11.2 The unitary group
Orthogonal transformations between Euclidean spaces correspond to so-called unitary transformations between unitary spaces:
Definition 11.19 • Unitary transformation
Let \((V,\langle\cdot{,}\cdot\rangle)\) and \((W,\langle\!\langle\cdot{,}\cdot\rangle\!\rangle)\) be unitary spaces. An isomorphism \(f : V \to W\) is called a unitary transformation if \[\langle u,v\rangle=\langle\!\langle f(u),f(v)\rangle\!\rangle\] for all \(u,v \in V.\)
With this definition, all the statements about orthogonal transformations have corresponding statements for unitary transformations.
Definition 11.20 • Unitary group & unitary matrices
Let \((V,\langle\cdot{,}\cdot\rangle)\) be a unitary space. The set of unitary transformations from \((V,\langle\cdot{,}\cdot\rangle)\) to itself is called the unitary group of \((V,\langle\cdot{,}\cdot\rangle)\) and denoted by \(\mathrm{U}(V,\langle\cdot{,}\cdot\rangle).\)
A matrix \(\mathbf{R}\in M_{n,n}(\mathbb{C})\) is called unitary if \(f_\mathbf{R}: \mathbb{C}^n \to \mathbb{C}^n\) is an unitary transformation of \((\mathbb{C}^n,\langle\cdot{,}\cdot\rangle),\) where \(\langle\cdot{,}\cdot\rangle\) denotes the standard Hermitian scalar product of \(\mathbb{C}^n.\) The set of unitary \(n\times n\)-matrices is denoted by \(\mathrm{U}(n)\) and called the unitary group.
Like the orthogonal group, the unitary group is indeed a group:
Proposition 11.21 Let \((V,\langle\cdot{,}\cdot\rangle)\) be a unitary space. Then the set \(\mathrm{U}(V,\langle\cdot{,}\cdot\rangle)\) is a group in the sense of Definition 8.4
Definition 8.4 • Group ➔
when the group operation is taken to be the composition of mappings. In particular, \(\mathrm{U}(n)\) is a group when the group operation is taken to be matrix multiplication.A group is a pair \((G,*_G)\) consisting of a set \(G\) together with a binary operation \(*_G : G \times G \to G,\) called group operation, so that the following properties hold:
The group operation \(*_G\) is associative, that is, \[(a*_G b)*_G c=a*_G(b*_Gc)\quad \text{for all }a,b,c \in G.\]
There exists an element \(e_G \in G\) such that \[e_G *_G a=a=a*_G e_G \quad \text{for all } a \in G.\] The element \(e_G\) is unique (see below) and is called the identity element of \(G\).
For each \(a \in G\) there exists an element \(b \in G\) such that \[a*_Gb=e_G=b*_G a.\] The element \(b\) is unique (see below) and called the inverse of \(a\) and is commonly denoted by \(a^{-1}.\)
We have the characterisation:
Lemma 11.22
For all \(n \in \mathbb{N}\) we have \[\mathrm{U}(n)=\left\{\mathbf{R}\in M_{n,n}(\mathbb{C}) | \overline{\mathbf{R}^T}\mathbf{R}=\mathbf{1}_{n}\right\}=\left\{\mathbf{R}\in \mathrm{GL}(n,\mathbb{C}) | \overline{\mathbf{R}^T}=\mathbf{R}^{-1}\right\}.\]
A unitary transformation has a unitary matrix representation with respect to an orthonormal basis:
Proposition 11.23
Let \(n \in \mathbb{N}\) and \((V,\langle\cdot{,}\cdot\rangle)\) be an \(n\)-dimensional unitary space equipped with an orthonormal ordered basis \(\mathbf{b}.\) Then an endomorphism \(f : V \to V\) is a unitary transformation if and only if its matrix representation \(\mathbf{R}=\mathbf{M}(f,\mathbf{b},\mathbf{b})\) with respect to \(\mathbf{b}\) is a unitary matrix.
