08. Inner Product Spaces

Bilinear Form

Let $V$ be a vector space over field $\mathbf{F}$

Bilinear Form on $V$ is a map

$$F:V\times V\to \mathbf{F}$$

such that for all $u,v,w\in V$ and $\lambda \in \mathbf{F}$ then

$$\begin{array}{rlrl}(i)& F(u+v,w)& & =F(u,w)+F(v,w)\\ (ii)& F(u,v+w)& & =F(u,v)+F(u,w)\\ (iii)& F(\lambda v,w)& & =\lambda F(v,w)=F(v,\lambda w)\end{array}$$

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Symmetric

$F$ is symmetric if

$$F(v,w)=F(w,v)\text{ for all }v,w\in V$$

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Non-Degenerate

$F$ is non-degenerate if

$$F(v,w)=0\text{ for all }v\in V⟹w=0$$

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Inner Product Space

Let $V$ be a real vector space then

$V$ with a bilinear, symmetric positive definite form $F(\cdot ,\cdot )$, more commonly written as

$$⟨\cdot ,\cdot ⟩$$

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Sesquilinear Form

Let $V$ be a vector space over $\mathbb{C}$

Sesquilinear Form on $V$ is a map

$$F:V\times V\to \mathbb{C}$$

such that for all $u,v,w\in V$ and $\lambda \in \mathbb{C}$ then

$$\begin{array}{rlrl}(i)& F(u+v,w)& & =F(u,w)+F(v,w)\\ (ii)& F(u,v+w)& & =F(u,v)+F(u,w)\\ (iii)& F(\overline{\lambda }v,w)& & =\lambda F(v,w)=F(v,\lambda w)\end{array}$$

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Conjugate Symmetric

$F$ is symmetric if

$$F(v,w)=F(w,v)\text{ for all }v,w\in V$$

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Complex Inner Product Space

Let $V$ be a complex vector space then

$V$ with a sesquilinear, conjugate symmetric, positive definite form $F(\cdot ,\cdot )$, more commonly written as

$$⟨\cdot ,\cdot ⟩$$

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Mutually Orthogonal Set

Let $V$ be a complex or real inner product space

$\{w_1,\cdots ,w_n\}$ is mutually orthogonal if

$$⟨w_i,w_j⟩=0\text{ for all }i\ne j$$

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Orthonormal Set

Let $V$ be a complex or real inner product space

$\{w_1,\cdots ,w_n\}$ is orthonormal if they are mutually orthogonal and

$$⟨w_i,w_i⟩=1\text{ for all }i$$

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Linearly Independence of an Orthogonal Set

Let $V$ be an inner product over $\mathbb{K}$ (equal $R$ or $C$)
Let $\{w_1,\cdots ,w_n\}\subset V$ be orthogonal with $w_i\ne 0$ for all $i$ then

$$w_1,\cdots ,w_n\text{ are linearly independent }$$

Proof

Assume ${\sum }_i{\lambda }_i{w}_i=0$ for some ${\lambda }_i\in \mathbb{K}$ then
For all $j$ then

$$⟨w_j,\sum _i{\lambda }_i{w}_i⟩=0$$

However

$$⟨w_j,\sum _i{\lambda }_i{w}_i⟩=\sum _i{\lambda }_i⟨w_j,w_i⟩={\lambda }_j⟨w_j,w_j⟩$$

Hence ${\lambda }_j=0$ for all $j$

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8.1 Gram-Schmidt Orthonormalisation Process

Gram-Schmidt Orthonormalisation Process

Let $\mathcal{B}=\{v_1,\cdots ,v_n\}$ be a basis of the inner product space $V$ over $\mathbb{K}=\mathbb{R},\mathbb{C}$
Define

$$\begin{array}{rl}{w}_1& =v_1\\ w_2& =v_2-\frac{⟨w_1,v_2⟩}{⟨w_1,w_1⟩}{w}_1\\ ⋮& \\ w_k& =v_k-\sum _{i=1}^{k-1}\frac{⟨w_i,v_k⟩}{⟨w_i,w_i⟩}{w}_i\\ ⋮& \end{array}$$

