3.1 KiB
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^*