We also have:
Corollary 11.24
Let \(n \in \mathbb{N}\) and \((V,\langle\cdot{,}\cdot\rangle)\) be an \(n\)-dimensional unitary space equipped with an orthonormal ordered basis \(\mathbf{b}.\) Then an ordered basis \(\mathbf{b}^{\prime}\) of \(V\) is orthonormal with respect to \(\langle\cdot{,}\cdot\rangle\) if and only if the change of basis matrix \(\mathbf{C}(\mathbf{b}^{\prime},\mathbf{b})\) is unitary.
The special unitary transformations are those with determinant one:
Definition 11.25 • Special unitary group & special unitary matrices
Let \((V,\langle\cdot{,}\cdot\rangle)\) be a unitary space. The subset of \(\mathrm{U}(V,\langle\cdot{,}\cdot\rangle)\) consisting of endomorphism whose determinant is \(1\) is called the special unitary group of \((V,\langle\cdot{,}\cdot\rangle)\) and is denoted by \(\mathrm{SU}(V,\langle\cdot{,}\cdot\rangle).\)
A matrix \(\mathbf{R}\in M_{n,n}(\mathbb{C})\) is called special unitary if \(\mathbf{R}\in \mathrm{U}(n)\) and \(\det \mathbf{R}=1.\) The set of special unitary \(n\times n\)-matrices is denoted by \(\mathrm{SU}(n)\) and called the special unitary group.
Again, we have indeed groups:
Example 11.26
While \(\mathrm{O}(1)\) just consists of the matrices \(\pm (1).\) The group \(\mathrm{U}(1)\) has infinitely many elements. Indeed \((z) \in \mathrm{U}(1)\) if and only if \(|z|^2=1\) so that \[\mathrm{U}(1)=\left\{(\mathrm{e}^{\mathrm{i}\vartheta}) | \vartheta \in \mathbb{R}\right\}.\]
11.3 Adjoint and normal endomorphisms
The notion of the adjoint for maps between unitary spaces is defined as in the case of Euclidean spaces. Given finite dimensional unitary spaces \((V,\langle\cdot{,}\cdot\rangle)\) and \((W,\langle\!\langle\cdot{,}\cdot\rangle\!\rangle)\) and a linear map \(f : V \to W,\) the adjoint of \(f\) is the unique map \(f^* : W \to V\) such that \[\langle\!\langle f(v),w\rangle\!\rangle=\langle v,f^*(w)\rangle\] for all \(v \in V\) and \(w \in W.\) The adjoint \(f^*\) is constructed by choosing an orthonormal basis \(\mathbf{b}\) of \(V\) and an orthonormal basis \(\mathbf{c}\) of \(W\) and by requesting that \[\tag{11.1} \mathbf{M}(f^*,\mathbf{c},\mathbf{b})=\overline{\mathbf{M}(f,\mathbf{b},\mathbf{c})}^T.\] The self-adjoint mappings of a unitary space \((V,\langle\cdot{,}\cdot\rangle)\) are then the linear maps \(f : V \to V\) satisfying \(f^*=f.\) If we equip \(V\) with an ordered orthonormal basis \(\mathbf{b},\) then a linear map \(f : V \to V\) is self-adjoint if and only if \(\mathbf{M}(f,\mathbf{b},\mathbf{b})\) is a Hermitian matrix.
Let \(\mathbf{A}\in M_{n,n}(\mathbb{C})\) and equip \(\mathbb{C}^n\) with the standard Hermitian scalar product \(\langle\cdot{,}\cdot\rangle.\) Then (11.1) implies that \((f_\mathbf{A})^*=f_{\overline{\mathbf{A}}^T}.\) This motivates the following definition:
Definition 11.27 • Adjoint matrix
For a matrix \(\mathbf{A}\in M_{n,n}(\mathbb{C})\) we define \[\mathbf{A}^*=\overline{\mathbf{A}}^T\] and call \(\mathbf{A}^*\) the adjoint matrix of \(\mathbf{A}.\)
The spectral theorem also holds in the unitary setting:
Theorem 11.28 • The spectral theorem for unitary spaces
Let \((V,\langle\cdot{,}\cdot\rangle)\) be a finite dimensional unitary space and \(f : V \to V\) a self-adjoint endomorphism. Then there exists an orthonormal basis of \(V\) consisting of eigenvectors of \(f.\) In particular, \(f\) is diagonalisable.
