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content/Math429/Math429_L27.md

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+# 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.