Hermitian and Unitary Operators

We start with two lemmas which will be useful in further considerations. \(\\\)

Lemma 0.

Let \(\ y\in V\,,\ \) where \(\ V\ \) is a unitary space. Then

\[\langle x,y\rangle = 0 \quad\text{for all}\quad x\in V \qquad\Leftrightarrow\qquad y = \theta\,.\]

Proof.

\(\ \Rightarrow\ :\ \) If \(\ \ \langle x,y\rangle = 0\,,\ \ x\in V\,,\ \) then putting \(\ x=y\ \) we obtain \(\ \langle y,y\rangle = 0\,,\ \) and thus \(\ y=\theta.\)

\(\ \Leftarrow\ :\ \) If \(\ y = \theta\,,\ \) then by properties of an inner product, \(\ \langle x,y\rangle = \langle x,\theta\rangle = 0\,.\)

If \(\ \,y_1,\,y_2\,\in\,V\,,\ \) where \(\ V\ \) is a unitary space, \(\,\) then

\[\langle x,y_1\rangle = \langle x,y_2\rangle \quad\text{for all}\quad x\in V \qquad\Leftrightarrow\qquad y_1 = y_2\,.\]

Indeed, if the condition \(\ \langle x,y_1\rangle = \langle x,y_2\rangle\ \) holds for all \(\ x\in V\,,\) then

\[ \begin{align}\begin{aligned}\langle\,x,\,y_1-y_2\,\rangle = 0\,,\quad x\in V \qquad\Leftrightarrow\qquad y_1-y_2=\theta \qquad\Leftrightarrow\qquad y_1=y_2\,.\\\;\end{aligned}\end{align} \]

The zero operator \(\ \mathcal{O}\ \) which occurs in the next lemma is defined by: \(\ \ \mathcal{O}(v)=\theta\,,\ \ v\in V. \\\)

Lemma 1.

Let \(\ F\ \) be a linear operator defined on a unitary space \(\,V.\ \) Then

\[\langle\,x,Fy\,\rangle\,=\,0 \quad\text{for all}\quad x,y\in V \qquad\Leftrightarrow\qquad F=\mathcal{O}\,.\]

Proof.

\(\ \Rightarrow\ :\ \) If \(\ \ \langle x,Fy\rangle = 0 \ \) for all \(\ \ x,y\in V\,,\ \) then putting \(\ x=Fy\ \) we obtain:

\(\ \langle Fy,Fy\rangle = 0\ \,\) and thus \(\ \,Fy=\theta\ \,\) for every \(\ y\in V\,,\ \,\) which means that \(\ \,F=\mathcal{O}\,.\)

\(\ \Leftarrow\ :\ \) If \(\ \,F=\mathcal{O}\,,\ \) then for arbitrary \(\ x,y\in V:\ \ \langle x,Fy\rangle = \langle x,\mathcal{O}y\rangle = \langle x,\theta\rangle = 0\,.\)

Corollary.

If \(\ F,\,G\ \) are linear operators defined on a unitary space \(\,V,\ \,\) then

\[\langle\,x,Fy\,\rangle\,=\,\langle\,x,G\,y\,\rangle \quad\text{for all}\quad x,y\in V \qquad\Leftrightarrow\qquad F=\,G\,.\]

Indeed, if the condition \(\ \langle\,x,Fy\,\rangle\,=\,\langle\,x,Gy\,\rangle\,\) holds for every \(\ x,y\in V\,,\) then

\[\langle\,x,(F-G)\,y\,\rangle\,=\,0\,,\quad x,y\in V\qquad\Leftrightarrow\qquad F-G=\mathcal{O} \qquad\Leftrightarrow\qquad F=G\,.\]

Hermitian Conjugation of a Linear Operator

Definition.

Hermitian conjugation \(\,\) of a linear operator \(\ F\in\,\text{End}(V)\) \(\\\) is a linear operator \(\ F^+\in\,\text{End}(V)\ \) which satisfies the condition

(1)\[\langle\,x,F^+y\,\rangle\,=\,\langle\,Fx,y\,\rangle \quad\text{for all}\quad x,y\in V\,.\]

Observe that if the condition (1) holds, \(\,\) then

(2)\[\langle\,F^+x,\,\,y\,\rangle\ =\ \langle\,y,\,F^+x\,\rangle^*\ =\ \langle\,Fy,\,x\,\rangle^*\ =\ \langle\,x,\,Fy\,\rangle\,.\]

The reverse implication is also true: the condition (2) implies (1). We may write then

Corollary.

