## Group Theory and Linear Algebra

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 $V×V ⟶ ℂ (v1,v2) ⟼ ⟨v1,v2⟩$ such that

 (a) If ${v}_{1},{v}_{2}\in V$ then $⟨{v}_{1},{v}_{2}⟩=\stackrel{‾}{⟨{v}_{2},{v}_{1}⟩},$ (b) If ${c}_{1},{c}_{2}\in ℂ$ and ${v}_{1},{v}_{2},{v}_{3}\in V$ then $⟨c1v1+c2v2,v3⟩= c1⟨v1,v3⟩+ c2⟨v2,v3⟩.$ (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 $‖v‖=⟨v,v⟩$ in ${ℝ}_{\ge 0},$ so that $‖ ‖:V\to {ℝ}_{\ge 0}$ is given by $‖v‖=\sqrt{⟨v,v⟩}\text{.}$

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 $\left\{{v}_{1},\dots ,{v}_{k}\right\}$ of $V$ such that $⟨vi,vj⟩ =δij,$ where ${\delta }_{ij}=\left\{\begin{array}{cc}1,& \text{if} i=j,\\ 0,& \text{if} i\ne j\text{.}\end{array}$

Let $V$ be a vector space over $ℂ$ with a positive definite Hermitian form $⟨,⟩: V×V→⟶ℂ.$ Let $B=\left\{{b}_{1},{b}_{2},\dots ,{b}_{}\right\}$ be a basis of $V\text{.}$ The matrix of $⟨,⟩$ with respect to $B$ is $A=(⟨bi,bj⟩).$ 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 $⟨v,w⟩ = ⟨ α1b1+⋯+ αkbk, γ1b1+ γ2b2+⋯+ γkbk ⟩ = α1γ1‾⟨b1,b1⟩+ α1γ2‾⟨b1,b2⟩+⋯+ α1γk‾⟨b1,bk⟩ +α2γ1‾⟨b2,b1⟩+⋯ +α2γ1‾⟨b2,b1⟩+⋯ +αkγk‾⟨bk,bk⟩ = ∑i,j=1k αiγj‾ ⟨bi,bj⟩ = ∑i,j=1k αi⟨bi,bj⟩ γj‾ = (α1,α2,…,αk) ( ⟨bi,bj⟩ ) ( γ1‾ γ2‾ ⋮ γk‾ ) = vtAw‾.$ Note: Since $\stackrel{‾}{⟨{b}_{i},{b}_{j}⟩}=⟨{b}_{j},{b}_{i}⟩,$ $\stackrel{‾}{{A}_{ij}}={A}_{ji}\text{.}$ So ${\stackrel{‾}{A}}^{t}=A\text{.}$

### Creating orthonormal bases: Gram-Schmidt

Let $V$ be a vector space with basis $B=\left\{{b}_{1},{b}_{2},{b}_{3}\right\}$ and $⟨,⟩:V×V\to ℂ$ having matrix $( 10-2 021 -213 )$ 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 $⟨v2,v1⟩ = ⟨12b2,b1⟩= 12⟨b2,b1⟩=0and ⟨v2,v2⟩ = ⟨12b2,12b2⟩= 12⟨b2,b2⟩= 12·2=1.$ Now, $⟨b3,v1⟩ = ⟨b3,b1⟩=-2and ⟨b3,v2⟩ = 12⟨b3,b2⟩=12.$ Let ${b}_{3}^{\prime }={b}_{3}-\left(-2\right){v}_{1}-\frac{1}{\sqrt{2}}{v}_{2}\text{.}$ Then $⟨b3′,v1⟩ = ⟨b3-(-2)v1-12v2,v1⟩= ⟨b3,v1⟩+2⟨v1,v1⟩-0 =-2+2=0, ⟨b3′,v2⟩ = ⟨b3-(-2)v1-12v2,v2⟩= ⟨b3,v2⟩+2·0- 12=12-12=0, ⟨b3′,b3′⟩ = ⟨b3′,b3-(-2)v1-12v2⟩= ⟨b3′,b3⟩+0+0 = ⟨b3-(-2)v1-12v2,b3⟩= 3+2(-2)-1212= -1-12=-32.$ Let $v3= 1-32b3′= 23ib3′= -2i3b3′= -23ib3-223iv1+i3v2.$ Then $⟨v3,v1⟩ = ⟨-2i3b3′,v1⟩=0, ⟨v3,v2⟩ = ⟨-2i3b3′,v2⟩=0,and ⟨v3,v3⟩ = ⟨1-32b3′,1-32b3′⟩= -1(-32)(-32)=-1.$

Let ${𝒫}_{2}\left(ℂ\right)=\left\{{a}_{1}+{a}_{2}x | {a}_{1},{a}_{2}\in ℂ\right\}$ with $⟨f,g⟩= ∫01f(x)g(x)‾ dx.$ Then $B=\left\{1,x\right\}$ is a basis of ${𝒫}_{2}\left(ℂ\right)\text{.}$ $⟨1,1⟩ = ∫01dx=x |01=1, ⟨1,x⟩ = ∫01x‾dx= ∫01xdx= x22|01=12, ⟨x,x⟩ = ∫01xx‾dx= ∫01x2dx= x33|01=13.$ So the matrix of $⟨,⟩$ with respect to the basis $\left\{1,x\right\}$ is $( 112 1213 ) .$ Let ${v}_{1}=1\text{.}$ Then $⟨{v}_{1},{v}_{1}⟩=⟨1,1⟩=1\text{.}$ $⟨x,v1⟩= ⟨x,1⟩=12.$ Let ${b}_{2}^{\prime }=x-⟨x,{v}_{1}⟩{v}_{1}=x-\frac{1}{2}{v}_{1}\text{.}$ Then $⟨b2′,v1⟩ = ⟨x-12v1,v1⟩= ⟨x,v1⟩-12 ⟨v1,v1⟩=12-12=0. ⟨b2′,b2′⟩ = ⟨b2′,x-12v1⟩= ⟨b2′,x⟩-0 = ⟨x-12v1,x⟩= 13-12·12=112.$ Let ${v}_{2}=\frac{1}{1/\sqrt{12}}{b}_{2}^{\prime }=\sqrt{12}{b}_{2}^{\prime }=2\sqrt{3}\left(x-\frac{1}{2}{v}_{1}\right)=2\sqrt{3}x-\sqrt{3}\text{.}$ Then $⟨v2,v2⟩= ⟨ 11/12b2′, 112b2′ ⟩ =11/12 ⟨b2′,b2′⟩ =1/121/12=1$ and $⟨v2,v1⟩= ⟨ 11/12b2′,v1 ⟩ =11/12·0=0.$ So $\left\{{v}_{1},{v}_{2}\right\}$ 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.