4.0 KiB
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.