---
title: Ch5-6
tags: Linear algebra
GA: G-77TT93X4N1
---

# 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
$$
\begin{align}
&V=\mathbb{P}_1, \quad \langle f, g\rangle = \int^1_{-1}f(x)g(x)\,dx,\\
&W=\mathbb{R}^2, \quad \langle (x, y), (u, v)\rangle = 2xu + 3yv.
\end{align}
$$
Let $T:V\to W$ be a linear transformation defined by
$$
\tag{1}
T(p) = \left(\int^1_0 p(x)\,dx, \int^1_{-1} p(x)\,dx\right).
$$

In the following we aim to determine the adjoint linear transformation $T^*$ of $T$.

Based on the definition of adjoint transformation, we should have, for all $p(x)\in\mathbb{P}_1$ and $(u,v)\in\mathbb{R}^2$,
$$
\tag{2}
\langle T(p), (u, v)\rangle = \langle p, T^*((u, v))\rangle.
$$

> The goal is to find $T^*:W\to V$.
> Since $W$ have a natural basis $e_1=(1, 0)$ and $e_2=(0, 1)$, if we can determine $T^*(e_1)$ and $T^*(e_2)$, we should have
> $$
> T^*((u, v)) = uT^*(e_1) + vT^*(e_2).
> $$

$T^*(e_1)$ should satisfy
$$
\tag{3}
\langle T(p), e_1\rangle = \langle p, T^*(e_1)\rangle = \int^1_{-1}p(x)T^*(e_1)\,dx.
$$

> 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 $\langle p(x), q(x)\rangle=\langle p(x), T^*(e_1)\rangle$ for all $p$ and we can deduce that $T^*(e_1)=q(x)$.

The left hand side of (3) is an inner product in $W$ that gives
$$
\begin{align}
\langle T(p), (1, 0)\rangle &= 2\int^1_0 p(x)\,dx\\
&=\int^1_{-1}p(x)(1+\frac{3}{2}x)\,dx\\
&=\langle p, 1+\frac{3}{2}x\rangle.\tag{4}
\end{align}
$$

> Exercise:
> Find $q(x)\in\mathbb{P}_1$ such that
> $$
> \int^1_0 2p(x)\,dx = \int^1_{-1}p(x)q(x)\,dx, \quad \forall p\in\mathbb{P}_1.
> $$
>
> > It turns out that $q(x)=1 + \frac{3}{2}x$.

Similarly, $T^*(e_2)$ should satisfy
$$
\langle T(p), e_2\rangle = \langle p, T^*(e_2)\rangle,
$$
and we easily obtain $T^*(e_2)=3$.

Therefore,
$$
T^*(e_1) = 1+\frac{3}{2}x, \quad T^*(e_2)=3,
$$
and
$$
\tag{5}
T^*((u, v)) = u + 3v+\frac{3u}{2}x.
$$

---

### Isometry

:::info
**Definition:**
Let $V, W$ be *normed spaces*, a linear transformation $U:V\to W$ is called an **isometry** if
$$
\tag{6}
\|U({\bf v})\| = \|{\bf v}\|, \quad \forall{\bf v}\in V.
$$
:::

---

**Remark:**
1. Let ${\bf u}, {\bf v}\in V$, where $V$ is a ***real*** inner product space, we have
$$
\tag{7}
\|{\bf u} + {\bf v}\|^2 = \|{\bf u}\|^2 + \|{\bf v}\|^2 + 2\langle {\bf u}, {\bf v}\rangle.
$$
2. Let ${\bf u}, {\bf v}\in V$, where $V$ is an inner product space, we have
$$
\tag{8}
\|{\bf u} + {\bf v}\|^2 = \|{\bf u}\|^2 + \|{\bf v}\|^2 + 2\text{Real}\left(\langle {\bf u}, {\bf v}\rangle\right).
$$

**Lemma: (polarization identity)**
1. Let ${\bf u}, {\bf v}\in V$, where $V$ is a ***real*** inner product space, we have
$$
\tag{9}
\langle {\bf u}, {\bf v}\rangle = \frac{1}{4}\left(\|{\bf u} + {\bf v}\|^2 - \|{\bf u} - {\bf v}\|^2\right)
, \quad \forall {\bf u,v}\in V.
$$
2. Let ${\bf u}, {\bf v}\in V$, where $V$ is an inner product space, we have, for all ${\bf u,v}\in V$,
$$
\tag{10}
\langle {\bf u}, {\bf v}\rangle = \frac{1}{4}\left(\|{\bf u} + {\bf v}\|^2 - \|{\bf u} - {\bf v}\|^2 + i\|{\bf u} + i{\bf v}\|^2 - i\|{\bf u} - i{\bf v}\|^2\right).
$$