Again we have a matrix version:
Theorem 11.29
Let \(n \in \mathbb{N}\) and \(\mathbf{A}\in M_{n,n}(\mathbb{C})\) be a Hermitian matrix. Then there exists a unitary matrix \(\mathbf{R}\in M_{n,n}(\mathbb{C})\) such that \(\mathbf{R}\mathbf{A}\mathbf{R}^*\) is a diagonal matrix.
As in the real case we call an endomorphism \(f : V \to V\) of a unitary space \((V,\langle\cdot{,}\cdot\rangle)\) normal if \(f\circ f^*=f^*\circ f.\) Normal endomorphisms can be characterised in terms of the following lemma:
Lemma 11.30
Let \((V,\langle\cdot{,}\cdot\rangle)\) be a unitary space and \(f : V \to V\) an endomorphism. Then \(f\) is normal if and only if \[\Vert f(v)\Vert=\Vert f^*(v)\Vert\] for all \(v \in V.\)
Before we give a proof, we remark:
Remark 11.31
Let \(V\) be a finite dimensional \(\mathbb{C}\)-vector space and \(\langle\cdot{,}\cdot\rangle\) and Hermitian form on \(V.\) Similar to the real case, writing \(q(v)=\langle v,v\rangle,\) we obtain for all \(v,w \in V\) \[\begin{aligned} 4\operatorname{Re}\langle v,w\rangle&=2(\langle v,w\rangle+\overline{\langle v,w\rangle})=2(\langle v,w\rangle+\langle w,v\rangle)\\ &=\langle v+w,v+w\rangle -\langle v-w,v-w\rangle=q(v+w)-q(v-w), \end{aligned}\] so that the real part of \(\langle\cdot{,}\cdot\rangle\) is determined by \(q.\) On the other hand, we have for all \(v,w \in V\) \[\operatorname{Re}(\langle \mathrm{i}v,w\rangle)=-\operatorname{Re}(\mathrm{i}\langle v,w\rangle)=\operatorname{Im}(\langle v,w\rangle),\] so that the imaginary part of \(\langle\cdot{,}\cdot\rangle\) is determined by \(q\) as well. It follows that two Hermitian forms \(\langle\cdot{,}\cdot\rangle\) and \(\langle\!\langle\cdot{,}\cdot\rangle\!\rangle\) on \(V\) satisfy \(\langle\cdot{,}\cdot\rangle=\langle\!\langle\cdot{,}\cdot\rangle\!\rangle\) if and only if \(\langle v,v\rangle=\langle\!\langle v,v\rangle\!\rangle\) for all \(v \in V.\)
Proof. Suppose \(f\) is normal, then we have for all \(v \in V\) \[\begin{aligned} \Vert f(v)\Vert^2&=\langle f(v),f(v)\rangle=\langle v,f^*(f(v))\rangle=\langle v,(f^*\circ f)(v)\rangle=\langle v,(f\circ f^*)(v)\rangle\\ &=\langle f^*(v),f^*(v)\rangle=\Vert f^*(v)\Vert^2. \end{aligned}\] Taking the square root implies that \(\Vert f(v)\Vert=\Vert f^*(v)\Vert\) for all \(v \in V.\)
Conversely, suppose \(\Vert f(v)\Vert=\Vert f^*(v)\Vert\) for all \(v \in V,\) then the previous calculation implies that \[\langle v,(f^*\circ f)(v)\rangle=\langle v,(f\circ f^*)(v)\rangle\] for all \(v \in V.\) We define Hermitian forms \(\varphi_1\) and \(\varphi_2\) on \(V\) by the rule \[\varphi_1(v,w)=\langle w,(f^*\circ f)(v)\rangle \quad \text{and} \quad \varphi_2(v,w)=\langle w,(f\circ f^*)(v)\rangle\] for all \(v,w \in V.\) We have \(\varphi_1(v,v)=\varphi_2(v,v)\) for all \(v \in V.\) By the previous remark this implies that \(\varphi_1=\varphi_2.\) Hence for all \(v,w \in V,\) we have \[\langle w,(f^*\circ f-f\circ f^*)(v)\rangle=0.\] Taking \(w=(f^*\circ f-f\circ f^*)(v),\) we conclude that \(\Vert (f^*\circ f-f\circ f^*)(v)\Vert=0\) for all \(v \in V.\) It follows that \(f\) is normal.
