# Lecture 27

## Chapter VI Inner Product Spaces

### Orthonormal basis 6B

#### Theorem 6.32 Gram-Schmidt

Suppose $v_1,...,v_m$ is a linearly independent list. Let $f_k\in V$ by $f_1=v_1$, and $f_k=v_k-\sum_{j=1}^{k-1}\frac{\langle v_k,f_j\rangle }{||f_j||^2}f_j$. Then set $e_k=\frac{f_k}{||f_k||}$, then $e_1,...,e_m$ is orthonormal with $Span(e_1,...,e_k)=Span(v_1,...,v_k)$ for each $k=1,...,m$

Proof:  note is suffice to show that $f_1,...,f_m$ is orthogonal and that $Span(e_1,...,e_m)=Span(v_1,...,v_m)$ Induct on $m$.

When $m=1$: clear

When $m>1$: Suppose we know the result for values $< m$. Need to show that $\langle f_m,f_k\rangle =0$ for $k<m$.

$$
\begin{aligned}
\langle f_m, f_k \rangle &=\langle v_m, f_k \rangle-\sum_{j=1}^{k-1}\frac{\langle v_m,f_j\rangle }{||f_j||^2} \langle f_j, f_k \rangle\\
&=\langle v_m, f_k \rangle-\frac{\langle v_m,f_k \rangle }{||f_j||^2} \langle f_k, f_k \rangle\\
&=\langle v_m, f_k \rangle-\langle v_m,f_k \rangle\\
&=0
\end{aligned}
$$

Then we want to test if $Span(f_1,...,f_m)=Span(v_1,...,v_m)$, given that $Span(f_1,...,f_{m-1})=Span(v_1,...,v_{m-1})$ (by induction)

Since $f_m=v_m-\sum_{j=1}^{m-1}\frac{\langle v_m,f_j\rangle }{||f_j||^2}f_j$, and $\sum_{j=1}^{m-1}\frac{\langle v_m,f_j\rangle }{||f_j||^2}f_j \in Span(v_1,...,v_{m-1})$, then $f_m\in Span(v_1,...,v_m)$

Since $v_m=f_m-\sum_{j=1}^{m-1}\frac{\langle v_m,f_j\rangle }{||f_j||^2}f_j \in Span(f_1,...,f_m)$, then $v_m\in Span(f_1,...,f_m)$

Example: Find an orthonormal basis for $\mathscr{P}_2(\mathbb{R})$ with $\langle p,q \rangle=\int^1_{-1}pq$.

Start with $1,x,x^2$, apply Gram-Schimidt procedure.
  
$f_1=1$,

$f_2=x-\frac{\langle x,1 \rangle}{||1||^2}1=x-\frac{0}{2}\cdot 1 =x$,

$f_3=x^3-\frac{\langle x^2,1 \rangle}{||1||^2}1-\frac{\langle x^2,x \rangle}{||1||^2}x=x^2-\frac{2/3}{2}1=x^2-\frac{1}{3}$

Convert it to orthonormal basis we have $\sqrt{\frac{1}{2}}, \sqrt{\frac{3}{2}}x,\sqrt{\frac{45}{8}}(x^2-\frac{1}{3})$
 
#### Theorem 6.35

Every finite dimensional inner product space has an orthonormal basis

Proof:

take any basis and apply Gram-Schmidt procedure.

#### Theorem 6.36

Every orthonormal list extends to an orthonormal basis.

Proof:

extend the basis and apply Gram-Schmidt procedure.

#### Theorem 6.37

$V$ be a finite dimensional $T\in \mathscr{L}(V)$. Then $T$ has an upper triangular matrix with respect to an orthonormal basis $\iff$ if the minimal polynomial is of the form $(z-\lambda_1)\dots (z-\lambda_m)$

Proof:

The critical step is $T$ upper triangular with respect to $v_1,...,v_n\iff Tv_k\in Span(v_1,...,v_k)$

IMportantly, if $e_1,...,e_n$ is the result of Gram-Schmidt, then the $Span(v_1,...,v_k)=Span(e_1,...,e_k)$ for all $k$.

$Tv_k\in Span(e_1,...,e_k)$ using the same work $Te_k\in Span(e_1,...,e_k)$

#### Corollary 6.37 (Schur's Theorem)

If $V$ is finite dimensional complex vector space and $T\in \mathscr{L}(V)$, then there exists an orthonormal basis where $T$ is upper triangular.

### Linear Functionals on Inner Product Spaces

Example: $\varphi\in (\mathbb{R}^3)'=\mathscr{L}(\mathbb{R}^3,\mathbb{R})$ given by $\varphi(x,y,z)=2x+3y-z$. note $\varphi(V)=\langle v,(2,3,-1)\rangle$ where $\langle,\rangle$ is the Euclidean inner product.

#### Theorem 6.42 (Riesz Representation Theorem)

Suppose that $\varphi\in V'=\mathscr{L}(V,\mathbb{F})$ on an inner product space $V$. Then there exists an unique vector $v\in V$ such that $\varphi(u)=\langle u, v\rangle$

Proof:

Fix an orthonormal basis $e_1,...,e_n$, 

$\varphi(u)=\varphi(\langle u,e_1 \rangle e_1+...\langle u,e_n \rangle e_n)$

Use linearity 

$\varphi(u)=\langle u,e_1 \rangle\varphi( e_1)+...\langle u,e_n \rangle \varphi(e_n)$

Use the conjugates

$\varphi(u)=\langle u,\overline{\varphi( e_1)} e_1 \rangle+...\langle u,\overline{\varphi( e_n)} e_n \rangle=\langle u,\overline{\varphi( e_1)} e_1 +...\overline{\varphi( e_n)} e_n \rangle$

Set $v=\overline{\varphi( e_1)} e_1 +...\overline{\varphi( e_n)} e_n$, thus $v$ exists.

uniqueness $\langle v_1-v_2,v_1-v_2 \rangle=0$ for any $v_1,v_2$ satifsying the conditions.