---

**Theorem:**
Let $V, W$ be *inner product spaces*. The operator $U:V\to W$ is an isometry if and only if
$$
\tag{11}
\langle U({\bf u}), U({\bf v})\rangle = \langle {\bf u}, {\bf v}\rangle, \quad \forall {\bf u,v}\in V.
$$
* Proof:
> ($\Leftarrow$)
> Choose ${\bf u}={\bf v}$, then
> $$
> \|U({\bf v})\|^2 = \langle U({\bf v}), U({\bf v})\rangle = \langle {\bf v}, {\bf v}\rangle = \|{\bf v}\|^2.
> $$
>
>
> ($\Rightarrow$)
> > Here we assume ***real*** inner product space. The derivation of the general version is similar.
>
> $$
> \begin{align}
> \langle {\bf u}, {\bf v}\rangle &\xlongequal[\text{polarization identity}]{} \frac{1}{4}\left(\|{\bf u} + {\bf v}\|^2 - \|{\bf u} - {\bf v}\|^2\right)\\[.2cm]
> &\xlongequal[\text{isometry}]{} \frac{1}{4}\left(\|U({\bf u} + {\bf v})\|^2 - \|U({\bf u} - {\bf v})\|^2\right)\\[.2cm]
> &\xlongequal[\text{linearity}]{}\frac{1}{4}\left(\|U({\bf u}) + U({\bf v})\|^2 - \|U({\bf u}) - U({\bf v})\|^2\right)\\[.2cm]
> &\xlongequal[\text{polarization identity}]{}\langle U({\bf u}), U({\bf v})\rangle.
> \end{align}
> $$

**Lemma:**
Let $V, W$ be *inner product spaces*. The operator $U:V\to W$ is an isometry if and only if $U^*U=I$, where $I$ is the identity transformation.
* Proof:
> ($\Leftarrow$)
> $$
> \|U{\bf v}\|^2=\langle U{\bf v}, U{\bf v}\rangle = \langle {\bf v}, U^*U{\bf v}\rangle = \langle {\bf v}, {\bf v}\rangle=\|{\bf v}\|^2.
> $$
>
> ($\Rightarrow$)
> Suppose $U$ is an isometry, we want to show that
> $$
> U^*U{\bf v}={\bf v}, \quad \forall{\bf v}\in V.
> $$
>
> > Fix ${\bf v}\in V$, given ${\bf u}\in V$, since $U$ is an isometry, we have (11) and
> > $$
> > \langle {\bf u}, {\bf v}\rangle = \langle U{\bf u}, U{\bf v}\rangle= \langle {\bf u}, U^*U{\bf v}\rangle.
> > $$
> > Therefore, $\langle {\bf u}, {\bf v}\rangle = \langle {\bf u}, U^*U{\bf v}\rangle$ for all ${\bf u}\in V$, that gives ${\bf v}=U^*U{\bf v}$.

**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.

:::info
**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 $\text{Ker}(U)=\{{\bf 0}\}$ and $\text{Ran}(U)=W$.
>
> Based on the [Rank-Nullity Theorem](https://hackmd.io/@teshenglin/2024LA2_ch5_5) 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, $\text{Ker}(U)=\{{\bf 0}\}$ 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 $\{{\bf w}_1, \cdots, {\bf w}_n\}\subset\text{Ran}(U)$ that is linearly independent such that $\text{Ran}(U) = \text{span}\{{\bf w}_1, \cdots, {\bf w}_n\}$. Since $\text{dim}(W)=n$, by [replacement Theorem](https://hackmd.io/@teshenglin/2024LA2_ch1_7) we can conclude that $\{{\bf w}_1, \cdots, {\bf w}_n\}$ 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. $U^{-1}=U^*$;
2. $U^*$ is unitary;
3. If $\{{\bf v}_1, \cdots, {\bf v}_n\}\subset V$ is an orthonormal basis, $\{U({\bf v}_1), \cdots, U({\bf v}_n)\}\subset W$ is an orthonormal basis.

---

### Rigid motion

:::info
**Definition:**
A rigid motion in a *real* inner product space $V$ is a *function* $f:V\to V$ such that
$$
\tag{12}
\|f({\bf u})-f({\bf v})\| = \|{\bf u}-{\bf v}\|.
$$
:::

**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 $T({\bf v})=f({\bf v})-f({\bf 0})$, then for all ${\bf u}, {\bf v}\in V$, we have
1.
$$
\tag{13}
\|T({\bf v})\| = \|{\bf v}\|;
$$
2.
$$
\tag{14}
\|T({\bf u})-T({\bf v})\| = \|{\bf u} - {\bf v}\|;
$$
3.
$$
\tag{15}
\langle T({\bf u}), T({\bf v})\rangle = \langle{\bf u}, {\bf v}\rangle.
$$
* Proof:
> (1.)
> $$
> \|T({\bf v})\| = \|f({\bf v})-f({\bf 0})\|\xlongequal[\text{rigid motion}]{}\|{\bf v}-{\bf 0}\|=\|{\bf v}\|.
> $$
>
> (2.)
> $$
> \begin{align}
> \|T({\bf u})-T({\bf v})\| &= \|f({\bf u})-f({\bf 0})-(f({\bf v})-f({\bf 0}))\| \\
> &= \|f({\bf u})-f({\bf v})\| \\
> &= \|{\bf u} - {\bf v}\|.
> \end{align}
> $$
>
> (3.)
>
> By direct computation and use (1.) we have
> $$
> \begin{align}
> \|T({\bf u})-T({\bf v})\|^2 &= \|T({\bf u})\|^2+\|T({\bf v})\|^2 -2\langle T({\bf u}), T({\bf v})\rangle\\
> &=\|{\bf u}\|^2+\|{\bf v}\|^2 -2\langle T({\bf u}), T({\bf v})\rangle,
> \end{align}
> $$
> and, use (2.), we obtain
> $$
> \begin{align}
> \|T({\bf u})-T({\bf v})\|^2 &= \|{\bf u}-{\bf v}\|^2\\
> &=\|{\bf u}\|^2+\|{\bf v}\|^2 -2\langle {\bf u}, {\bf v}\rangle.
> \end{align}
> $$
> Hence, $\langle T({\bf u}), T({\bf v})\rangle = \langle{\bf u}, {\bf v}\rangle$.
>