Hermitian conjugation \(\,F^+\,\) of the operator \(\,F\in\text{End}(V)\ \) \(\\\) satisfies two equivalent conditions:

(3)\[\begin{split}\begin{array}{l} \langle\,x,F^+y\,\rangle\,=\,\langle\,Fx,y\,\rangle\,, \\ \\ \langle\,F^+x,\,\,y\,\rangle\ =\ \langle\,x,\,Fy\,\rangle \end{array} \qquad\text{for all}\quad x,y\in V\,.\end{split}\]

We will explain the following issues that arise from using such a definition:

1. do the formulae (3) define the operator \(\,F^+\,,\ \) i.e., for a given operator \(F\,\) is it possible to effectively determine the image \(\,F^+y\ \) of any vector \(\,y\in V\,?\)

2. is the operator \(\,F^+\ \) linear?

3. is the operator \(\,F^+\ \) uniquely defined?

For the sake of an answer, assume that \(\,\dim V=n\ \) and \(\ \mathcal{B}=(u_1,u_2,\ldots,u_n)\ \) is an orthonormal basis.

1. Substitution \(\ x=u_i\ \) in (1) gives a formula for the \(\,i\)-th coordinate of a vector \(\,F^+y\,\) in terms of basis vectors:

   \[(F^+y)_{\,i}\ =\ \langle u_i,F^+y\rangle\ =\ \langle Fu_i,y\rangle\,, \qquad i=1,2,\dots,n.\]

   Hence, the vector \(\,F^+y\ \) is determined by its coordinates in basis \(\ \mathcal{B}.\)

2. Properties of an inner product imply that for arbitrary \(\,x\in V:\)

   \[\begin{split}\begin{array}{rcl} \langle\,x,\,F^+(\alpha_1\,y_1+\alpha_2\,y_2)\,\rangle & = & \langle\,Fx,\,\alpha_1\,y_1\,+\,\alpha_2\,y_2\,\rangle\,= \\ \\ & = & \alpha_1\,\langle Fx,y_1\rangle\,+\,\alpha_2\:\langle Fx,y_2\,\rangle\,= \\ \\ & = & \alpha_1\,\langle x,F^+y_1\rangle\,+\,\alpha_2\:\langle x,F^+y_2\,\rangle\,= \\ \\ & = & \langle\,x,\,\alpha_1\,F^+y_1+\alpha_2\,F^+y_2\,\rangle\,. \end{array}\end{split}\]

   Corollary to Lemma 0. implies linearity of the operator \(\,F^+:\)

   \[F^+(\alpha_1\,y_1+\alpha_2\,y_2)\,=\ \alpha_1\,F^+y_1+\alpha_2\,F^+y_2\,, \qquad\alpha_1,\alpha_2\in C,\ \ y_1,y_2\in V\,.\]

3. In order to prove that the operator \(\,F^+\ \) is uniquely defined assume that there exists an operator \(\,G\ \) which satisfies the condition (1):

   \[\langle\,x,F^+y\,\rangle\,=\,\langle\,Fx,y\,\rangle \quad\text{and}\quad \langle\,x,Gy\,\rangle\,=\,\langle\,Fx,y\,\rangle \quad\text{for all}\quad x,y\in V\,.\]

   This means that \(\ \langle\,x,F^+y\,\rangle\,=\,\langle\,x,Gy\,\rangle\ \) for all \(\ x,y\in V\,.\ \) Lemma 1. implies that in this case \(\,G=F^+.\ \) Hence, the conditions (3) define the operator \(\,F^+\ \) uniquely.

The issue of existence and uniqueness of the operator \(\,F^+\ \) may be independently explained by

Theorem 8.