Assuming that $⟨w_1,cdost,w_{k-1}⟩=⟨v_1,\cdots ,v_{k-1}⟩$, the general form from above shows that

$$⟨w_1,\cdots ,w_k⟩=⟨w_1,\cdots w_{k-1},v_k⟩=\cdots =⟨v_1,\cdots ,v_k⟩$$

Assuming that $\{w_1,\cdots ,w_{k-1}\}$ are orthogonal then for $j<k$

$$⟨w_j,w_k⟩=⟨w_j,v_k⟩-\frac{⟨w_j,v_k⟩}{⟨w_j,w_j⟩}⟨w_j,w_j⟩=0$$

So by induction $D=\{w_1,\cdots ,w_n\}$ is orthogonal, spanning
Hence by Linear Independence of an Orthogonal Set then

$$\mathbf{D}\text{ is an orthogonal basis}$$

Define

$$u_i=\frac{w_i}{||w_i||}\text{ where }||w_i||=\sqrt{⟨w_i,w_i⟩}$$

Then

$$\mathcal{E}=\{u_i,\cdots ,u_n\}\text{ is a orthonormal basis }$$

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Existence of Orthonormal Bases corollary

Every finite dimensional inner product space $V$ over $\mathbb{K}=\mathbb{R},\mathbb{C}$ has an orthonormal basis

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8.2 Orthogonal Complements and Duals of Inner Product Space

25 - Inner Product-Dual Isomorphism

Inner Product-Dual Isomorphism

Let $V$ be an inner product space over $\mathbb{K}=\mathbb{R},\mathbb{C}$ then
For all $v\in V$

$$\begin{array}{rl}⟨v,\cdot ⟩:V& \to \mathbb{K}\\ \text{ in other words }w& \mapsto ⟨v,w⟩\end{array}$$

is a linear functional as $⟨,⟩$ is linear in the second co-ordinate

Map $v\mapsto ⟨v,\cdot ⟩$ is a natural injective $\mathbb{R}-$linear map

$$\varphi :V\to V^{\prime }$$

which is an isomorphism when $V$ is finite dimensional

Note that every complex vector space $V$ is a real vector space and if it is finite dimensional then

$$2{\dim }_{\mathbb{C}}(V)={\dim }_{\mathbb{R}}(V)$$

Proof

Note $\varphi :v\mapsto ⟨v,\cdot ⟩$ so need to show $\varphi (v+\lambda w)+\varphi (v)+\lambda \varphi (w)$ for all $v,w\in V$, $\lambda \in \mathbb{R}$
So

$$⟨v+\lambda w,\cdot ⟩=⟨v,\cdot ⟩+\lambda ⟨w,\cdot ⟩$$

which is true by linearity of an inner product space
Hence $\varphi$ is $\mathbb{R}-$linear (conjugate linear for $\lambda \in \mathbb{C}$ )

As $⟨\cdot ,\cdot ⟩$ is non-degenerate then

$$⟨v,\cdot ⟩=\overline{⟨\cdot ,v⟩}\text{ is not the zero functional unless }v=0$$

Hence $\varphi$ is injective

If $V$ is finite-dimensional then ${\dim }_{\mathbb{R}}V={\dim }_{\mathbb{R}}{V}^{\prime }$ hence $\mathrm{Im}\varphi =V^{\prime }$
So $\varphi$ is surjective and hence

$$\varphi \text{ is a }\mathbb{R}\text{-linear isomorphism}$$

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Orthogonal Complement

Let $U\subseteq V$ be a subspace of an inner product space $V$
Orthogonal Complement $U^{\perp }$ of $U$ is defined as

$$U^{\perp }:=\{v\in V\mid ⟨u,v⟩=0\text{ for all }u\in U\}$$

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Orthogonal Complement Subspace Property

