (v), f (w) \ rangle = (A_ {f} x) ^ {H} (A_ {f} y) = x ^ {H} A_ {f} ^ {H} A_ {f} y = x ^ {H} y = \ langle v, w \ rangle}$ ,

where and are. ${\ displaystyle v = x_ {1} e_ {1} + \ ldots + x_ {n} e_ {n}}$ ${\ displaystyle w = y_ {1} e_ {1} + \ ldots + y_ {n} e_ {n}}$ ## Unitary operators

A bijective unitary mapping between two Hilbert spaces is also called a unitary operator. Unitary operators are always constrained and, if so , normal . The inverse operator of a unitary operator is equal to its adjoint operator , that is, it holds ${\ displaystyle T \ colon V \ to W}$ ${\ displaystyle V = W}$ ${\ displaystyle T ^ {- 1} = T ^ {\ ast}}$ .

Important examples of unitary operators between function spaces are the Fourier transform and the time evolution operators of quantum mechanics .