A linear operator \(\,\widetilde{F}\ \) is a Hermitian conjugation of the linear operator \(\,F\ \) \(\\\) if and only if its matrix in an orthonormal basis \(\,\mathcal{B}\ \) \(\\\) is a Hermitian congujate of the matrix of the operator \(\,F\ \) in this basis:

\[\widetilde{F}=F^+\quad\Leftrightarrow\quad M_{\mathcal{B}}(\widetilde{F})\,=\,[\,M_{\mathcal{B}}(F)\,]^+\,, \qquad F,\,\widetilde{F}\,\in\,\text{End}(V)\,.\]

Proof. \(\,\) Let \(\ \ \mathcal{B}=(u_1,u_2,\ldots,u_n),\ \ M_{\mathcal{B}}(F)=\boldsymbol{F}=[\,\varphi_{ij}\,]_{n\times n}\,,\ \ M_{\mathcal{B}}(\widetilde{F})=\widetilde{\boldsymbol{F}}= [\,\widetilde{\varphi}_{ij}\,]_{n\times n}\,.\)

\(\ \Rightarrow\ :\ \) Assume that \(\ \widetilde{F}=F^+,\ \) that is, the operator \(\ \widetilde{F}\ \) satisfies the condition (1):

(4)\[\langle\,x,\widetilde{F}y\,\rangle\,=\,\langle\,Fx,y\,\rangle \quad\text{for all}\quad x,y\in V\,.\]

In particular, for \(\,x=u_i,\,y=u_j\ \) we obtain:

\[\widetilde{\varphi}_{ij}\,=\, \langle\,u_i\,,\widetilde{F}u_j\,\rangle\,=\,\langle\,Fu_i,u_j\,\rangle\,=\, \langle\,u_j,Fu_i\,\rangle^*\,=\, \varphi_{ji}^*\,=\,\varphi_{ij}^+\,,\qquad i,j=1,2,\ldots,n.\]

Equality of the corresponding matrix elements implies equality of matrices:

\[\widetilde{\boldsymbol{F}}=\boldsymbol{F}^+\,, \qquad\text{and thus}\qquad M_{\mathcal{B}}(\widetilde{F})\ =\ \left[\,M_{\mathcal{B}}(F)\,\right]^+.\]

\(\ \Leftarrow\ :\ \) Assume that \(\ M_{\mathcal{B}}(\widetilde{F})\ =\ \left[\,M_{\mathcal{B}}(F)\,\right]^+,\ \,\) that is, \(\ \widetilde{\boldsymbol{F}}=\boldsymbol{F}^+.\)

We have to show that the operator \(\,\widetilde{F}\ \) satisfies the condition (4), \(\,\) which will mean that \(\ \widetilde{F}=F^+.\)

Let \(\ \ x\,=\,\displaystyle\sum_{i\,=\,1}^n\ \alpha_i\,u_i\,,\ \ y=\displaystyle\sum_{j\,=\,1}^n\ \beta_j\,u_j\,.\ \) Then

\[\begin{split}\begin{array}{rcl} \langle\,x,\widetilde{F}y\,\rangle & =\ & \left\langle\ \;\displaystyle\sum_{i\,=\,1}^n\ \alpha_i\,u_i\,,\ \widetilde{F} \left(\:\sum_{j\,=\,1}^n\ \beta_j\,u_j\right)\;\right\rangle\ \ = \\ \\ & =\ \ & \displaystyle\sum_{i,\,j=1}^n\,\alpha_i^*\,\beta_j\ \langle\,u_i,\widetilde{F}u_j\,\rangle \ \,=\ \, \sum_{i,\,j=1}^n\,\alpha_i^*\,\beta_j\ \widetilde{\varphi}_{ij}\ \ = \\ & =\ \ & \displaystyle\sum_{i,\,j=1}^n\,\alpha_i^*\,\beta_j\ \varphi_{ij}^+\ \,=\ \, \sum_{i,\,j=1}^n\,\alpha_i^*\,\beta_j\ \varphi_{ji}^*\ \,= \\ & =\ \ & \displaystyle\sum_{i,\,j=1}^n\,\alpha_i^*\,\beta_j\ \langle\,u_j,Fu_i\,\rangle^*\ \,=\ \, \sum_{i,\,j=1}^n\,\alpha_i^*\,\beta_j\ \langle\,Fu_i,u_j\,\rangle\ \,= \\ \\ & =\ \ & \left\langle\ F\,\left(\,\displaystyle\sum_{i\,=\,1}^n\ \alpha_i\,u_i\right)\,,\ \ \displaystyle\sum_{j\,=\,1}^n\ \beta_j\,u_j\,\right\rangle\ \ = \ \ \langle\,Fx,y\,\rangle\,. \end{array}\end{split}\]

If we use the notion of Hermitian conjugation also for the operation of Hermitian conjugation, we may write

Corollary.

