# Lecture 30

## Chapter VII Operators on Inner Product Spaces

**Assumption: $V,W$ are finite dimensional inner product spaces.**

### Self adjoint and Normal Operators 7A

#### Definition 7.1

Suppose $T\in \mathscr{L}(V,W)$. The adjoint is the function $T^*:W\to V$ such that 

$$
\langle Tv,w \rangle=\langle v,T^*w \rangle, \forall v\in V, w\in W
$$

#### Theorem 7.4

Suppose $T\in \mathscr{L}(V,W)$ then $T^*\in \mathscr{L}(W,V)$

Proof:

Additivity, let $w_1,w_2\in W$. We want to show $T^*(w_1+w_2)T^*w_1+T^*w_2$

Let $v\in V$, then

$$
\begin{aligned}
    \langle v,T^*(w_1+w_2) \rangle &=\langle v,T^*(w_1+w_2) \rangle\\
    &=\langle Tv,w_1+w_2 \rangle\\
    &=\langle Tv,w_1 \rangle+\langle Tv,w_2 \rangle\\
    &=\langle v,T^*w_1 \rangle+\langle v,T^* w_2 \rangle\\
    &=\langle v,T^*w_1 +T^* w_2 \rangle\\
\end{aligned}
$$

Note: If $\langle v,u \rangle=\langle v,u'\rangle$, forall $v\in V$ then $u=u'$

Homogeneity: same as idea above.

#### Theorem 7.5

Suppose $S,T\in \mathscr{L}(V,W)$, and $\lambda\in \mathbb{F}$, then

(a) $(S+T)^*=S^*+T^*$  
(b) $(ST)^*=T^* S^*$  
(c) $(\lambda T)^*=\bar{\lambda}S^*$  
(d) $I^*=I$  
(e) $(T^*)^*=T$  
(f) If $T$ is invertible, then $(T^*)^{-1}=(T^{-1})^*$

Proof:

(d) $\langle (ST)v,u \rangle=\langle S(Tv),u \rangle=\langle Tv,S^*u \rangle=\langle v,T^*S^*u \rangle$

#### Theorem 7.6

Suppose $T\in\mathscr{L}(V,W)$, then

(a) $null\ T^*=(range\ T)^\perp$  
(b) $null\ T=(range\ T^*)^\perp$  
(c) $range\ T^*=(null\ T)^\perp$  
(d) $range\ T=(null\ T^*)^\perp$  

Proof:

$(a)\iff (c)$ since we can use **Theorem 7.5** (c) while replacing $T$ with $T^*$. Same idea give $(b)\iff (d)$. Also $(a)\iff (d)$ Since $(V^\perp)^\perp=V$

Now we prove (a). Suppose $w\in null\ T^*\iff T^*w=0$

$T^*w=0\iff \langle v,T^* w\rangle=0\forall v\in V\iff \langle Tv,w\rangle =0\forall v\in V\iff w\in (range\ T)^\perp$

#### Definition 7.7

The **conjugate transpose** of a $m\times n$ matrix $A$ is the $n\times m$ matrix denoted $A^*$ given the conjugate of the transpose.

ie. $(A^*)_{j,k}=A_{j,k}$

#### Theorem 7.9

Let $T\in \mathscr{L}(V,W)$ and $e_1,..,e_n$ an orthonormal basis of $V$, $f_1,...,f_m$ be an orthonormal basis of $W$. Then $M(T^*,(f_1,...,f_m),(e_1,..,e_n))=M(T,(f_1,...,f_m),(e_1,..,e_n))^*$

Proof:

The k-th column of $T$ is given by writing $Te_k$. in the basis $f_1,...,f_m$. ie. the $k,j$ entry of $M(T)$ is $\langle Te_k,f_j \rangle$, but then the $j,k$ entry of $M(T^*)$ is $\langle T^*f,e_k \rangle$. But $\langle Te_k,f_j\rangle=\langle e_k,T^*f_j\rangle=\overline{\langle T^*f_j,e_k\rangle}$

Example:

Suppose $T(x_1,x_2,x_3)=(x_2+3x_3,2x_1)$

$$
\begin{aligned}
    \langle T(x_1,x_2,x_3),(y_1,y_2)\rangle&=\langle (x_2+3x_3,2x_1),(y_1,y_2)\rangle\\
    &=(x_2+3x_3,2x_1)\bar{y_1},(x_2+3x_3,2x_1)\bar{y_2}\\
    &=\bar{y_1}x_2+3\bar{y_1}x_3+2\bar{y_2}x_1\\
    &=\langle (x_1,x_2,x_3),(2y_2,y_1,3y_1)\rangle
\end{aligned}
$$

So $T^*(y_1,y_2)=(2y_2,y_1,3y_1)$

Idea: Reisz Representation gives a function from $V$ to $V'$ (3.118) tells us given $T\in \mathscr{L}(V,W)$, we have