109 lines
3.0 KiB
Markdown
109 lines
3.0 KiB
Markdown
# Lecture 32
|
|
|
|
## Chapter VII Operators on Inner Product Spaces
|
|
|
|
**Assumption: $V,W$ are finite dimensional inner product spaces.**
|
|
|
|
### Spectral Theorem 7B
|
|
|
|
Recall
|
|
|
|
#### Definition 7.10
|
|
|
|
An operator $T\in \mathscr{L}(V)$ is self adjoint if $T=T^*$
|
|
|
|
#### Definition 7.18
|
|
|
|
AN operator $T\in\mathscr{L}(V)$ is normal if $TT^*=T^*T$
|
|
|
|
#### Theorem 7.20
|
|
|
|
Suppose $T\in \mathscr{L}(V)$, $T$ is normal $\iff ||Tv||=||T^*v||$
|
|
|
|
#### Lemma 7,26
|
|
|
|
Suppose $T\in\mathscr{L}(V)$ is self adjoint operator and $b,c\in \mathbb{R}$ such that $b^2<4c$, then
|
|
|
|
$$
|
|
T^2+bT+cI
|
|
$$
|
|
|
|
is invertible.
|
|
|
|
Proof:
|
|
|
|
Prove $T^2+bT+cI$ is injective by showing $\langle(T^2+bT+cI),v\rangle\neq 0$ (for $v\neq 0$)
|
|
|
|
$$
|
|
\begin{aligned}
|
|
\langle(T^2+bT+cI),v\rangle&=\langle T^2v,v\rangle+\langle bTv,v\rangle+c\langle v,v\rangle\\
|
|
&=\langle Tv,Tv\rangle+b\langle Tv,v\rangle +c||v||^2\\
|
|
&\geq ||Tv||^2-|b|\ ||Tv||\ ||v||+c||v||^2 \textup{ by cauchy schuarz}\\
|
|
&=\left(||Tv||-\frac{b||v||}{2}\right)^2+\left(c-\frac{b^2}{4}\right)||v||^2>0
|
|
\end{aligned}
|
|
$$
|
|
|
|
#### Theorem 7.27
|
|
|
|
Suppose $T\in \mathscr{L}(V)$ is self adjoint. Then the minimal polynomial is of the form $(z-\lambda_1)...(z-\lambda_m)$ for some $\lambda_1,...,\lambda_m\in\mathbb{R}$
|
|
|
|
Proof:
|
|
|
|
$\mathbb{F}=\mathbb{C}$ clear from previous results
|
|
|
|
$\mathbb{F}=\mathbb{R}$ assume for contradiction $q(z)$, where $b^2\leq 4c$. Then $P(T)=0$ but $q(T)\neq 0$. So let $v\in V$ such that $q(T)v\neq 0$.
|
|
|
|
then $(T^2+bT+cI)(q(T)v)=0$ but $T^2+bT+cI$ is invertible so $q(T)v=0$ this is a contradiction so $p(z)=(z-\lambda_1)...(z-\lambda_m)$
|
|
|
|
#### Theorem 7.29 Real Spectral theorem
|
|
|
|
Suppose $V$ is a finite dimensional real inner product space and $T\in \mathscr{L}(V)$ then the following are equivalent.
|
|
|
|
(a) $T$ is self adjoint.
|
|
(b) $T$ has a diagonal matrix with respect to same orthonormal basis.
|
|
(c) $V$ has an orthonormal basis of eigenvectors of $T$
|
|
|
|
Proof:
|
|
|
|
$b\iff c$ clear by definition
|
|
|
|
$b\implies a$ because the transpose of a diagonal matrix is itself.
|
|
|
|
$a\implies b$ by (**Theorem 7.27**) there exists an orthonormal basis such that $M(T)$ is upper triangular. But $M(T^*)=M(T)$ and $M(T^*)=(M(T))^*$
|
|
|
|
but this $M(T)$ is both upper and lower triangular, so $M(T)$ is diagonal.
|
|
|
|
#### Theorem 7.31 Complete Spectral Theorem
|
|
|
|
Suppose $V$ is a complex finite dimensional inner product space. $T\in \mathscr{L}(V)$, then the following are equivalent.
|
|
|
|
(a) $T$ is normal
|
|
(b) $T$ has a diagonal matrix with respect to an orthonormal basis
|
|
(c) $V$ has an orthonormal basis of eigenvectors of $T$.
|
|
|
|
$a\implies b$
|
|
|
|
$$
|
|
M(T)=\begin{pmatrix}
|
|
a_{1,1}&\dots&a_{1,n}\\
|
|
&\ddots &\vdots\\
|
|
0& & a_{n,n}
|
|
\end{pmatrix}
|
|
$$
|
|
|
|
with respect to an appropriate basis $e_1,...,e_n$
|
|
|
|
Then $||Te_1||^2=|a_{1,1}|^2$, $||Te_1||^2=||T^*e_1||^2=|a_{1,1}|^2+|a_{1,2}|^2+...+|a_{1,n}|^2$. So $a_{1,2}=...=a_{1,n}=0$, without loss of generality, $||Te_2||^2=0$. Repeating this procedure we have $M(T)$ is diagonal.
|
|
|
|
Example:
|
|
|
|
$T\in \mathscr{L}(\mathbb{C}^2)$ $M(T)=\begin{pmatrix}
|
|
2&-3\\
|
|
3&2
|
|
\end{pmatrix}$
|
|
|
|
$M(T,(f_1,f_2))=\begin{pmatrix}
|
|
2+3c&0\\
|
|
0&2-3c
|
|
\end{pmatrix}$
|