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content/Math429/Math429_L29.md

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+# 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.