# 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.

$$
(T^*T-TT^*)^*=(T^*T)^*-(TT^*)^*=T^*T-TT^*
$$