Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 21 September 2014

Lecture 15: Inner products and Gram-Schmidt

Let $V$ be a vector space over $ℂ\text{.}$ A positive definite Hermitian Form, or inner product, on $V$ is a function $$\begin{array}{ccc}V\times V& ⟶& ℂ\\ (v_1,v_2)& ⟼& ⟨v_1,v_2⟩\end{array}$$ such that

(a)	If $v_1,v_2\in V$ then $⟨v_1,v_2⟩=\overline{⟨v_2,v_1⟩},$
(b)	If $c_1,c_2\in ℂ$ and $v_1,v_2,v_3\in V$ then $$⟨c_1{v}_1+c_2{v}_2,v_3⟩=c_1⟨v_1,v_3⟩+c_2⟨v_2,v_3⟩\text{.}$$
(c)	If $v\in V$ then $⟨v,v⟩\in {ℝ}_{\ge 0}\text{.}$
(d)	If $v\in V$ and $⟨v,v⟩=0$ then $v=0\text{.}$

Let $v\in V\text{.}$ The length of $v$ is $$\Vert v\Vert =\sqrt{⟨v,v⟩}$$ in ${ℝ}_{\ge 0},$ so that $\Vert \hspace{0.17em}\Vert :V\to {ℝ}_{\ge 0}$ is given by $\Vert v\Vert =\sqrt{⟨v,v⟩}\text{.}$

Let $u,w\in V\text{.}$ The elements $u,w$ are orthogonal if $⟨u,w⟩=0\text{.}$

An orthonormal basis of $V$ is a basis $\{v_1,\dots ,v_k\}$ of $V$ such that $$⟨v_i,v_j⟩={\delta }_{ij},$$ where ${\delta }_{ij}=\{\begin{array}{cc}1,& \text{if}\hspace{0.17em}i=j,\\ 0,& \text{if}\hspace{0.17em}i\ne j\text{.}\end{array}$

Let $V$ be a vector space over $ℂ$ with a positive definite Hermitian form $$⟨,⟩:V\times V\to ⟶ℂ\text{.}$$ Let $B=\{b_1,b_2,\dots ,b_{}\}$ be a basis of $V\text{.}$ The matrix of $⟨,⟩$ with respect to $B$ is $$A=\left(⟨b_i,b_j⟩\right)\text{.}$$ If $v={\alpha }_1{b}_1+{\alpha }_2{b}_2+\cdots +{\alpha }_k{b}_k$ and $w={\gamma }_1{b}_1+{\gamma }_2{b}_2+\cdots +{\gamma }_k{b}_k$ then $$\begin{array}{ccc}{\displaystyle ⟨v,w⟩}& =& {\displaystyle ⟨{\alpha }_1{b}_1+\cdots +{\alpha }_k{b}_k,{\gamma }_1{b}_1+{\gamma }_2{b}_2+\cdots +{\gamma }_k{b}_k⟩}\\ & =& {\displaystyle {\alpha }_1\overline{{\gamma }_1}⟨b_1,b_1⟩+{\alpha }_1\overline{{\gamma }_2}⟨b_1,b_2⟩+\cdots +{\alpha }_1\overline{{\gamma }_k}⟨b_1,b_k⟩}\\ & & {\displaystyle +{\alpha }_2\overline{{\gamma }_1}⟨b_2,b_1⟩+\cdots }\\ & & {\displaystyle \phantom{+{\alpha }_2\overline{{\gamma }_1}⟨b_2,b_1⟩+}\cdots +{\alpha }_k\overline{{\gamma }_k}⟨b_k,b_k⟩}\\ & =& {\displaystyle \sum _{i,j=1}^k{\alpha }_i\overline{{\gamma }_j}⟨b_i,b_j⟩}\\ & =& {\displaystyle \sum _{i,j=1}^k{\alpha }_i⟨b_i,b_j⟩\overline{{\gamma }_j}}\\ & =& {\displaystyle ({\alpha }_1,{\alpha }_2,\dots ,{\alpha }_k)\left(\begin{array}{ccc}\hspace{0.17em}& \hspace{0.17em}& \hspace{0.17em}\\ \hspace{0.17em}& ⟨b_i,b_j⟩& \hspace{0.17em}\\ \hspace{0.17em}& \hspace{0.17em}& \hspace{0.17em}\end{array}\right)\left(\begin{array}{c}\overline{{\gamma }_1}\\ \overline{{\gamma }_2}\\ ⋮\\ \overline{{\gamma }_k}\end{array}\right)}\\ & =& {\displaystyle v^{t}A\bar{w}\text{.}}\end{array}$$ Note: Since $\overline{⟨b_i,b_j⟩}=⟨b_j,b_i⟩,$ $\overline{A_{ij}}=A_{ji}\text{.}$ So ${\bar{A}}^t=A\text{.}$