Similar to the real case, every unitary endomorphism is normal. In addition, we mention the following properties of normal endomorphisms:
Proposition 11.32 • Properties of normal endomorphisms
Let \((V,\langle\cdot{,}\cdot\rangle)\) be a finite dimensional unitary space and \(f : V \to V\) a normal endomorphism. Then
\(\operatorname{Ker}f=\operatorname{Ker}f^*\);
\(\lambda\) is an eigenvalue of \(f\) if and only if \(\overline{\lambda}\) is an eigenvalue of \(f^*\);
the eigenspaces of \(f\) are orthogonal. That is, for eigenvalues \(\lambda\neq \mu\) of \(f\) we have \(\langle u,v\rangle=0\) for all \(u \in \operatorname{Eig}_f({\lambda})\) and for all \(v \in \operatorname{Eig}_f({\mu})\);
if \(f\) is self-adjoint, then the eigenvalues of \(f\) are real;
if \(f\) is unitary, then the eigenvalues of \(f\) are complex numbers of modulus \(1.\)
Proposition 10.12 • Properties of the norm ➔
. Since \(\Vert f(v)\Vert=\Vert f^*(v)\Vert\) for all \(v \in V\) by the previous lemma, we conclude that \(\operatorname{Ker}f=\operatorname{Ker}f^*.\)Let \((V,\langle\cdot{,}\cdot\rangle)\) be a Euclidean space with induced norm \(\Vert \cdot \Vert : V \to \mathbb{R}.\) Then
for all \(v \in V\) we have \(\Vert v \Vert \geqslant 0\) and \(\Vert v \Vert=0\) if and only if \(v=0_V\);
for all \(s\in \mathbb{R}\) and all \(v \in V\) we have \(\Vert sv\Vert=|s|\Vert v\Vert\);
for all vectors \(v_1,v_2 \in V,\) we have the so-called triangle inequality \[\Vert v_1+v_2\Vert\leqslant \Vert v_1\Vert+\Vert v_2\Vert.\]
(ii) Observe that if \(f\) is normal then \(f-s\mathrm{Id}_V\) is normal as well for all \(s\in \mathbb{C}.\) Indeed, using the normality of \(f,\) we compute \[\begin{aligned} (s\mathrm{Id}_V-f)\circ (s\mathrm{Id}_V-f)^*&=(s\mathrm{Id}_V-f)\circ (\overline{s}\mathrm{Id}_V-f^*)=f\circ f^*-\overline{s}f-sf^*+|s|^2\mathrm{Id}_V\\ &=f^*\circ f-sf^*-\overline{s}f+|s|^2 \mathrm{Id}_V=(\overline{s}\mathrm{Id}_V-f^*)\circ (s\mathrm{Id}_V-f)\\& =(s\mathrm{Id}_V-f)^*\circ (s\mathrm{Id}_V-f). \end{aligned}\] Using (i), we conclude that for all \(s\in \mathbb{C}\) we have \[\operatorname{Eig}_f(s)=\operatorname{Ker}(s\mathrm{Id}_V-f)=\operatorname{Ker}(s\mathrm{Id}_V-f)^*=\operatorname{Ker}(\overline{s}\mathrm{Id}_V-f^*)=\mathrm{Eig}_{f^*}(\overline{s}).\]
(iii) Let \(\lambda\) be an eigenvalue of \(f\) with eigenvector \(u\) and \(\mu\) be an eigenvalue of \(f\) with eigenvector \(v.\) Then, using (ii) and the conjugate linearity of \(\langle\cdot{,}\cdot\rangle\) in the first argument, we obtain \[\lambda\langle u,v\rangle=\langle \overline{\lambda}u,v\rangle=\langle f^*(u),v\rangle=\langle u,f(v)\rangle=\langle u,\mu v\rangle=\mu \langle u,v\rangle\] If \(\lambda \neq \mu,\) it follows that \(\langle u,v\rangle=0,\) as claimed.