Hermitian conjugation of a linear operator \(\,F\ \) is equivalent to Hermitian conjugation of the matrix of this operator in every orthonormal basis \(\,\mathcal{B}:\)

(5)\[M_{\mathcal{B}}(F^+)\ =\ \left[\,M_{\mathcal{B}}(F)\,\right]^+\,.\]

\(\;\)

Properties of Hermitian conjugation. \(\\\)

1. Hermitian conjugation of sum of two operators is equal to sum of their Hermitian conjugations:

   \[(F+G)^+\,=\;F^++\:G^+\,,\qquad F,\,G\,\in\,\text{End}(V)\,.\]

2. Multiplication of an operator by a complex number \(\ \alpha\ \) multiplies its Hermitian conjugation by \(\ \alpha^*:\)

   \[(\alpha\,F)^+\ =\ \;\alpha^*\,F^+\,,\qquad\alpha\in C,\ \ F\in\text{End}(V)\,.\]

3. Hermitian conjugate of product (i.e., composition) of operators is equal to product of Hermitian conjugations with reverse order of the factors:

   \[(F\,G)^+\ =\ \;G^+\,F^+\,,\qquad F,\,G\,\in\,\text{End}(V)\,.\]

4. Double Hermitian conjugation returns the initial operator :

   \[(F^+)^+\,=\ F\,,\qquad F\in\text{End}(V)\,.\]

Proof of the properties bases on Lemma 1. preceding this section. \(\\\)

1. Definition of a sum of two linear operators implies the equalities:

   \[ \begin{align}\begin{aligned}\begin{split}\begin{array}{lcl} \langle\,x,\,(F+G)^+\,y\,\rangle & \ = & \ \langle\,(F+G)\,x,\,y\,\rangle\ \ = \\ \\ & \ = & \ \langle\,Fx+Gx,\,y\,\rangle\ \ = \\ \\ & \ = & \ \langle\,Fx,y\,\rangle + \langle\,Gx,\,y\,\rangle\ \ = \\ \\ & \ = & \ \langle\,x,F^+y\,\rangle + \langle\,x,G^+y\,\rangle\ \ = \\ \\ & \ = & \ \langle\,x,F^+y+G^+y\,\rangle\quad=\quad\langle\,x,(F^+\!+G^+)\,y\,\rangle\,; \\ & & \end{array}\end{split}\\\langle\,x,\,(F+G)^+\,y\,\rangle = \langle\,x,(F^+\!+G^+)\,y\,\rangle, \ \ x,y\in V \quad\Rightarrow\quad (F+G)^+\ =\ F^++\,G^+ .\end{aligned}\end{align} \]

2. A proof of this property proceeds in a similar way as in the point 1.

3. By definition of composition of two linear operators:

   \[ \begin{align}\begin{aligned}\begin{split}\begin{array}{rclcl} \langle\,x,\,(F\,G)^+\,y\,\rangle & = & \langle\,(F\,G)\,x,\,y\,\rangle\ \ =\ \ \langle\,F(Gx),\,y\,\rangle & = & \\ \\ & = & \langle\,Gx,F^+y\,\rangle\ \ =\ \ \langle\,x,G^+(F^+y)\,\rangle & = & \langle\,x,(G^+F^+)\,y\,\rangle\,; \\ & & & & \end{array}\end{split}\\\langle\,x,\,(F\,G)^+\,y\,\rangle = \langle\,x,(G^+F^+)\,y\,\rangle,\ \ x,y\in V \qquad\Rightarrow\qquad (F\,G)^+\ =\ G^+F^+\,.\end{aligned}\end{align} \]

4. Formulae (3) imply:

   \[\begin{split}\begin{array}{c} \langle\,x,\,(F^+)^+\,y\,\rangle\ =\ \langle\,F^+x,\,\,y\,\rangle\ =\ \langle\,x,\,Fy\,\rangle\,; \\ \\ \langle\,x,\,(F^+)^+\,y\,\rangle\ =\ \langle\,x,\,Fy\,\rangle\,,\quad x,y\in V \qquad\Rightarrow\qquad (F^+)^+\ =\ F\,. \end{array}\end{split}\]

Corollary.

