# Lecture 28

## Chapter VI Inner Product Spaces

### Orthonormal basis 6B

Example: 

Find a polynomial $q\in \mathscr{P}_2(\mathbb{R})$ such that

$$
\int^1_{-1}p(t)cos(\pi t)dt=\int^1_{-1}p(t)q(t)dt
$$

for $p\in \mathscr{P}_2(\mathbb{R})$

note that $\varphi(p)=\int^1_{-1}p(t)cos(\pi t)cos(\pi t)dt$ is a linear functional. Thus by **Riesz Representation Theorem**, $\exists$ unique $q$ such that $\varphi (p)=\langle p,q \rangle=\int^1_{-1}pq$

$$
q=\overline{\varphi(e_0)}e_0+\overline{\varphi(e_z)}e_z
$$

where $e_0,e_1,e_z$ is an orthonormal basis.

and $q=\frac{15}{2\pi^2}(1-3x^2)$

#### Orthogonal Projection and Minimization

#### Definition 6.46

If $U$ is a subset of $V$, then the **orthogonal complement** of $U$ denoted $U^\perp$

$$
U^\perp=\{v\in V\vert \langle u,v\rangle =0,\forall u\in U\}
$$

The set of vectors orthogonal to every vector in $U$.

#### Theorem 6.48

Let $U$ be a subset of $V$.

(a) $U^\perp$ is a subspace of $V$.  
(b) $\{0\}^\perp=V$  
(c) $V^\perp =\{0\}$  
(d) $U\cap U^\perp\subseteq\{0\}$  
(e) If $G,H$ subsets of $V$ with $G\subseteq H$, then $H^\perp\subseteq G^\perp$

Example: 

Two perpendicular line in 2D plane. 

Let $e_1,...,e_m$ be an orthonormal list, let $u=Span(e_1,...,e_m)$ How do I find $U^\perp$?

Extend to an orthonormal basis $e_1,...,e_m,f_1...,f_n$. $U^\perp=Span(f_1,...,f_n)$

#### Theorem 6.40

Suppose $U$ is finite dimensional subspace of $V$, then $V=U\oplus U^\perp$

Proof:

Note $U\cap U^\perp=\{0\}$, so it suffices to show $U+U^\perp=V$. Fix an orthonormal basis $e_1,...,e_m$ of $U$. Let $v\in V$, let $u=\langle v,e_1\rangle e_1+...+\langle v,e_m\rangle e_m$ let $w=v-u$, then $v=u+W$ we need to check that $w\in U^\perp$

$\langle w,e_k \rangle=\langle v,e_k\rangle-\langle u,e_k\rangle=\langle v,e_k\rangle-\langle v,e_k\rangle=0$

So $w\in U^\perp$

#### Corollary 6.51

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

#### Theorem 6.52

Let $U$ be a finite dimensional of a vector space $V$. Then $(U^\perp)^\perp=U$

Proof:

First let $u\in U$ we want to show $u\in (U^\perp)^\perp$, then $\langle u,w \rangle=0$ for all $w\in U^\perp$ but then $u\in (U^\perp)^\perp$

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#### Corollary 6.54

$U^\vert=\{0\}\iff U=V$

Proof: 

$(U^\perp)^\perp=\{0\}^\perp\implies U=V$

#### Definition 6.55

Given $U$ a finite dimensional subspace of $V$. The **orthogonal projection of $V$ onto $U$** is the operator $P_u\in \mathscr{L}(V)$ defined by: For each $v$ write $v=u+w$ where $u\in U$ and $w\in U^\perp$ then $P_u v=u$

Formula: 

Let $e_1,...,e_n$ an orthonormal basis of $U$.

$P_u v=\langle v,e_1\rangle e_1+...+\langle v,e_m\rangle e_m$

#### Theorem 6.57

(a) $P_u$ is linear.  
(b) $P_u u=U,\forall u\in U$  
(c) $P_u w=0,\forall w\in U^\perp$  
(d) $range\ P_u=U$  
(e) $null\ P_u=U^\vert$  
(f) $v-P_u v\in U^\perp$  
(g) $P_u^2=P_u$  
(h) $||P_u v||\leq ||v||$

Proof: 

(a) Let $v,v'\in V$ and suppose $v=u+w,v'=u'+w'$, then $v+u'=(u+u')+(w+w')$ this implies that $P_u(v+v')=u+u'=P_u v+ P_u v'$

...

#### Theorem 6.58 Riesz Representation Theorem

Let $V$ be a finite dimensional vector space for $v\in V$ define $\varphi_v\in V'$ by $\varphi_v(u)=\langle u,v \rangle$. Then the map $v\to \varphi_v$ is a bijection.

Proof:

Surjectivity Ideal is let $w\in (null\ \varphi)^\perp$

$$
v=\frac{\varphi(w)}{||w||^2}w,\varphi(v)=||v||^2
$$

make sense $\varphi_v=\varphi$