Let $U\subseteq V$ be a subspace of an inner product space $V$ then

$$U^{\perp }\text{ is a subspace of }V$$

Proof

We have $0\in U^{\perp }$ by definition of $0$
For $v,w\in U^{\perp }$ and $\lambda \in \mathbb{K}$ then for all $u\in U$

$$⟨u,v+\lambda w⟩=⟨u,v⟩+\lambda ⟨u,w⟩=0+\lambda \cdot 0=0$$

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Properties of Orthogonal Complement

Let $U\subseteq V$ be a subspace of an inner product space $V$

$$U\cap U^{\perp }=\{0\}$$

$$U\oplus U^{\perp }=V\text{ if }V\text{ is finite dimensional}$$

so

$$\dim U^{\perp }=\dim V-\dim U$$

$$(U+W{)}^{\perp }=U^{\perp }\cap W^{\perp }$$

$$(U\cap W{)}^{\perp }\supseteq U^{\perp }+W^{\perp }$$

with equality if $\dim V<\infty$

$$U\subseteq (U^{\perp }{)}^{\perp }$$

with equality if $V$ is finite dimensional

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Isomorphism between Orthogonal Complement and Annihilators

Let $V$ be finite dimensional then

Under $R$-linear isomorphism $\varphi :V\to V^{\prime }$ given by $v\mapsto ⟨v,\cdot ⟩$
Space $U^{\perp }$ maps is0oomorphically to $U^0$ (considered as $\mathbb{R}$ vector spaces)

Proof

Let $v\in U^{\perp }$ then for all $u\in U$

$$⟨u,v⟩=\overline{⟨v,u⟩}=0$$

Hence $⟨v,\cdot ⟩\in U^0$ so $\mathrm{Im}(\varphi {\mid }_{U^{\perp }})\subseteq U^0$ and

$$\dim U^{\perp }=\dim V-\dim U=\dim U^0$$

So

$$\varphi (U^{\perp })=U^0$$

As the kernel of $\varphi$ is trivial then we only need to check equality of dimensions

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8.3 Adjoints of Maps

Adjoint of map

Let $V$ be a inner product space over $\mathbb{K}=\mathbb{R},\mathbb{C}$

Given a linear map $T:V\to V$ then
Linear map $T^{\ast }:V\to V$ is the adjoint of $T$ if

$$⟨v,T(w)⟩=⟨T^{\ast }(v),w⟩\text{ for all }v,w\in V$$

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Adjoint Map is Unique lemma

Given a linear map $T:V\to V$ then

$$\text{if }{T}^{\ast }\text{ exists it is unique}$$

Proof

Let $\tilde{T}$ be another map satisfying the adjoint definition then
For all $v,w\in V$

$$\begin{array}{rl}⟨T^{\ast }(v)-\tilde{T}(v),w⟩& =⟨T^{\ast }(v),w⟩-⟨\tilde{T}(v),w⟩\\ & =⟨v,T(w)⟩-⟨v,T(w)⟩\\ & =0\end{array}$$

As $⟨⟩$ is non-degenerate then for all $v\in V$

$$T^{\ast }(v)-\tilde{T}(v)=0$$

Hence $T^{\ast }=\tilde{T}$

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26 - Existence and Linearity of an Adjoint Map

Existence of an Adjoint Map

Let $T:V\to V$ be linear where $V$ is finite dimensional then

$$\text{adjoint }{T}^{\ast }\text{ exists and is linear }$$

Proof

Let $v\in V$ and consider map $V\to \mathbb{K}$ given by

$$w\mapsto ⟨v,T(w)⟩$$

then

$$⟨v,T(\cdot )⟩$$

is a linear functional as $T$ is linear so $⟨⟩$ is linear in second coordinate

As $V$ is finite dimensional then $\varphi :V\to V^{\prime }$ given by $\varphi (u)=⟨u,\cdot ⟩$
is a $R$-linear isomorphism and in particular a surjective map

