An operator, $A$, is _unitary_ if and only if it preserves the [complex inner product](Inner%20products.md#Complex%20inner%20products). That is, an operator is unitary, if and only if $\langle A x, Ay \rangle = \langle x, y \rangle$ where $x$ and $y$ are elements of a [complex inner product space.](Complex%20vector%20spaces.md#Complex%20inner%20product%20spaces) ^219d76
By definition, unitary transformations on complex vectors preserve their [vector length.](Linear%20Algebra%20and%20Matrix%20Theory%20(index).md#Vector%20length) This length preserving quality means that unitary transformations are [linear isometries](Linear%20isometry.md) on complex inner product spaces.
%%At some point the concepts of transformation and operator needs to be defined and separated. This is one statement that hints at that.%%
It follows from the fact that unitary operators preserve the inner product and vector lengths that unitary operators also preserve [orthonormality](Orthonormal%20vectors.md) and more generally preserve [vector normalization](Vector%20normalization.md)s in complex vector spaces.
%%Additional proofs of these definitions in Linear algebra done right!%%
Equivalently an operator, $A,$ is unitary if and only if [$A^{-1}=A^{\dagger}$](Unitary%20operators.md#^332d8e) where $A^{-1}$ is its [inverse](Operator%20inverse.md) and $A^{\dagger}$ is its [adjoint.](Unitary%20operators.md#Adjoint%20properties) ^37f781
# [[Adjoint]] properties
Consider an [invertible operator](Operator%20inverse.md) $A$ and its adjoint $A^{\dagger}.$
1) if $A$ is a [unitary operator](Unitary%20operators.md) then $A^{-1}= A^{\dagger}$ ^332d8e
2) If $A$ and $B$ are both [unitary operators](Unitary%20operators.md) then $(AB)^{\dagger}$ is also unitary. ^64e49c
Property [1.](Unitary%20operators.md#^332d8e) is true by the [definition for unitary operators](Unitary%20operators.md#^219d76) since $\langle A x, Ay \rangle = \langle A^\dagger A x, y \rangle = \langle x, y \rangle$
We know [2.](Unitary%20operators.md#^64e49c) must be true by [property 1](Adjoint.md#^01011f) of the [adjoint](Adjoint.md) where we find that if $A$ and $B$ are both unitary then $(AB)^{\dagger}(AB) = B^{\dagger}A^{\dagger}AB = \mathbb{1}$
Properties [1.](Unitary%20operators.md#^332d8e) and [2.](Unitary%20operators.md#^64e49c) are consistent since $(AA^{-1})^\dagger = \mathbb{1}$ and $(A^{\dagger})^{-1}A^{-1} = (AA^\dagger)^{-1} = \mathbb{1}.$
# Unitary matrices
A [matrix](Matrices.md), $A$ is unitary if
$\sum_{l=1}^n A_{ij}^* A_{jk} = \delta_{jk} = \sum_{l=1}^n (A^\dagger)_{ij}(A)_{ik}$
where $^\dagger$ is the [conjugate transpose](Adjoint.md#conjugate%20transpose). This definition is equivalent to saying that the column vectors form an [orthonormal basis](Orthonormal%20bases.md). Furthermore, the eigenvectors also form an orthonormal basis ([[Eigenvalues and eigenvectors]]).
## Unitary matrix groups
.md#^ea2bd5)
## [Determinant](Determinants.md) properties
1) $\det{(A^\dagger A)} = |\det(A)|^2 =\det(I) = 1$
2) $\det(A) = \pm 1$
# Construction of a unitary operator from a [Hermitian operator](Hermitian%20operators.md)
Given a Hermitian operator $X$, we may construct a unitary operator as a [Complex matrix exponential](Complex%20matrix%20exponentials.md) where the unitary operator is of the form $U=e^{itX}\,\,\,\, \mbox{for}\,\, t\in\mathbb{R}$ ^8e52f6
Indeed _Any unitary operator may be expressed as a [complex matrix exponential](Complex%20matrix%20exponentials.md)_. ^932ae0
## Derivation of the matrix exponential form of unitary operators
Since we can express a [complex matrix exponential](Complex%20matrix%20exponentials.md) [as a limit of a sequence](Complex%20matrix%20exponentials.md#Complex%20matrix%20exponentials%20as%20limits%20of%20sequences) we may construct an operator of form $U=(\mathbb{1}+itX)$which is unitary for $t<<1$ since $UU^{\dagger}=(\mathbb{1}-itX)(\mathbb{1}+itX)=\mathbb{1}-itX+itX-i^2O(t^2)$$= \mathbb{1}+O(t^2)$ and would be represented by a matrix approaching identity element as $t\rightarrow 0.$^7dab13
## Why unitary operators are matrix exponentials
One way of understanding why unitary operators are exponentials is to consider results from [Lie group theory.](Lie%20groups.md) [[U(n)]], the group containing all $n\times n$ unitary operator only contains [Lie group](Lie%20groups.md) elements, which are [matrix exponentials.](Matrix%20exponentials.md)
## Generators
In this context we say that $X$ is a _generator_ of a [unitary operator](Unitary%20operators.md). An example of such a generator is the [Infinitessimal generator](Infinitessimal%20generators.md) which is considered a generator for [strongly continuous](Strong%20continuity.md) unitary matrices in the [one-parameter unitary group](One-parameter%20unitary%20groups.md). The exact relationship between unitary operators and Hermitian operators on [Hilbert space](Hilbert%20Space.md) is established by [Stone's theorem](Stone's%20theorem.md).
#MathematicalFoundations/Algebra/AbstractAlgebra/LinearAlgebra/Operators/Matrices
#MathematicalFoundations/Analysis/FunctionalAnalysis