Hermitian conjugation is an antilinear operation:

\[(\alpha\,F+\beta\,G)^+\ =\ \, \alpha^*\,F^+\,+\,\beta^*\,G^+\,,\qquad \alpha,\beta\in C\,,\quad F,\,G\in\,\text{End}(V)\,.\]

In view of Theorem 8. and its corollary, an anlogy between properties of Hermitian conjugation of matrices and linear operators is not coincidental. \(\\\)

Hermitian Operators

Definition.

A linear operator \(\,F\in\text{End}(V)\ \) is a \(\,\) Hermitian operator \(\\\) if it is equal to its Hermitian conjugation:

\[F=F^+\qquad\text{that is}\qquad\langle\,x,Fy\,\rangle\ =\ \langle\,Fx,y\,\rangle\quad \text{for all}\quad x,y\in V\,.\]

In particular, for a Hermitian operator \(\,F\,:\)

(6)\[\langle\,x,Fx\,\rangle\ =\ \langle\,Fx,x\,\rangle\quad \text{for all}\quad x\in V\,.\]

A corollary to Theorem 8. implies immediately

Theorem 9.

A linear operator \(\,F\in\text{End}(V)\ \) is Hermitian if and only if in every orthonormal basis \(\,\mathcal{B}\ \) of the space \(\ V\ \) its matrix is Hermitian:

\[F=F^+\quad\Leftrightarrow\quad M_{\mathcal{B}}(F)\,=\,[\,M_{\mathcal{B}}(F)\,]^+\,, \qquad F\in\text{End}(V)\,.\]

In what follows we will make use of a criterion for a complex number \(\,z\,\) to be real:

\[z\in R\quad\Leftrightarrow\quad z^*=\,z\,,\qquad z\in C\,.\]

Properties of Hermitian operators. \(\\\)

Let \(\,F\in\text{End}(V)\ \) be a Hermitian operator. Then: \(\\\)

1. For every \(\,x\in V\,\) an expression \(\,\langle x,Fx\rangle\ \) is a real number.

   Indeed, according to the definition of an inner product and the formula (6), we have

   \[\langle\,x,Fx\,\rangle^*\ =\ \langle\,Fx,x\,\rangle\ =\ \langle\,x,Fx\,\rangle \qquad\Rightarrow\qquad\langle\,x,Fx\,\rangle\in R.\]

   One can prove that the condition \(\,\forall_{x\in V}\langle x,Fx\rangle\in R\,\) is not only necessary, but also sufficient for an operator \(\,F\ \) to be Hermitian. Hence,

   Corollary.

   If \(\,F\in\text{End}(V)\,,\ \) then \(\qquad F\ =\ F^+\quad\Leftrightarrow\quad \langle\,x,Fx\,\rangle\in R\,,\quad x\in V\,.\)

\(\;\)

2. Eigenvalues of the operator \(\,F\ \) are real.

   Proof. \(\,\) Assume that \(\quad Fv\,=\,\lambda\,v\,,\quad v\in V\!\smallsetminus\!\{\theta\},\quad\lambda\in C\,.\ \ \) By the formula (6),

   \begin{eqnarray*} \langle\,v,Fv\,\rangle & \! = \! & \langle\,Fv,v\,\rangle\,, \\ \langle\,v,\,\lambda\,v\,\rangle & \! = \! & \langle\,\lambda\,v,v\,\rangle\,, \\ \lambda\ \langle v,v\rangle & \! = \! & \lambda^*\;\langle v,v\rangle\,, \quad\text{where}\quad\langle v,v\rangle>0\,; \\ \lambda & \! = \! & \lambda^* \quad\ \ \Leftrightarrow\quad\ \ \,\lambda\in R\,. \end{eqnarray*}

3. Eigenvectors of the operator \(\,F\ \) associated with different eigenvalues are orthogonal.

   Proof. \(\,\) Assume that \(\quad Fv_1\,=\,\lambda_1\,v_1\,,\ \ Fv_2\,=\,\lambda_2\,v_2\,,\quad v_1,v_2\in V\!\smallsetminus\!\{\theta\}\,,\quad\lambda_1\neq\lambda_2\,.\)

