Chapter 5 extra note 6

Selected lecture notes

isometry transformation
polarization identity
unitary transformation
rigid motion

Adjoint linear transformation

Example

Consider inner product spaces

$V$ and

$W$ defined as

$$\begin{array}{rl}& V={\mathbb{P}}_1,⟨f,g⟩={\int }_{-1}^{1}f(x)g(x)dx,\\ & W={\mathbb{R}}^2,⟨(x,y),(u,v)⟩=2xu+3yv.\end{array}$$
Let

$T:V\to W$ be a linear transformation defined by

$$\begin{array}{}\text{(1)}& T(p)=({\int }_0^{1}p(x)dx,{\int }_{-1}^{1}p(x)dx).\end{array}$$

In the following we aim to determine the adjoint linear transformation

$T^{\ast }$ of

$T$.

Based on the definition of adjoint transformation, we should have, for all

$p(x)\in {\mathbb{P}}_1$ and

$(u,v)\in {\mathbb{R}}^2$,

$$\begin{array}{}\text{(2)}& ⟨T(p),(u,v)⟩=⟨p,T^{\ast }((u,v))⟩.\end{array}$$

The goal is to find

$T^{\ast }:W\to V$.
Since

$W$ have a natural basis

$e_1=(1,0)$ and

$e_2=(0,1)$, if we can determine

$T^{\ast }(e_1)$ and

$T^{\ast }(e_2)$, we should have

$$T^{\ast }((u,v))=u{T}^{\ast }(e_1)+v{T}^{\ast }(e_2).$$

$T^{\ast }(e_1)$ should satisfy

$$\begin{array}{}\text{(3)}& ⟨T(p),e_1⟩=⟨p,T^{\ast }(e_1)⟩={\int }_{-1}^{1}p(x)T^{\ast }(e_1)dx.\end{array}$$

We want to rewrite the left hand side to be also in the form of

${\int }_{-1}^{1}p(x)q(x)dx$, then we have

$⟨p(x),q(x)⟩=⟨p(x),T^{\ast }(e_1)⟩$ for all

$p$ and we can deduce that

$T^{\ast }(e_1)=q(x)$.

The left hand side of (3) is an inner product in

$W$ that gives

$$\begin{array}{rl}⟨T(p),(1,0)⟩& =2{\int }_0^{1}p(x)dx\\ & ={\int }_{-1}^{1}p(x)(1+\frac{3}{2}x)dx\\ \text{(4)}& & =⟨p,1+\frac{3}{2}x⟩.\end{array}$$

Exercise:
Find

$q(x)\in {\mathbb{P}}_1$ such that

$${\int }_0^{1}2p(x)dx={\int }_{-1}^{1}p(x)q(x)dx,\mathrm{\forall }p\in {\mathbb{P}}_1.$$

It turns out that

$q(x)=1+\frac{3}{2}x$.

Similarly,

$T^{\ast }(e_2)$ should satisfy

$$⟨T(p),e_2⟩=⟨p,T^{\ast }(e_2)⟩,$$
and we easily obtain

$T^{\ast }(e_2)=3$.

Therefore,

$$T^{\ast }(e_1)=1+\frac{3}{2}x,T^{\ast }(e_2)=3,$$
and

$$\begin{array}{}\text{(5)}& T^{\ast }((u,v))=u+3v+\frac{3u}{2}x.\end{array}$$

---

Isometry

Definition:
Let

$V,W$ be normed spaces, a linear transformation

$U:V\to W$ is called an isometry if

$$\begin{array}{}\text{(6)}& \Vert U(\mathbf{v})\Vert =\Vert \mathbf{v}\Vert ,\mathrm{\forall }\mathbf{v}\in V.\end{array}$$

---

Remark:

1. Let
   $\mathbf{u},\mathbf{v}\in V$, where
   $V$ is a real inner product space, we have
   
   $$\begin{array}{}\text{(7)}& \Vert \mathbf{u}+\mathbf{v}{\Vert }^2=\Vert \mathbf{u}{\Vert }^2+\Vert \mathbf{v}{\Vert }^2+2⟨\mathbf{u},\mathbf{v}⟩.\end{array}$$
2. Let
   $\mathbf{u},\mathbf{v}\in V$, where
   $V$ is an inner product space, we have
   
   $$\begin{array}{}\text{(8)}& \Vert \mathbf{u}+\mathbf{v}{\Vert }^2=\Vert \mathbf{u}{\Vert }^2+\Vert \mathbf{v}{\Vert }^2+2\text{Real}(⟨\mathbf{u},\mathbf{v}⟩).\end{array}$$

Lemma: (polarization identity)

1. Let
   $\mathbf{u},\mathbf{v}\in V$, where
   $V$ is a real inner product space, we have
   
   $$\begin{array}{}\text{(9)}& ⟨\mathbf{u},\mathbf{v}⟩=\frac{1}{4}(\Vert \mathbf{u}+\mathbf{v}{\Vert }^2-\Vert \mathbf{u}-\mathbf{v}{\Vert }^2),\mathrm{\forall }\mathbf{u}\mathbf{,}\mathbf{v}\in V.\end{array}$$
2. Let
   $\mathbf{u},\mathbf{v}\in V$, where
   $V$ is an inner product space, we have, for all
   $\mathbf{u}\mathbf{,}\mathbf{v}\in V$,
   
   $$\begin{array}{}\text{(10)}& ⟨\mathbf{u},\mathbf{v}⟩=\frac{1}{4}(\Vert \mathbf{u}+\mathbf{v}{\Vert }^2-\Vert \mathbf{u}-\mathbf{v}{\Vert }^2+i\Vert \mathbf{u}+i\mathbf{v}{\Vert }^2-i\Vert \mathbf{u}-i\mathbf{v}{\Vert }^2).\end{array}$$

---

Theorem:
Let

$V,W$ be inner product spaces. The operator

$U:V\to W$ is an isometry if and only if

$$\begin{array}{}\text{(11)}& ⟨U(\mathbf{u}),U(\mathbf{v})⟩=⟨\mathbf{u},\mathbf{v}⟩,\mathrm{\forall }\mathbf{u}\mathbf{,}\mathbf{v}\in V.\end{array}$$

• Proof:

  (

  $\Leftarrow$)
  Choose

  $\mathbf{u}=\mathbf{v}$, then
  

  $$\Vert U(\mathbf{v}){\Vert }^2=⟨U(\mathbf{v}),U(\mathbf{v})⟩=⟨\mathbf{v},\mathbf{v}⟩=\Vert \mathbf{v}{\Vert }^2.$$

  (

  $\Rightarrow$)

  Here we assume real inner product space. The derivation of the general version is similar.

  $$\begin{array}{rl}⟨\mathbf{u},\mathbf{v}⟩& \text{xlongequal}[\text{polarization identity}]\frac{1}{4}(\Vert \mathbf{u}+\mathbf{v}{\Vert }^2-\Vert \mathbf{u}-\mathbf{v}{\Vert }^2)\\ & \text{xlongequal}[\text{isometry}]\frac{1}{4}(\Vert U(\mathbf{u}+\mathbf{v}){\Vert }^2-\Vert U(\mathbf{u}-\mathbf{v}){\Vert }^2)\\ & \text{xlongequal}[\text{linearity}]\frac{1}{4}(\Vert U(\mathbf{u})+U(\mathbf{v}){\Vert }^2-\Vert U(\mathbf{u})-U(\mathbf{v}){\Vert }^2)\\ & \text{xlongequal}[\text{polarization identity}]⟨U(\mathbf{u}),U(\mathbf{v})⟩.\end{array}$$

Lemma:
Let

$V,W$ be inner product spaces. The operator

$U:V\to W$ is an isometry if and only if

$U^{\ast }U=I$, where

$I$ is the identity transformation.

