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content/Math429/Math429_L31.md

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+# Lecture 31
+
+## Chapter VII Operators on Inner Product Spaces
+
+**Assumption: $V,W$ are finite dimensional inner product spaces.**
+
+### Self adjoint and Normal Operators 7A
+
+#### Definition 7.10
+
+An operator $T\in\mathscr{L}(V)$ is **self adjoint** if $T=T^*$. ie. $\langle Tv,u\rangle=\langle v,Tu \rangle$ for $u,v\in V$.
+
+Example: 
+
+Consider $M(T)=\begin{pmatrix}
+    2 & i\\
+    -i& 3
+\end{pmatrix}$, Then
+
+$$
+M(T^*)=M(T)^*=\begin{pmatrix}
+    \bar{2},\bar{-i}\\
+    \bar{i},\bar{3}
+\end{pmatrix}=\begin{pmatrix}
+    2 &i\\
+    -i& 3
+\end{pmatrix}=M(T)
+$$
+
+So $T=T^*$ so $T$ is self adjoint
+
+#### Theorem 7.12
+
+Every eigenvalue of a self adjoint operator $T$ is real.
+
+Proof:
+
+Suppose $T$ is self adjoint and $\lambda$ is an eigenvalue of $T$, and $v$ is an eigenvector with eigenvalue $\lambda$.
+
+Consider $\langle Tv,v\rangle$
+
+$$
+\langle Tv, v\rangle=
+\langle v, Tv\rangle=
+\langle v,\lambda v\rangle=
+\bar{\lambda}\langle v,v\rangle=\bar{\lambda}||v||^2
+$$
+$$
+\langle Tv, v\rangle=
+\langle \lambda v, v\rangle=
+\langle v, v\rangle=
+\lambda\langle v,v\rangle=\lambda||v||^2\\
+$$
+
+So $\lambda=\bar{\lambda}$, so $\lambda$ is real.
+
+NoteL (7.12) is only interesting for complex vector spaces.
+
+#### Theorem 7.13
+
+Suppose $V$ is a complex inner product space and $T\in\mathscr{L}(V)$, then 
+
+$$
+\langle Tv, v\rangle =0 \textup{ for every }v\in V\iff T=0
+$$
+
+Note: (7.13) is **False** over real vector spaces. The counterexample is $T$ the rotation by $90\degree$ operator. ie. $M(T)=\begin{pmatrix}
+    0&-1\\
+    1&0
+\end{pmatrix}$
+
+Proof:
+
+$\Rightarrow$ Suppose $u,w\in V$
+
+$$
+\begin{aligned}
+\langle Tu,w \rangle&=\frac{\langle T(u+w),u+w\rangle -\langle T(u-w),u-w\rangle}{4}+\frac{\langle T(u+iw),u+iw\rangle -\langle T(u-iw),u-iw\rangle}{4}i\\
+&=0
+\end{aligned}
+$$
+
+Since $w$ is arbitrary $\implies Tu=0, \forall u\in V\implies T=0$.
+
+#### Theorem 7.14
+
+Suppose $V$ is a complex inner product space and $T\in \mathscr{L}(V)$ thne 
+
+$$
+T \textup{ is self adjoint }\iff \langle Tv, v\rangle \in \mathbb{R} \textup{ for every} v \in V
+$$
+
+Proof:
+
+$$
+\begin{aligned}
+    T\textup{ is self adjoint}&\iff T-T^*=0\\
+    &\iff \langle (T-T^*)v,v\rangle =0 (\textup{ by \textbf{7.13}})\\
+    &\iff \langle Tv, v\rangle -\langle T^*v,v \rangle =0\\
+    &\iff \langle Tv, v\rangle -\overline{\langle T,v \rangle} =0\\
+    &\iff\langle Tv,v\rangle \in \mathbb{R}
+\end{aligned}
+$$
+
+#### Theorem 7.16
+
+Suppose $T$ is a self adjoint operator, then $\langle Tv, v\rangle =0,\forall v\in V\iff T=0$
+
+Proof:
+
+Note the complex case is **Theorem 7.13**, so assume $V$ is a real vector space. Let $u,w\in V$ consider 
+
+$\Rightarrow$ 
+
+$$
+\langle Tu,w\rangle=\frac{\langle T(u+w),u+w\rangle -\langle T(u-w),u-w\rangle}{4}=0
+$$
+
+We set $\langle Tw,u\rangle=\langle w,Tu\rangle =\langle Tu,w\rangle$
+
+#### Normal Operators
+
+#### Definition 7.18
+
+An operator $T\in \mathscr{L}(V)$ on an inner product space is **normal** if $TT^*=T^*T$ ie. $T$ commutes with its adjoint
+
+#### Theorem (7.20) 
+
+An operator $T$ is normal if and only if
+
+$$
+||Tv||=||T^*v||,\forall v\in V
+$$
+
+Proof:
+
+The key idea is that $T^*T-TT^*$ is self adjoint.
+
+$$
+(T^*T-TT^*)^*=(T^*T)^*-(TT^*)^*=T^*T-TT^*
+$$
+