(iv) if there exists a non-zero vector \(v \in V\) and a scalar \(\lambda \in \mathbb{C}\) such that \(f(v)=\lambda v,\) then we obtain \[\langle v,f(v)\rangle=\langle v,\lambda v\rangle=\lambda \langle v,v\rangle=\langle f(v),v\rangle=\langle \lambda v,v\rangle=\overline{\lambda}\langle v,v\rangle.\] Since \(\langle v,v\rangle\neq 0,\) this implies that \(\lambda=\overline{\lambda}\) and hence \(\lambda\) is real.
(v) Suppose \(\lambda\) is an eigenvalue with non-zero eigenvector \(v\) of the unitary endomorphism \(f,\) then \[|\lambda|^2\langle v,v\rangle=\overline{\lambda}\lambda \langle v,v\rangle=\langle \lambda v, \lambda v\rangle=\langle f(v),f(v)\rangle=\langle v,f^*(f(v))\rangle=\langle v,v\rangle,\] where we use the conjugate linearity of \(\langle\cdot{,}\cdot\rangle\) in the first argument and that \(f^*=f^{-1}\) for a unitary endomorphism. Since \(\langle v,v\rangle \neq 0,\) it follows that \(|\lambda^2|=1.\)
It turns out that an endomorphism of a unitary space is diagonalisable with a orthonormal basis if and only if it is normal. This is a statement which is not true in the real setting. For instance, a rotation around the origin in \(\mathbb{R}^2\) is a normal endomorphism with respect to the standard scalar product of \(\mathbb{R}^2,\) but rotations have in general no eigenvectors.
Theorem 11.33 • Spectral theorem for normal endomorphisms
Let \((V,\langle\cdot{,}\cdot\rangle)\) be a finite dimensional unitary space and \(f : V \to V\) an endomorphism. Then there exists a basis of \(V\) consisting of orthonormal eigenvectors of \(f\) if and only if \(f\) is normal.
Theorem 11.33 • Spectral theorem for normal endomorphisms ➔
.Let \((V,\langle\cdot{,}\cdot\rangle)\) be a finite dimensional unitary space and \(f : V \to V\) an endomorphism. Then there exists a basis of \(V\) consisting of orthonormal eigenvectors of \(f\) if and only if \(f\) is normal.