---

**Corollary:**
$T:V\to W$ is a linear transformation if and only if
$$
\tag{16}
T(\alpha{\bf u}+ {\bf v}) = \alpha T({\bf u})+ T({\bf v}), \quad \forall {\bf u}, {\bf v}\in V, \quad \alpha\in F.
$$
* Proof:
> ($\Rightarrow$)
> Trivial!
>
> ($\Leftarrow$)
> Let $\alpha=1$, $T({\bf u}+{\bf v}) = T({\bf u})+ T({\bf v})$.
> Let $\alpha=1$, ${\bf u}={\bf v}={\bf 0}$, $T({\bf 0}) = 2T({\bf 0})$, so $T({\bf 0})={\bf 0}$.
> Let ${\bf v}={\bf 0}$, $T(\alpha{\bf u}) = \alpha T({\bf u})$.

---

**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 $T({\bf v}) = f({\bf v})-f({\bf 0})$. Then $T$ is a linear orthogonal transformation.
* Proof:
> claim: $\|\alpha T({\bf u})+ T({\bf v}) - T(\alpha{\bf u}+ {\bf v})\|=0$, for all ${\bf u}, {\bf v}\in V$, $\alpha\in F$.
>
> > We use (13), (14) and (15) to have
> > $$
> > \begin{align}
> > &\quad \|\alpha T({\bf u})+ T({\bf v}) - T(\alpha{\bf u}+ {\bf v})\|^2\\
> > &= \|\alpha T({\bf u})\|^2 + \|T({\bf v}) - T(\alpha{\bf u}+ {\bf v})\|^2 + 2\langle\alpha T({\bf u}), T({\bf v}) - T(\alpha{\bf u}+ {\bf v})\rangle\\
> > &= \|\alpha T({\bf u})\|^2 + \|T({\bf v}) - T(\alpha{\bf u}+ {\bf v})\|^2 + 2\alpha\langle T({\bf u}), T({\bf v})\rangle- 2\alpha\langle T({\bf u}), T(\alpha{\bf u}+ {\bf v})\rangle\\
> > &= |\alpha|^2\|{\bf u}\|^2 + \|{\bf v} - (\alpha{\bf u}+{\bf v})\|^2 + 2\alpha\langle {\bf u}, {\bf v}\rangle-2\alpha\langle {\bf u}, \alpha{\bf u}+ {\bf v}\rangle\\
> > &=0.
> > \end{align}
> > $$
> 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 ${\bf v}_0\in V$ and an unique $U:V\to V$ that is an orthogonal transformation such that $f({\bf v}) = U{\bf v} + {\bf v}_0$, for all ${\bf v}\in V$.
* Proof:
> Define $U({\bf v})=f({\bf v})-f({\bf 0})$, then $U$ is an orthogonal transformation.
> Define ${\bf v}_0=f({\bf 0})$, then $f({\bf v}) = U({\bf v}) + {\bf v}_0$, for all ${\bf v}\in V$.
>
> (uniqueness)
> Suppose there exists ${\bf u}_0\in V$ and $T:V\to V$ that is an orthogonal transformation such that $f({\bf v}) = T({\bf v}) + {\bf u}_0$, for all ${\bf v}\in V$.
> Then,
> $$
> f({\bf v}) = T({\bf v}) + {\bf u}_0 = U({\bf v}) + {\bf v}_0.
> $$
> Since $T$ and $U$ are both linear, we have
> $$
> f({\bf 0}) = T({\bf 0}) + {\bf u}_0 = {\bf u}_0 ={\bf v}_0= U({\bf 0}) + {\bf v}_0.
> $$
> So ${\bf u}_0={\bf v}_0$.
> We then have $T({\bf v})=U({\bf v})$ for all ${\bf v}\in V$ that leads to $T=U$.

**Corollary:**
We also have, for all ${\bf v}\in V$, $f({\bf v}) = U({\bf v} + \tilde{{\bf v}}_0)$, where $\tilde{{\bf v}}_0 =U^{-1}{\bf v}_0$.