   Starting from a definition of the Hermitian operator, we obtain

   \begin{eqnarray*} \langle\,v_1,Fv_2\,\rangle & = & \langle\,Fv_1,v_2\,\rangle \\ \langle\,v_1,\lambda_2\,v_2\,\rangle & = & \langle\,\lambda_1\,v_1,v_2\,\rangle \\ \lambda_2\ \langle v_1,v_2\rangle & = & \lambda_1^*\ \langle v_1,v_2\rangle \\ \lambda_2\ \langle v_1,v_2\rangle & = & \lambda_1\ \langle v_1,v_2\rangle \\ (\lambda_2-\lambda_1)\ \langle v_1,v_2\rangle & = & 0\,. \end{eqnarray*}

   Since by an assumption \(\ \lambda_1\neq\lambda_2\,,\ \) so we must have \(\ \langle v_1,v_2\rangle=0\,,\ \) as required. \(\\\)

Hence, eigenvectors of the Hermitian operator \(\,F\ \) associated with different eigenvalues comprise an orthogonal system. Recall that one may normalise any ortoghonal set of vectors and thus obtain an orthonormal system. This implies

Corollary.

If a Hermitian operator \(\,F\ \) defined on an \(\,n\)-dimensional unitary space \(\,V\ \) has \(\,n\,\) distinct eigenvalues, then one may choose an orthonormal basis for the space \(\,V\ \) consisting of eigenvectors of this operator.

Unitary Operators

Definition.

A linear operator \(\,U\ \) defined on a unitary space \(\,V\ \) is unitary \(\,\) if

\[U^+U\ =\ I\,,\]

where \(\,I\ \) is an identity operator defined by the condition: \(\,I(v)=v\,,\ v\in V.\)

Unitary operators are closely related with unitary matrices. Namely,

Theorem 10.

A linear operator \(\ U\ \) defined on an \(\,n\)-dimensional unitary space \(\,V\ \) is unitary if and only if in every orthonormal basis \(\ \mathcal{B}\ \) its matrix is unitary:

\[U^+U\ =\ I\qquad\Leftrightarrow\qquad \boldsymbol{B}^+\boldsymbol{B}\ =\ \boldsymbol{I}_n\,,\]

where \(\ \ \boldsymbol{B}\,=\,M_{\mathcal{B}}(U)\,,\ \ \boldsymbol{I}_n\ \) - \(\,\) an identity matrix of size \(\,n.\)

Proof.

The mapping \(\,M_{\mathcal{B}}:\,\text{End}(V)\rightarrow M_n(C)\,\) which assigns matrices to linear operators is - as an algebra isomorphism - bijective and multiplicative. \(\\\) Hence, and also by the equation (5), the following equalities are equivalent:

\begin{eqnarray*} U^+U & = & I\,, \\ M_{\mathcal{B}}(U^+U) & = & M_{\mathcal{B}}(I)\,, \\ M_{\mathcal{B}}(U^+)\,M_{\mathcal{B}}(U) & = & M_{\mathcal{B}}(I)\,, \\ \left[\,M_{\mathcal{B}}(U)\,\right]^+M_{\mathcal{B}}(U) & = & M_{\mathcal{B}}(I)\,, \\ \boldsymbol{B}^+\boldsymbol{B} & = & \boldsymbol{I}_n\,. \end{eqnarray*}

Properties of unitary operators.

Consider a unitary operator \(\ U\ \) defined on a unitary space \(\,V:\)

(7)\[U^+U\ =\ I\,.\]

0. The condition (7) implies existence of the inverse operator \(\ U^{-1}=U^+\ \) and an identity

   \[UU^+\ =\ \left(U^+\right)^+\,U^+\ =\ I\,,\]

   which means that if \(\ U\ \) is a unitary operator, then both the conjugate operator \(\ U^+\ \) and \(\,\) the inverse operator \(\ U^{-1}\,\) are unitary.