Lemma 11.34
Let \((V,\langle\cdot{,}\cdot\rangle)\) be a finite dimensional unitary space equipped with an orthonormal ordered basis \(\mathbf{b}\) and \(f : V \to V\) an endomorphism. Then \(f\) is normal if and only if \(\mathbf{A}\mathbf{A}^*=\mathbf{A}^*\mathbf{A},\) where \(\mathbf{A}=\mathbf{M}(f,\mathbf{b},\mathbf{b}).\)
Corollary 3.101 ➔
and that \(\mathbf{M}(f^*,\mathbf{b},\mathbf{b})=\mathbf{M}(f,\mathbf{b},\mathbf{b})^*\) by (11.1). Applying Proposition 2.20Let \(V_1,V_2,V_3\) be finite dimensional \(\mathbb{K}\)-vector spaces and \(\mathbf{b}_i\) an ordered basis of \(V_i\) for \(i=1,2,3.\) Let \(g_1 : V_1 \to V_2\) and \(g_2 : V_2 \to V_3\) be linear maps. Then \[\mathbf{M}(g_2\circ g_1,\mathbf{b}_1,\mathbf{b}_3)=\mathbf{M}(g_2,\mathbf{b}_2,\mathbf{b}_3)\mathbf{M}(g_1,\mathbf{b}_1,\mathbf{b}_2).\]
Proposition 2.20 ➔
, we conclude that \(f \circ f^*=f^*\circ f\) if and only if \(\mathbf{A}\mathbf{A}^*=\mathbf{A}^*\mathbf{A}.\)Let \(\mathbf{A},\mathbf{B}\in M_{m,n}(\mathbb{K}).\) Then \(f_\mathbf{A}=f_\mathbf{B}\) if and only if \(\mathbf{A}=\mathbf{B}.\)
Theorem 11.33 • Spectral theorem for normal endomorphisms ➔
. \(\Rightarrow\) Suppose there exists an ordered orthonormal basis \(\mathbf{b}\) of \((V,\langle\cdot{,}\cdot\rangle)\) consisting of eigenvectors of \(f.\) Hence \(\mathbf{A}=\mathbf{M}(f,\mathbf{b},\mathbf{b})\) is diagonal, that is, \(\mathbf{A}=\sum_{i=1}^n \lambda_i \mathbf{E}_{i,i},\) where \(\lambda_1,\ldots,\lambda_n\) denote the eigenvalues of \(f\) and \(\{\mathbf{E}_{i,j}\}_{1\leqslant i,j\leqslant n}\) the standard basis of \(M_{n,n}(\mathbb{C}).\) We thus have that \(\mathbf{A}^*=\sum_{j=1}^n \overline{\lambda}_j \mathbf{E}_{j,j}\) and \[\mathbf{A}\mathbf{A}^*=\sum_{i=1}^n \lambda_i \mathbf{E}_{i,i}\sum_{j=1}^n \overline{\lambda}_j \mathbf{E}_{j,j}=\sum_{i=1}^n\sum_{j=1}^n\lambda_i\overline{\lambda}_j\mathbf{E}_{i,i}\mathbf{E}_{j,j}=\sum_{i=1}^n|\lambda_i|^2\mathbf{E}_{i,i},\] where we use Lemma 4.4Let \((V,\langle\cdot{,}\cdot\rangle)\) be a finite dimensional unitary space and \(f : V \to V\) an endomorphism. Then there exists a basis of \(V\) consisting of orthonormal eigenvectors of \(f\) if and only if \(f\) is normal.
Lemma 4.4 ➔
. Likewise we compute that \(\mathbf{A}^*\mathbf{A}=\sum_{i=1}^n |\lambda_i|^2\mathbf{E}_{i,i}\) and applying Lemma 11.34Let \(m \in \mathbb{N}.\) For \(1\leqslant k,l,p,q \leqslant m,\) we have \[\mathbf{E}_{k,l}\mathbf{E}_{p,q}=\left\{\begin{array}{cc}\mathbf{E}_{k,q} & p=l \\ \mathbf{0}_{m,m} & p\neq l \end{array}\right.\]
Lemma 11.34 ➔
we conclude that \(f\) is normal.Let \((V,\langle\cdot{,}\cdot\rangle)\) be a finite dimensional unitary space equipped with an orthonormal ordered basis \(\mathbf{b}\) and \(f : V \to V\) an endomorphism. Then \(f\) is normal if and only if \(\mathbf{A}\mathbf{A}^*=\mathbf{A}^*\mathbf{A},\) where \(\mathbf{A}=\mathbf{M}(f,\mathbf{b},\mathbf{b}).\)
\(\Leftarrow\) We use induction. For \(n=1\) every endomorphism is diagonal, hence there is nothing to show and the statement is anchored.
