Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last updated: 21 September 2014

Lecture 15: Inner products and Gram-Schmidt

Let V be a vector space over . A positive definite Hermitian Form, or inner product, on V is a function V×V (v1,v2) v1,v2 such that

(a) If v1,v2V then v1,v2=v2,v1,
(b) If c1,c2 and v1,v2,v3V then c1v1+c2v2,v3= c1v1,v3+ c2v2,v3.
(c) If vV then v,v0.
(d) If vV and v,v=0 then v=0.

Let vV. The length of v is v=v,v in 0, so that :V0 is given by v=v,v.

Let u,wV. The elements u,w are orthogonal if u,w=0.

An orthonormal basis of V is a basis {v1,,vk} of V such that vi,vj =δij, where δij= { 1, ifi=j, 0, ifij.

Let V be a vector space over with a positive definite Hermitian form ,: V×V. Let B={b1,b2,,bmi>k} be a basis of V. The matrix of , with respect to B is A=(bi,bj). If v=α1b1+α2b2++αkbk and w=γ1b1+γ2b2++γkbk then v,w = α1b1++ αkbk, γ1b1+ γ2b2++ γkbk = α1γ1b1,b1+ α1γ2b1,b2++ α1γkb1,bk +α2γ1b2,b1+ +α2γ1b2,b1+ +αkγkbk,bk = i,j=1k αiγj bi,bj = i,j=1k αibi,bj γj = (α1,α2,,αk) ( bi,bj ) ( γ1 γ2 γk ) = vtAw. Note: Since bi,bj=bj,bi, Aij=Aji. So At=A.

Creating orthonormal bases: Gram-Schmidt

Let V be a vector space with basis B={b1,b2,b3} and ,:V×V having matrix ( 10-2 021 -213 ) with respect to B. Then b1,b1=1. Let v1=b1. So v1,v1=1. Then b2,v1=b2,b1=0 and b2,b2=. Let v2=12b2 so that v2,v1 = 12b2,b1= 12b2,b1=0and v2,v2 = 12b2,12b2= 12b2,b2= 12·2=1. Now, b3,v1 = b3,b1=-2and b3,v2 = 12b3,b2=12. Let b3=b3-(-2)v1-12v2. Then b3,v1 = b3-(-2)v1-12v2,v1= b3,v1+2v1,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 𝒫2()={a1+a2x|a1,a2} with f,g= 01f(x)g(x) dx. Then B={1,x} is a basis of 𝒫2(). 1,1 = 01dx=x |01=1, 1,x = 01xdx= 01xdx= x22|01=12, x,x = 01xxdx= 01x2dx= x33|01=13. So the matrix of , with respect to the basis {1,x} is ( 112 1213 ) . Let v1=1. Then v1,v1=1,1=1. x,v1= x,1=12. Let b2=x-x,v1v1=x-12v1. 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 v2=11/12b2=12b2=23(x-12v1)=23x-3. 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 {v1,v2} 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.

page history