1. Product (i.e., composition) of two unitary operators is a unitary operator.

   Indeed, if \(\,U_1^+U_1=U_2^+U_2=I\,,\ \ \) then by properties of Hermitian conjugation \(\\\) of operators and by associativity of composition of operators, we obtain

   \[(U_1\,U_2)^+(U_1\,U_2)\ =\ (U_2^+\,U_1^+)(U_1\,U_2)\ =\ U_2^+\,(U_1^+U_1)\,U_2\ =\ U_2^+\,I\ U_2\ =\ U_2^+\,U_2\ =\ I\,.\]

   Hence, composition is an operation on the set of unitary operators. \(\\\) Further, since the identity operator \(\,I\,\) is unitary and so is an inverse of a unitary operator, we may write

   Corollary.

   Unitary operators defined on the space \(\,V\ \) together with their composition comprise a (nonabelian) group.

   \(\,\)

2. The operator \(\ U\ \) preserves an inner product:

   (8)\[\langle\,Ux,\,Uy\,\rangle\ =\ \langle x,y\rangle\,,\qquad x,y\in V\,,\]

   since \(\quad\langle\,Ux,\,Uy\,\rangle\ =\ \langle\,U^+U\,x,\,y\,\rangle\ =\ \langle\,Ix,y\,\rangle\ =\ \langle x,y\rangle\,.\)

   In particular, \(\ U\ \) preserves a vector norm:

   (9)\[\|\,Ux\,\|\ =\ \|x\|\,,\qquad x\in V\,,\]

   because \(\quad\|\,Ux\,\|^{\,2}\ =\ \langle\,Ux,Ux\,\rangle\ =\ \langle\,U^+U\,x,\,x\,\rangle\ =\ \langle x,x\rangle\,.\)

   Preservation of a norm (which is a generalised length) of a vector by the operator \(\ U\ \) allows to interpret its action as a generalised rotation of a vector in the space \(\ V.\)

   One may prove that the conditions \(\,\) (7), \(\,\) (8) \(\,\) and \(\,\) (9) \(\,\) are equivalent, \(\\\) and thus each of them may serve as a definition of a unitary operator. \(\\\)

3. Eigenvalues of the operator \(\ U\ \) are complex numbers with modulus \(\,1.\)

   Proof. \(\,\)

   Assume that \(\ v\ \) is an eigenvector of the operator \(\ U\ \) associated with an eigenvalue \(\ \lambda\in C.\ \) Then

   \begin{eqnarray*} Uv & = & \lambda\,v\,,\quad v\neq\theta\,, \\ \|\,Uv\,\| & = & \|\,\lambda\,v\,\|\,, \\ \|v\| & = & |\lambda|\ \|v\|\,, \\ (|\lambda|-1)\ \|v\| & = & 0\,,\quad\|v\|\neq 0\,, \\ |\lambda|-1 & = & 0\,, \\ |\lambda| & = & 1\,. \end{eqnarray*}

4. Eigenvectors of the operator \(\ U\ \) associated with different eigenvalues are orthogonal.

   Proof. \(\,\) Assume that \(\quad Uv_1\,=\,\lambda_1\,v_1\,,\ \ Uv_2\,=\,\lambda_2\,v_2\,,\quad v_1,v_2\in V\!\smallsetminus\!\{\theta\}\,,\quad\lambda_1\neq\lambda_2\,.\)

   We already know that \(\quad|\lambda_1|=|\lambda_2|=1\,,\quad\) and thus \(\quad|\lambda_1|^2=\lambda_1^*\,\lambda_1=1\,,\quad\lambda_1^*=1/\lambda_1\,.\ \) Hence,

   \(\langle v_1,v_2\rangle\ =\ \langle\,Uv_1,\,Uv_2\,\rangle\ =\ \langle\,\lambda_1\,v_1,\,\lambda_2\,v_2\,\rangle\ =\ \lambda_1^*\;\lambda_2\ \langle v_1,v_2\rangle\ =\ \displaystyle\frac{\lambda_2}{\lambda_1}\ \ \langle v_1,v_2\rangle\,,\)

   \(\left(\,1\ -\ \displaystyle\frac{\lambda_2}{\lambda_1}\;\right)\ \langle v_1,v_2\rangle\ =\ 0\,,\quad\text{and}\ \ \text{so}\quad\lambda_1\neq\lambda_2\,, \quad\text{implies}\quad\langle v_1,v_2\rangle\ =\ 0\,.\)