• Proof:

  (

  $\Leftarrow$)
  

  $$\Vert U\mathbf{v}{\Vert }^2=⟨U\mathbf{v},U\mathbf{v}⟩=⟨\mathbf{v},U^{\ast }U\mathbf{v}⟩=⟨\mathbf{v},\mathbf{v}⟩=\Vert \mathbf{v}{\Vert }^2.$$

  (

  $\Rightarrow$)
  Suppose

  $U$ is an isometry, we want to show that
  

  $$U^{\ast }U\mathbf{v}=\mathbf{v},\mathrm{\forall }\mathbf{v}\in V.$$

  Fix

  $\mathbf{v}\in V$, given

  $\mathbf{u}\in V$, since

  $U$ is an isometry, we have (11) and
  

  $$⟨\mathbf{u},\mathbf{v}⟩=⟨U\mathbf{u},U\mathbf{v}⟩=⟨\mathbf{u},U^{\ast }U\mathbf{v}⟩.$$
  Therefore,

  $⟨\mathbf{u},\mathbf{v}⟩=⟨\mathbf{u},U^{\ast }U\mathbf{v}⟩$ for all

  $\mathbf{u}\in V$, that gives

  $\mathbf{v}=U^{\ast }U\mathbf{v}$.

Corollary:

1. An isometry is left invertible.
2. An isometry is a one-to-one map.
3. The kernel of an isometry contains only the zero vector.

Definition:
An invertible isometry is called unitary.

Remark:
An unitary transformation in a real inner product space is often called an orthogonal transformation.

Proposition:
An isometry

$U:V\to W$ is unitary if and only if

$\text{dim}(V)=\text{dim}(W)$.

Let

$U:V\to W$ be an isometry.

$U$ is unitary if and only if

$\text{dim}(V)=\text{dim}(W)$.

• Proof:

  (

  $\Rightarrow$)
  

  $U$ is unitary, so

  $U$ is invertible, one-to-one and onto. That means

  $\text{Ker}(U)=\{\mathbf{0}\}$ and

  $\text{Ran}(U)=W$.

  Based on the Rank-Nullity Theorem we can deduce that that
  

  $$\text{dim}(V)=\text{dim}(\text{Ker}(U))+\text{dim}(\text{Ran}(U))=\text{dim}(\text{Ran}(U))=\text{dim}(W).$$

  (

  $\Leftarrow$)
  Let

  $U:V\to W$ be an isometry and assume

  $\text{dim}(V)=\text{dim}(W)$.

  Since

  $U$ is an isometry,

  $\text{Ker}(U)=\{\mathbf{0}\}$ and we can deduce that
  

  $$\text{dim}(W)=\text{dim}(V)=\text{dim}(\text{Ker}(U))+\text{dim}(\text{Ran}(U))=\text{dim}(\text{Ran}(U)),$$
  where the second equality comes from the Rank-Nullity Theorem.

  We also knew that

  $\text{Ran}(U)\subseteq W$.

  Assume

  $\text{dim}(\text{Ran}(U))=n$, there exists

  $\{{\mathbf{w}}_1,\cdots ,{\mathbf{w}}_n\}\subset \text{Ran}(U)$ that is linearly independent such that

  $\text{Ran}(U)=\text{span}\{{\mathbf{w}}_1,\cdots ,{\mathbf{w}}_n\}$. Since

  $\text{dim}(W)=n$, by replacement Theorem we can conclude that

  $\{{\mathbf{w}}_1,\cdots ,{\mathbf{w}}_n\}$ is also a generating set of

  $W$, so

  $\text{Ran}(U)=W$ and

  $U$ is onto.

Proposition:
Let

$U:V\to W$ be an unitary transformation, then

1. $U^{-1}=U^{\ast }$;
2. $U^{\ast }$ is unitary;
3. If
   $\{{\mathbf{v}}_1,\cdots ,{\mathbf{v}}_n\}\subset V$ is an orthonormal basis,
   $\{U({\mathbf{v}}_1),\cdots ,U({\mathbf{v}}_n)\}\subset W$ is an orthonormal basis.

---

Rigid motion

Definition:
A rigid motion in a real inner product space

$V$ is a function

$f:V\to V$ such that

$$\begin{array}{}\text{(12)}& \Vert f(\mathbf{u})-f(\mathbf{v})\Vert =\Vert \mathbf{u}-\mathbf{v}\Vert .\end{array}$$

Remark:
We do not assume

$f$ is linear.

