# Lecture 29

## Chapter VI Inner Product Spaces

### Orthogonal Complements and Minimization Problems 6C

#### Minimization Problems

#### Theorem 6.61

Suppose $U$ is a finite dimensional subspace of $V$. Let $v\in V$, $u\in U$. Then $||v-P_u v||\leq|| v-u||$. with equality if and only if $u=P_u v$

Proof:

Using triangle inequality

$$
\begin{aligned}
||v-P_u v||^2 &\leq ||v-P_u v||^2+||P_u v-u||^2\\
&=||(v-P_u v)+(P_u v-u)||^2\\
&=||v-u||^2
\end{aligned}
$$

Example:

Find $u(x)\in \mathscr{P}_5(\mathbb{R}) minimizing

$$
\int^{\pi}_{-\pi}|sin(x)-u(x)|^2 dx
$$

$V=C([-\pi,\pi])=$ continuous (real valued) function on $[-\pi,\pi]$

$u=\mathscr{P}_5(\mathbb{R})$. Note $U\subseteq V$ and $u$ is finite dimensional.

$\langle f,g \rangle=\int^{\pi}_{-\pi}fg$ gives an inner product on $V$.

Minimize $||sin-u||^2$, choose an orthonormal basis $e_0,...,e_5$ of $\mathscr{P}_5(\mathbb{R})$, so $u=P_u(sin)=\langle e_0,sin\rangle e_0+...+\langle e_5,sin \rangle e_5$

#### Pseudo inverses

Idea: Want to (approximately) solve $Tx=b$.

- If $T$ is invertible $x=T^{-1}b$
- If $T$ is not invertible, want $T^{T}$ such that $y=T^{T}b$ is the "best solution"

#### Lemma 6.67

If $V$ is a finite dimensional vector space, $T\in \mathscr{L}(V,W)$ then $T\vert_{{null\ T}^\perp}$ is one to one onto $range\ T$.

Proof:

Note $(null\ T)^\perp \simeq V/(null\ T)$

Exercise, prove this...

If $v\in null(T\vert_{{null}^\perp})\implies v\in null\ T,$ and $v\in (null\ T)^\perp\implies v=0$

If $w\in range\ T$ so $\exists v\in V$ such that $Tv=w$ write $v$ as $v=u+x$ sor $u\in null\ T,x\in (null\ T)^\perp$.

#### Definition 6.68

V is a finite dimensional space $T\in \mathscr{L}(V,W)$. The **pseudo-inverse** denoted $T^\dag\in \mathscr{L}(W,V)$ is given by

$$
T^\dag w=(T\vert_{{null\ T}^\perp})^{-1}P_{range\ T}w
$$

Some explanation:

_Let $T\in \mathscr{L}(V,W)$.Since there exists isomorphism between $(null\ T)^\perp\subseteq V$ and $range\ T\subseteq W$.We can always map $W$ to $V$ using $T^\dag\in \mathscr{L}(W,V)$. $P_{range\ T}$ is the map that $W\mapsto range\ T$ and $(T\vert_{{null\ T}^\perp})^{-1}$ is a linear map that map $w\in W$_

#### Proposition 6.69

$V$ is a finite dimensional vector space. $T\in\mathscr{L}(V,W)$, then 

(a) If $T$ is invertible, then $T^\dag=T^{-1}$.  
(b) $TT^\dag=P_{range\ T}$.  
(c) $T^\dag T=P_{(null\ T)^\perp}$.

#### Theorem 6.70

$V$ is a finite dimensional vector space. $T\in\mathscr{L}(V,W)$, for $b\in W$, then

(a) If $x\in V$, then $||T(T^* b)-b||\leq ||Tx-b||$ with equality if and only if $x\in T^\dag b+null\ T$ (_$T^\dag$ is the best solution we can have as "inverse" for non-invertible linear map_)

(b) If $x\in T^\dag b+null\ T$ then

$$
||T^\dag b ||\leq ||x||
$$

Proof:

(a) $Tx-b=(Tx-TT^\dag b)+(TT^\dag b-b)$

Using pythagorean theorem, we have

$||Tx-b||\geq ||TT^\dag b-b||$

## Chapter VII Operators on Inner Product Spaces

### Self adjoint and Normal Operators 7A

#### Definition 7.1

Let $T\in \mathscr{L}(V,W)$, then the **adjoint** of $T$ denoted $T^*$ is the function $T^*:W\to V$ such that $\langle Tv,w \rangle =\langle v,T^* w \rangle$

For euclidean inner product $T^*$ is given by the conjugate transpose.