So there exists $u\in V$ such that

$$⟨v,T(\cdot )⟩=⟨u,\cdot ⟩$$

By defining $T^{\ast }(v):=u$ then

$$⟨v,T(\cdot )⟩=⟨T^{\ast }(v),\cdot ⟩$$

In other words then

$$⟨v,T(w)⟩=⟨T^{\ast }(v),w⟩\text{ for all }w\in V$$

To see that $T^{\ast }$ is linear, note that for all $v_1,v_2,w\in V$ and $\lambda \in \mathbb{K}$ then

$$\begin{array}{rl}⟨T^{\ast }(v_1+\lambda v_2),w⟩& =⟨v_1+\lambda v_2,T(w)⟩\\ & =⟨v_1,T(w)⟩+\overline{\lambda }⟨v_2,T(w)⟩\\ & =⟨T^{\ast }(v_1),w⟩+\overline{\lambda }⟨T^{\ast }(v_2),w⟩\\ & =⟨T^{\ast }(v_1)+\lambda T^{\ast }(v_2),w⟩\end{array}$$

As $⟨⟩$ is non-degenerate then

$$T^{\ast }(v_1+v_2)=T^{\ast }(v_1)+\lambda T^{\ast }(v_2)$$

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Matrix Representation of the Adjoint

Let $T:V\to V$ be linear and let $\mathcal{B}=\{e_1,\cdots ,e_n\}$ be an orthonormal basis for $V$ then

$${}_{\mathcal{B}}[T^{\ast }{]}_{\mathcal{B}}={\overline{{}_{\mathcal{B}}[T{]}_{\mathcal{B}}}}^T$$

Proof

Let $A={}_{\mathcal{B}}[T{]}_{\mathcal{B}}$ then

$$a_{ij}=⟨e_i,T(e_j)⟩$$

So let $B={}_{\mathcal{B}}[T^{\ast }{]}_{\mathcal{B}}$ then

$$b_{ij}=⟨e_i,T^{\ast }(e_j)⟩=\overline{⟨T^{\ast }(e_j),e_i⟩}=\overline{⟨e_j,T^{\ast }(e_i)⟩}=\overline{a_{ji}}$$

Hence $B={\overline{A}}^T$

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Properties of Adjoint Maps

Let $S,T:V\to V$ be linear
Let $V$ be finite dimensional

Let $\lambda \in \mathbb{K}$ then

$$(S+T{)}^{\ast }=S^{\ast }+T^{\ast }$$

$$(\lambda T{)}^{\ast }=\overline{\lambda }{T}^{\ast }$$

$$(ST{)}^{\ast }=T^{\ast }{S}^{\ast }$$

$$(T^{\ast }{)}^{\ast }=T$$

5. If $m_T$ is the minimal polynomial of $T$ then

$$m_{T^{\ast }}=\overline{m_T}$$

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Self-Adjoint Map

Let $T:V\to V$ be a linear map then

$$T\text{ is self-adjoint}\text{ if }T=T^{\ast }$$

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Eigenvalues of Self-Adjoint Linear Operators lemma

If $\lambda$ is an eigenvalue of a self-adjoint linear operator then

$$\lambda \in \mathbb{R}$$

Proof

Assume $w\ne 0$ and $T(w)=\lambda w$ for some $\lambda \in \mathbb{C}$ then

$$\begin{array}{rl}\lambda ⟨w,w⟩& =⟨w,\lambda w⟩=⟨w,T(w)⟩=⟨T^{\ast }(w),w⟩\\ & =⟨T(w),w⟩=⟨\lambda w,w⟩=\overline{\lambda }⟨w,w⟩\end{array}$$

As $⟨w,w⟩\ne 0$ then $\lambda =\overline{\lambda }$
Hence

$$\lambda \in \mathbb{R}$$

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Orthogonal Complements of Invariant Subspaces lemma