Creating orthonormal bases: Gram-Schmidt

Let $V$ be a vector space with basis $B=\{b_1,b_2,b_3\}$ and $⟨,⟩:V\times V\to ℂ$ having matrix $$\left(\begin{array}{ccc}1& 0& -2\\ 0& 2& 1\\ -2& 1& 3\end{array}\right)$$ with respect to $B\text{.}$ Then $⟨b_1,b_1⟩=1\text{.}$ Let $v_1=b_1\text{.}$ So $⟨v_1,v_1⟩=1\text{.}$ Then $⟨b_2,v_1⟩=⟨b_2,b_1⟩=0$ and $⟨b_2,b_2⟩=\text{.}$ Let $v_2=\frac{1}{\sqrt{2}}{b}_2$ so that $$\begin{array}{ccc}{\displaystyle ⟨v_2,v_1⟩}& =& {\displaystyle ⟨\frac{1}{\sqrt{2}}{b}_2,b_1⟩=\frac{1}{\sqrt{2}}⟨b_2,b_1⟩=0\text{and}}\\ {\displaystyle ⟨v_2,v_2⟩}& =& {\displaystyle ⟨\frac{1}{\sqrt{2}}{b}_2,\frac{1}{\sqrt{2}}{b}_2⟩=\frac{1}{2}⟨b_2,b_2⟩=\frac{1}{2}·2=1\text{.}}\end{array}$$ Now, $$\begin{array}{ccc}{\displaystyle ⟨b_3,v_1⟩}& =& {\displaystyle ⟨b_3,b_1⟩=-2\text{and}}\\ {\displaystyle ⟨b_3,v_2⟩}& =& {\displaystyle \frac{1}{\sqrt{2}}⟨b_3,b_2⟩=\frac{1}{\sqrt{2}}\text{.}}\end{array}$$ Let $b_3^{\prime }=b_3-(-2)v_1-\frac{1}{\sqrt{2}}{v}_2\text{.}$ Then $$\begin{array}{ccc}{\displaystyle ⟨b_3^{\prime },v_1⟩}& =& {\displaystyle ⟨b_3-(-2)v_1-\frac{1}{\sqrt{2}}{v}_2,v_1⟩=⟨b_3,v_1⟩+2⟨v_1,v_1⟩-0=-2+2=0,}\\ {\displaystyle ⟨b_3^{\prime },v_2⟩}& =& {\displaystyle ⟨b_3-(-2)v_1-\frac{1}{\sqrt{2}}{v}_2,v_2⟩=⟨b_3,v_2⟩+2·0-\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=0,}\\ {\displaystyle ⟨b_3^{\prime },b_3^{\prime }⟩}& =& {\displaystyle ⟨b_3^{\prime },b_3-(-2)v_1-\frac{1}{\sqrt{2}}{v}_2⟩=⟨b_3^{\prime },b_3⟩+0+0}\\ & =& {\displaystyle ⟨b_3-(-2)v_1-\frac{1}{\sqrt{2}}{v}_2,b_3⟩=3+2(-2)-\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}=-1-\frac{1}{2}=-\frac{3}{2}\text{.}}\end{array}$$ Let $$v_3=\frac{1}{\sqrt{-\frac{3}{2}}}{b}_3^{\prime }=\frac{\sqrt{2}}{\sqrt{3}i}{b}_3^{\prime }=-\frac{\sqrt{2}i}{\sqrt{3}}{b}_3^{\prime }=-\frac{\sqrt{2}}{3}i{b}_3-\frac{2\sqrt{2}}{\sqrt{3}}i{v}_1+\frac{i}{\sqrt{3}}{v}_2\text{.}$$ Then $$\begin{array}{ccc}{\displaystyle ⟨v_3,v_1⟩}& =& {\displaystyle ⟨-\frac{\sqrt{2}i}{\sqrt{3}}{b}_3^{\prime },v_1⟩=0,}\\ {\displaystyle ⟨v_3,v_2⟩}& =& {\displaystyle ⟨-\frac{\sqrt{2}i}{\sqrt{3}}{b}_3^{\prime },v_2⟩=0,\text{and}}\\ {\displaystyle ⟨v_3,v_3⟩}& =& {\displaystyle ⟨\frac{1}{\sqrt{-\frac{3}{2}}}{b}_3^{\prime },\frac{1}{\sqrt{-\frac{3}{2}}}{b}_3^{\prime }⟩=\frac{-1}{(-\frac{3}{2})}(-\frac{3}{2})=-1\text{.}}\end{array}$$