Theorem 6.49 • Existence of eigenvalues ➔
to conclude that \(f : V \to V\) admits an eigenvalue \(\lambda \in \mathbb{C}.\) Let \(W=\operatorname{Eig}_f(\lambda).\) We will argue next that the orthogonal complement \(W^{\perp}\) of \(W\) is stable under \(f.\) Let \(w_1 \in W^{\perp}\) and \(w_2 \in W=\operatorname{Eig}_f(\lambda)=\operatorname{Eig}_{f^*}(\overline{\lambda}).\) Then, we have \[\langle f(w_1),w_2\rangle=\langle w_1,f^*(w_2)\rangle=\langle w_1,\overline{\lambda}w_2\rangle=\overline{\lambda}\langle w_1,w_2\rangle=0,\] where the last equality follows since \(w_1 \in W^{\perp}\) and \(w_2 \in W.\) It follows that \(f(w_1) \in W^{\perp},\) hence \(W^{\perp}\) is stable under \(f.\) Let \(g=f|_{W^{\perp}} : W^{\perp} \to W^{\perp}\) denote the restriction of \(f\) to \(W^{\perp}.\) We want to show that \(g\) is normal with respect to the restriction of \(\langle\cdot{,}\cdot\rangle\) to \(W^{\perp}.\) Using Lemma 11.30Let \(g : V \to V\) be an endomorphism of a complex vector space \(V\) of dimension \(n\geqslant 1.\) Then \(g\) admits at least one eigenvalue. Moreover, the sum of the algebraic multiplicities of the eigenvalues of \(g\) is equal to \(n.\) In particular, if \(\mathbf{A}\in M_{n,n}(\mathbb{C})\) is a matrix, then there is at least one eigenvalue of \(\mathbf{A}.\)
Lemma 11.30 ➔
, we have for all \(w \in W^{\perp}\) \[\Vert g(w)\Vert=\Vert f(w)\Vert=\Vert f^*(w)\Vert=\Vert g^*(w)\Vert\] and hence \(g\) is normal. By the induction hypothesis, there exists an orthonormal basis of \(W^{\perp}\) consisting of eigenvectors of \(g.\) As in the real case, we can complement this basis with an orthonormal basis of \(W=\operatorname{Eig}_f(\lambda)\) to obtain an orthonormal basis of \(V=W\oplus W^{\perp}\) consisting of eigenvectors of \(f.\)Let \((V,\langle\cdot{,}\cdot\rangle)\) be a unitary space and \(f : V \to V\) an endomorphism. Then \(f\) is normal if and only if \[\Vert f(v)\Vert=\Vert f^*(v)\Vert\] for all \(v \in V.\)
Exercises
Exercise 11.35
Let \(n \in \mathbb{N}\) and \(\mathbf{A}\in M_{n,n}(\mathbb{C}).\) Show that \(\mathbf{A}\) is unitary if and only if its column vectors form an orthonormal basis of \(\mathbb{C}^n\) with respect to the standard Hermitian scalar product \(\langle\cdot{,}\cdot\rangle.\)
Solution
We write \(\mathbf{A}= \begin{pmatrix} \vec a_1 & \cdots & \vec a_n\end{pmatrix}.\) So that \[\overline{\mathbf{A}}^T = \begin{pmatrix}\overline{\vec a_1}^T\\ \vdots \\ \overline{\vec a_n}^T\end{pmatrix}.\] Then it holds that \[\overline{\mathbf{A}}^T\mathbf{A}= (\langle \vec a_i,\vec a_j\rangle)_{1\leqslant i,j \leqslant n}.\] If \(\mathbf{A}\) is unitary, we find \((\langle \vec a_i,\vec a_j\rangle)_{1\leqslant i,j \leqslant n}=\mathbf{1}_{n}\) and we conclude that \(\langle \vec a_i,\vec a_j\rangle = \delta_{ij},\) which shows that \(\mathbf{b}= (\vec a_1,\ldots,\vec a_n)\) is an orthonormal basis of \(\mathbb{C}^n\) (with respect to the standard Hermitian scalar product \(\langle\cdot{,}\cdot\rangle\)). Conversely, if \(\mathbf{b}=(\vec a_1,\ldots,\vec a_n)\) is an orthonormal basis with respect to \(\langle\cdot{,}\cdot\rangle,\) we find (using the same computation) that \[\overline{\mathbf{A}}^T\mathbf{A}= (\langle \vec a_i,\vec a_j\rangle)_{1\leqslant i,j \leqslant n}=\mathbf{1}_{n}\] and hence \(\mathbf{A}\) is unitary.