Let $T$ be a self-adjoint map
Let $U\subseteq V$ and $U$ is $T$-invariant then

$$U^{\perp }\text{ is }T-\text{invariant}$$

Proof

Let $w\in U^{\perp }$ then for all $u\in U$

$$⟨u,T(w)⟩=⟨T^{\ast }(u),w⟩=⟨T(u),w⟩=0$$

as $T(u)\in U$ and $w\in U^{\perp }$ hence $T(w)\in U^{\perp }$

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27 - Orthonormal Basis for Self-Adjoint Maps

Orthonormal Basis for Self-Adjoint Maps

Let $T:V\to V$ be a linear self-adjoint map and $V$ be a finite dimensional vector space then

There exists orthonormal basis of eigenvectors for $T$

Proof

By Eigenvalues of Adjoint Maps being real then there exists $\lambda \in \mathbb{R}$ and $v\ne 0$ such that

$$T(v)+=\lambda v$$

Consider $U=⟨v⟩$ then $U$ is $T$-invariant and by Orthogonal Complements of Invariant Subspaces then

$$T{\mid }_{{⟨V⟩}^{\perp }}:⟨v{⟩}^{\perp }\to ⟨v{⟩}^{\perp }$$

is well-defined and self-adjoint

So by induction on $n:=\dim V$ then there exists orthonormal basis set

$$\{e_2,\cdots ,e_n\}\text{ of eigenvectors for }T{\mid }_{{⟨v⟩}^{\perp }}$$

Define $e_1=\frac{v}{||v||}$ then

$$\{e_1,\cdot ,e_n\}\text{ is a orthonormal basis of eigenvectors for }T$$

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8.4 Orthogonal and Unitary Transformations

Orthogonal and Unitary

Let $V$ be a finite dimensional inner product space
Let $T:V\to V$ be a linear transformation then

If $T^{\ast }=T^{-1}$ then $T$ is called

$$\begin{array}{r}\text{orthogonal}\text{ when }\mathbb{K}=\mathbb{R}\\ \text{unitary}\text{ when }\mathbb{K}=\mathbb{C}\end{array}$$

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Matrix Equivalence of Orthogonal and Unitary

Let $\mathcal{B}$ be an orthonormal basis for $V$
Let $T$ be an orthogonal / unitary transformation of $V$ then

$${}_{\mathcal{B}}[T{]}_{\mathcal{B}}\text{ is a orthogonal / unitary matrix}$$

using matrix representation of adjoint

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28 - Equivalent Statements of Adjoint Map

Equivalent Statements of Adjoint Map

Let $T$ be a linear map on $V$ then

We have the following equivalent statements

1. $T$ adjoint is the inverse

$$T^{\ast }=T^{-1}$$

2. $T$ preserves inner products so

$$⟨v,w⟩=⟨Tv,Tw⟩\text{ for all }v,w\in V$$

3. $T$ preserves lengths so

$$||v||=||Tv||\text{ for all }v\in V$$

Proof

$$\begin{array}{rlrlrl}(1)⟹(2)& ⟨v,w⟩& & =⟨\mathrm{Id}v,w⟩& & \\ & & & =⟨T^{\ast }Tv,w⟩& & \\ & & & =⟨Tv,Tw⟩,& & \text{ for all }v,w\in V\\ & & & & & \\ (2)⟹(3)& ||v|{|}^2& & =⟨v,v⟩& & \\ & & & =⟨Tv,Tv⟩& & \\ & & & =||Tv|{|}^2,& & \text{ for all }v\in V\\ & & & & & \\ (2)⟹(1)& ⟨v,w⟩& & =⟨Tv,Tw⟩& & \\ & & & =⟨T^{\ast }Tv,w⟩,& & \text{ for all }v,w\in V\\ & & & ⟹T^{\ast }Tv=v& & \text{ by non-degeneracy of }⟨,⟩,\text{ for all }v\in V\\ & & & ⟹T^{\ast }T=\mathrm{Id}& & \end{array}$$

For $(3)⟹(2)$ refer to length determines inner product

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Length Determines Inner Product