Lemma:
Let

$f:V\to V$ be a rigid motion and

$V$ is an real inner product space. We can define

$T(\mathbf{v})=f(\mathbf{v})-f(\mathbf{0})$, then for all

$\mathbf{u},\mathbf{v}\in V$, we have
1.

$$\begin{array}{}\text{(13)}& \Vert T(\mathbf{v})\Vert =\Vert \mathbf{v}\Vert ;\end{array}$$
2.

$$\begin{array}{}\text{(14)}& \Vert T(\mathbf{u})-T(\mathbf{v})\Vert =\Vert \mathbf{u}-\mathbf{v}\Vert ;\end{array}$$
3.

$$\begin{array}{}\text{(15)}& ⟨T(\mathbf{u}),T(\mathbf{v})⟩=⟨\mathbf{u},\mathbf{v}⟩.\end{array}$$

• Proof:

  (1.)
  

  $$\Vert T(\mathbf{v})\Vert =\Vert f(\mathbf{v})-f(\mathbf{0})\Vert \text{xlongequal}[\text{rigid motion}]\Vert \mathbf{v}-\mathbf{0}\Vert =\Vert \mathbf{v}\Vert .$$

  (2.)
  

  $$\begin{array}{rl}\Vert T(\mathbf{u})-T(\mathbf{v})\Vert & =\Vert f(\mathbf{u})-f(\mathbf{0})-(f(\mathbf{v})-f(\mathbf{0}))\Vert \\ & =\Vert f(\mathbf{u})-f(\mathbf{v})\Vert \\ & =\Vert \mathbf{u}-\mathbf{v}\Vert .\end{array}$$

  (3.)

  By direct computation and use (1.) we have
  

  $$\begin{array}{rl}\Vert T(\mathbf{u})-T(\mathbf{v}){\Vert }^2& =\Vert T(\mathbf{u}){\Vert }^2+\Vert T(\mathbf{v}){\Vert }^2-2⟨T(\mathbf{u}),T(\mathbf{v})⟩\\ & =\Vert \mathbf{u}{\Vert }^2+\Vert \mathbf{v}{\Vert }^2-2⟨T(\mathbf{u}),T(\mathbf{v})⟩,\end{array}$$
  and, use (2.), we obtain
  

  $$\begin{array}{rl}\Vert T(\mathbf{u})-T(\mathbf{v}){\Vert }^2& =\Vert \mathbf{u}-\mathbf{v}{\Vert }^2\\ & =\Vert \mathbf{u}{\Vert }^2+\Vert \mathbf{v}{\Vert }^2-2⟨\mathbf{u},\mathbf{v}⟩.\end{array}$$
  Hence,

  $⟨T(\mathbf{u}),T(\mathbf{v})⟩=⟨\mathbf{u},\mathbf{v}⟩$.

---

Corollary:

$T:V\to W$ is a linear transformation if and only if

$$\begin{array}{}\text{(16)}& T(\alpha \mathbf{u}+\mathbf{v})=\alpha T(\mathbf{u})+T(\mathbf{v}),\mathrm{\forall }\mathbf{u},\mathbf{v}\in V,\alpha \in F.\end{array}$$

• Proof:

  (

  $\Rightarrow$)
  Trivial!

  (

  $\Leftarrow$)
  Let

  $\alpha =1$,

  $T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})$.
  Let

  $\alpha =1$,

  $\mathbf{u}=\mathbf{v}=\mathbf{0}$,

  $T(\mathbf{0})=2T(\mathbf{0})$, so

  $T(\mathbf{0})=\mathbf{0}$.
  Let

  $\mathbf{v}=\mathbf{0}$,

  $T(\alpha \mathbf{u})=\alpha T(\mathbf{u})$.

---

Theorem:
Let

$f:V\to V$ be a rigid motion in a real inner product space

$V$ and let

$T:V\to V$ be defined as

$T(\mathbf{v})=f(\mathbf{v})-f(\mathbf{0})$. Then

$T$ is a linear orthogonal transformation.

