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