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

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 $\langle Tv,v\rangle$

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

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 $u,w\in V$

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

Proof:

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

$\Rightarrow$

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

Proof:

The key idea is that $T^*T-TT^*$ is self adjoint.