The length function determines the inner product

Given two inner products $⟨,{⟩}_1$ and $⟨,{⟩}_2$ then

$$⟨v,v{⟩}_1=⟨v,v{⟩}_2\forall v\in V⟺⟨v,w{⟩}_1=⟨v,w{⟩}_2\forall v,w\in V$$

Proof

Implication for $⟸$ is trivial
For $⟹$ note that

$$⟨v+w,v+w⟩=⟨v,v⟩+⟨v,w⟩+\overline{⟨v,w⟩}+⟨w,w⟩$$

For $\mathbb{K}=\mathbb{C}$ then

$$⟨v+iw,v+iw⟩=⟨v,v⟩+i⟨v,w⟩-i\overline{⟨v,w⟩}+⟨w,w⟩$$

Hence

$$\begin{array}{rlrl}\mathrm{Re}⟨v,w⟩& =& & \frac{1}{2}(||v+w|{|}^2-||v|{|}^2-||w|{|}^2)\\ \mathrm{Im}⟨v,w⟩& =-& & \frac{1}{2}(||v+iw|{|}^2-||v|{|}^2-||w|{|}^2)\end{array}$$

Hence the inner product is given in terms of the length function

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Eigenvalues of Orthogonal / Unitary Linear Transformation lemma

Let $\lambda$ be an eigenvalue of an orthogonal / unitary linear transformation $T:V\to V$ then

$$|\lambda |=1$$

Proof

Let $v\ne 0$ be a $\lambda$-eigenvector then

$$⟨v,v⟩=⟨Tv,Tv⟩$$

by Equivalent Statements of Adjoint Map then

$$⟨Tv,Tv⟩=⟨\lambda v,\lambda v⟩=\overline{\lambda }\lambda ⟨v,v⟩$$

Hence $1=\overline{\lambda }\lambda$ so $|\lambda |=1$

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Determinant of Orthogonal / Unitary Matrix corollary

Let $A$ be an orthogonal / unitary $n\times n$ matrix then

$$|\det A|=1$$

Proof

As we are working over $\mathbb{C}$ and $\det A$ is the product of all eigenvalues (with repetitions)
Hence

$$\det A=|{\lambda }_1{\lambda }_2\cdots {\lambda }_n|=|{\lambda }_1||{\lambda }_2|\cdots |{\lambda }_n|=1$$

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Orthogonal Complements of Invariant Subspaces for Unitary Operators lemma

Assume that $V$ is finite dimensional and $T:V\to V$ with $T^{\ast }T=\mathrm{Id}$ then

$$U\text{ is }T-\text{invariant}⟹U^{\perp }\text{ is }T-\text{invariant}$$

Proof

For $w\in U^{\perp }$ then for all $u\in U$ then

$$⟨u,Tw⟩=⟨T^{\ast }u,w⟩=⟨T^{-1}u,w⟩$$

As $U$ is invariant under $T$ then it must be invariant under $T^{-1}=T^{\ast }$
This can be seen by writing

$$m_T(x)=x^m+a_{m-1}{x}^{m-1}+\cdots +a_{1}x+a_0$$

into

$$T(T^{m-1}+a_{m-2}{T}^{m-2}+\cdots +a_1)=-a_{0}I$$

where $T^{-1}$ can be written as a polynomial in terms of $T$ so

So $T^{\ast }u\in U$ and $⟨T^{\ast }u,w⟩=0$ for all $u\in U$ hence
Hence

$$⟨u,Tw⟩=0\text{ for all }u\in U⟹Tw\in U^{\perp }$$

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29 - Orthonormal Basis for Unitary Maps

Orthonormal Basis for Unitary Maps

Assume $V$ is finite-dimensional and $T:V\to V$ is unitary then

$$\text{ then there exists an orthonormal basis of eigenvectors}$$

Proof

As $\mathbb{K}=\mathbb{C}$ is algebraically closed then
There exists a $\lambda$ and $v\ne 0\in V$ such that

$$Tv=\lambda v$$

Then $U=⟨v⟩$ is $T-$invariant so the complement $U^{\perp }$ is also $T$-invariant
Therefore restriction $T{\mid }_{U^{\perp }}$ is a map of $U^{\perp }$ to itself which satisfies the hypothesis