Exercise 11.36 Verify that \(\mathrm{SU}(V,\langle\cdot{,}\cdot\rangle)\) is a subgroup of \(\mathrm{U}(V,\langle\cdot{,}\cdot\rangle)\) in the sense of Definition 8.8
According to Definition 8.8
Definition 8.8 • Subgroup ➔
. In particular, \(\mathrm{SU}(V,\langle\cdot{,}\cdot\rangle)\) is indeed a group and hence so is \(\mathrm{SU}(n).\)A non-empty subset \(H\) of a group \(G\) is called a subgroup if for all \(a,b \in H,\) we have \(a*_G b \in H\) and for all \(a\in H,\) we have \(a^{-1} \in H.\)
Solution
Definition 8.8 • Subgroup ➔
, we need to show that for all \(f,g\in \mathrm{SU}(V,\langle\cdot{,}\cdot\rangle)\) we have \(f\circ g\in\mathrm{SU}(V,\langle\cdot{,}\cdot\rangle)\) and \(f^{-1}\in\mathrm{SU}(V,\langle\cdot{,}\cdot\rangle).\) We use Proposition 6.21A non-empty subset \(H\) of a group \(G\) is called a subgroup if for all \(a,b \in H,\) we have \(a*_G b \in H\) and for all \(a\in H,\) we have \(a^{-1} \in H.\)
Proposition 6.21 ➔
to compute \(\det(f\circ g) = \det(f)\det(g) = 1 \cdot 1 = 1\) and conclude that \(f\circ g\in\mathrm{SU}(V,\langle\cdot{,}\cdot\rangle).\) Since \(1 = \det(\mathrm{Id}_V) = \det(f\circ f^{-1})=\det(f)\det(f^{-1}) = 1 \cdot \det(f^{-1}),\) we conclude that \(\det(f^{-1})=1\) and hence \(f^{-1}\in\mathrm{SU}(V,\langle\cdot{,}\cdot\rangle).\)Let \(V\) be a finite dimensional \(\mathbb{K}\)-vector space. Then, for all endomorphisms \(f,g :V \to V\) we have
\(\operatorname{Tr}(f\circ g)=\operatorname{Tr}(g\circ f)\);
\(\det(f\circ g)=\det(f)\det(g).\)
Exercise 11.37
Show that \[\mathrm{SU}(2)=\left\{\begin{pmatrix}z & -\overline{w} \\ w & \overline{z} \end{pmatrix} | z,w \in \mathbb{C}, |z|^2+|w|^2=1\right\}.\]
Solution
If the matrix \[\mathbf{R}= \begin{pmatrix} z & \omega \\ w & \xi \end{pmatrix}\] is unitary, then \(|z|^2+|w|^2=|\omega|^2+|\xi|^2 = 1\) and \(\bar z \omega + \bar w\xi = 0.\) If it is moreover special unitary, then \(\det(\mathbf{R}) = z\xi-w\omega = 1.\)
Multiplying the second equation by \(z\) and using the third one leads to \[0 =|z|^2\omega + \bar w(1+w\omega) = \omega(|z|^2+|w|^2)+\bar w \Longrightarrow \omega = -\bar w.\] A similar argument with the multiplication by \(z\) replaced by multiplication with \(\bar \omega\) leads to \[0 = \bar z|\omega|^2+\bar w\xi\bar\omega = \bar z|\omega|^2 +(\bar z\bar \xi-1)\xi = \bar z(|\omega|^2+ |\xi|^2)-\xi \Longrightarrow \xi = \bar z\] and the claim follows.