Let ${𝒫}_2\left(ℂ\right)=\{a_1+a_{2}x\hspace{0.17em}|\hspace{0.17em}{a}_1,a_2\in ℂ\}$ with $$⟨f,g⟩={\int }_0^{1}f\left(x\right)\overline{g\left(x\right)}dx\text{.}$$ Then $B=\{1,x\}$ is a basis of ${𝒫}_2\left(ℂ\right)\text{.}$ $$\begin{array}{ccc}{\displaystyle ⟨1,1⟩}& =& {\displaystyle {\int }_0^{1}dx=x{|}_0^1=1,}\\ {\displaystyle ⟨1,x⟩}& =& {\displaystyle {\int }_0^1\bar{x}dx={\int }_0^{1}xdx=\frac{x^2}{2}{|}_0^1=\frac{1}{2},}\\ {\displaystyle ⟨x,x⟩}& =& {\displaystyle {\int }_0^{1}x\bar{x}dx={\int }_0^1{x}^{2}dx=\frac{x^3}{3}{|}_0^1=\frac{1}{3}\text{.}}\end{array}$$ So the matrix of $⟨,⟩$ with respect to the basis $\{1,x\}$ is $$\left(\begin{array}{cc}1& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{3}\end{array}\right)\text{.}$$ Let $v_1=1\text{.}$ Then $⟨v_1,v_1⟩=⟨1,1⟩=1\text{.}$ $$⟨x,v_1⟩=⟨x,1⟩=\frac{1}{2}\text{.}$$ Let $b_2^{\prime }=x-⟨x,v_1⟩v_1=x-\frac{1}{2}{v}_1\text{.}$ Then $$\begin{array}{ccc}{\displaystyle ⟨b_2^{\prime },v_1⟩}& =& {\displaystyle ⟨x-\frac{1}{2}{v}_1,v_1⟩=⟨x,v_1⟩-\frac{1}{2}⟨v_1,v_1⟩=\frac{1}{2}-\frac{1}{2}=0\text{.}}\\ {\displaystyle ⟨b_2^{\prime },b_2^{\prime }⟩}& =& {\displaystyle ⟨b_2^{\prime },x-\frac{1}{2}{v}_1⟩=⟨b_2^{\prime },x⟩-0}\\ & =& {\displaystyle ⟨x-\frac{1}{2}{v}_1,x⟩=\frac{1}{3}-\frac{1}{2}·\frac{1}{2}=\frac{1}{12}\text{.}}\end{array}$$ Let $v_2=\frac{1}{1/\sqrt{12}}{b}_2^{\prime }=\sqrt{12}{b}_2^{\prime }=2\sqrt{3}(x-\frac{1}{2}{v}_1)=2\sqrt{3}x-\sqrt{3}\text{.}$ Then $$⟨v_2,v_2⟩=⟨\frac{1}{1/\sqrt{12}}{b}_2^{\prime },\frac{1}{\sqrt{12}}{b}_2^{\prime }⟩=\frac{1}{1/12}⟨b_2^{\prime },b_2^{\prime }⟩=\frac{1/12}{1/12}=1$$ and $$⟨v_2,v_1⟩=⟨\frac{1}{\sqrt{1/12}}{b}_2^{\prime },v_1⟩=\frac{1}{\sqrt{1/12}}·0=0\text{.}$$ So $\{v_1,v_2\}$ is an orthonormal basis.

Notes and References

These are a typed copy of Lecture 15 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on August 24, 2011.

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