By induction on dimension $n:=\dim (V)$ and noting that $\dim U^{\perp }=n-1$
Then there exist orthonormal basis $\{e_2,\cdots ,e_n\}$ of $U^{\perp }$
Putting

$$e_1=\frac{v}{||v||}$$

then

$$\{e_1,e_2,\cdots ,e_n\}\text{ is an orthonormal basis of eigenvectors for }V$$

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Unitary Diagonalisation corollary

Let $A\in U(n)$ then

$$\exists P\in U(n)\text{ such that }{P}^{-1}AP\text{ is diagonal}$$

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30 - Canonical Form of Orthogonal Operators

Canonical Form of Orthogonal Operators

Let $T:V\to V$ be orthogonal and $V$ be a finite dimensional real vector space then

There exists orthogonal basis $\mathcal{B}$ such that

$${}_{\mathcal{B}}[T{]}_{\mathcal{B}}=\left[\begin{array}{ccccc}I& & & & \\ & -I& & & \\ & & R_{{\theta }_1}& & \\ & & & \ddots & \\ & & & & R_{{\theta }_l}\end{array}\right]\text{ where }{\theta }_i\ne 0,\pi$$

where

$$R_{\theta }=\left(\begin{array}{cc}\cos (\theta )& -\sin (\theta )\\ \sin (\theta )& \cos (\theta )\end{array}\right)$$

Proof

Let $S=T+T{1}^{-1}=T+T^{\ast }$ so $S^{\ast }=(T+T^{\ast }{)}^{\ast }=T^{\ast }+T=S$
Hence $S$ is self-adjoint so it has a Orthonormal Basis for Self-Adjoint Maps thus

$$V=V_1\oplus \cdots \oplus V_k$$

decomposes into orthogonal eigenspaces of $S$ with distinct eigen values ${\lambda }_1,\cdots ,{\lambda }_k$
As each $V_i$ is $T-$invariant for $v\in V_i$ so

$$S(T(V))=T(S(v))={\lambda }_{i}T(v)\text{ and so }T(v)\in V_i$$

So restrict to $T{\mid }_{V_i}$

By definition of $V_i$ for all $v\in V_i$ then

$$(T+T^{-1})v={\lambda }_{i}v$$

Hence $T^2-{\lambda }_{i}T+I=0$ hence the minimal polynomial of $T{\mid }_{V_i}$ divides $x^2-{\lambda }_{i}x+1$
So any eigenvalue of $T{\mid }_{V_i}$ is a root of it

If ${\lambda }_i=\pm 2$ then $(T+I{)}^2=0$ or $(T-I{)}^2=0$
Thus the eigenvalue of $T{\mid }_{V_i}$ is $-1$ or $+1$ respectively
Since we know $T{\mid }_{V_i}$ may be diagonalised over $\mathbb{C}$ then

$$T{\mid }_{V_i}=+I\text{ or }-I$$

If ${\lambda }_i\ne \pm 2$ then $T{\mid }_{V_i}$ does not have any real eigenvalues as they would to be $\pm 1$ by eigenvalues of unitary transformations
Hence this forces ${\lambda }_i=\pm 2$

So$\{v,T(v)\}$ are linearly independent over the reals for $v\ne 0\in V_i$

Consider plane $W=⟨v,T(v)⟩$ then $W$ is $T-$invariant as

$$v\mapsto T(v),T(v)\mapsto T^2(v)={\lambda }_{i}T(v)-v$$

Hence $W^{\perp }$ is $T$-invariant by Orthogonal Complement of Invariant Subspaces for Unitary Operators
So we see that $V_i$ splits into $2$-dimensional $T-$invariant subspaces hence

$$T{\mid }_W=R_{\theta }$$

for some orthonormal basis of $W$ and for some $\theta \ne 0,\pi$

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