• Proof:

  claim:

  $\Vert \alpha T(\mathbf{u})+T(\mathbf{v})-T(\alpha \mathbf{u}+\mathbf{v})\Vert =0$, for all

  $\mathbf{u},\mathbf{v}\in V$,

  $\alpha \in F$.

  We use (13), (14) and (15) to have
  

  $$\begin{array}{rl}& \Vert \alpha T(\mathbf{u})+T(\mathbf{v})-T(\alpha \mathbf{u}+\mathbf{v}){\Vert }^2\\ & =\Vert \alpha T(\mathbf{u}){\Vert }^2+\Vert T(\mathbf{v})-T(\alpha \mathbf{u}+\mathbf{v}){\Vert }^2+2⟨\alpha T(\mathbf{u}),T(\mathbf{v})-T(\alpha \mathbf{u}+\mathbf{v})⟩\\ & =\Vert \alpha T(\mathbf{u}){\Vert }^2+\Vert T(\mathbf{v})-T(\alpha \mathbf{u}+\mathbf{v}){\Vert }^2+2\alpha ⟨T(\mathbf{u}),T(\mathbf{v})⟩-2\alpha ⟨T(\mathbf{u}),T(\alpha \mathbf{u}+\mathbf{v})⟩\\ & =|\alpha {|}^2\Vert \mathbf{u}{\Vert }^2+\Vert \mathbf{v}-(\alpha \mathbf{u}+\mathbf{v}){\Vert }^2+2\alpha ⟨\mathbf{u},\mathbf{v}⟩-2\alpha ⟨\mathbf{u},\alpha \mathbf{u}+\mathbf{v}⟩\\ & =0.\end{array}$$
  Therefore,

  $T$ is a linear transformation.

  Since

  $T$ is linear and satisfies (13), so

  $T$ is an isometry transformation.

  Since

  $T:V\to V$,

  $\text{dim}(V)=\text{dim}(V)$, and is an isometry, based on the Proposition we shown before,

  $T$ is an orthogonal transformation.

Theorem:
Let

$f:V\to V$ be a rigid motion, then there exists an unique

${\mathbf{v}}_0\in V$ and an unique

$U:V\to V$ that is an orthogonal transformation such that

$f(\mathbf{v})=U\mathbf{v}+{\mathbf{v}}_0$, for all

$\mathbf{v}\in V$.

• Proof:

  Define

  $U(\mathbf{v})=f(\mathbf{v})-f(\mathbf{0})$, then

  $U$ is an orthogonal transformation.
  Define

  ${\mathbf{v}}_0=f(\mathbf{0})$, then

  $f(\mathbf{v})=U(\mathbf{v})+{\mathbf{v}}_0$, for all

  $\mathbf{v}\in V$.

  (uniqueness)
  Suppose there exists

  ${\mathbf{u}}_0\in V$ and

  $T:V\to V$ that is an orthogonal transformation such that

  $f(\mathbf{v})=T(\mathbf{v})+{\mathbf{u}}_0$, for all

  $\mathbf{v}\in V$.
  Then,
  

  $$f(\mathbf{v})=T(\mathbf{v})+{\mathbf{u}}_0=U(\mathbf{v})+{\mathbf{v}}_0.$$
  Since

  $T$ and

  $U$ are both linear, we have
  

  $$f(\mathbf{0})=T(\mathbf{0})+{\mathbf{u}}_0={\mathbf{u}}_0={\mathbf{v}}_0=U(\mathbf{0})+{\mathbf{v}}_0.$$
  So

  ${\mathbf{u}}_0={\mathbf{v}}_0$.
  We then have

  $T(\mathbf{v})=U(\mathbf{v})$ for all

  $\mathbf{v}\in V$ that leads to

  $T=U$.

Corollary:
We also have, for all

$\mathbf{v}\in V$,

$f(\mathbf{v})=U(\mathbf{v}+{\tilde{\mathbf{v}}}_0)$, where

${\tilde{\mathbf{v}}}_0=U^{-1}{\mathbf{v}}_0$.