Math 311 Lecture 20
Gram-Schmidt process orthonormalizaton
For easier reading, we again write uv instead of (u,v).
THEOREM. If {v1,v2, ..., vn} is an orthonormal basis, then
for any v,
v = c1v1+c2v2 + ... cnvn where ci = vvi .
PROOF. {v1,v2, ..., vn} orthonormal implies vivi = 1 (they are
unit vectors) and vivk= 0 if i k (they are orthogonal).
v = c1v1+ ...+civi + ... cnvn for some ci’s (vi’s are a basis).
vvi = (c1v1+c2v2 + ...+civi +...+ cnvn)vi =
= (c1v1vi)+(c2v2vi) + ... + (civivi) + ... +(cnvnvi)
= c1(v1vi)+ c2(v2vi) + ... + ci(vivi) + ... +cn(vnvi)
= c1(0)+ c2(0) + ... + ci(1) + ... +cn(0) = ci. E
Hence for an orthonormal basis S = {v1,v2, ..., vn}, the
coordinate vector [v]S is:
[v]S = [c1, c2, ..., cn]T = [vv1, vv2, ..., vvn]T.
If u and v are orthogonal i.e., (u,v) = 0, so are u and cv
since u(cv) = c(uv) = c(0) = 0. Multiplying by a
positive scalar does not change the direction or angle.
CGiven an orthogonal basis {v1,v2}={[½, ½, 0], [0, 0, -5]}
for a subspace W, find an simpler orthogonal basis.
Answer: {[1, 1, 0], [0, 0, -1]}. Note, don't factor out -1.
For any vector v, v/||v|| is the unit vector in the same
direction as v. We say v normalizes to v/||v||.
CGiven an orthogonal basis {v1, v2} = {[½,½,0], [0,0,-5]}
for a subspace W, find an orthonormal basis.
First simplify to {[1, 1, 0], [0, 0, -1]} as above.
Answer: {v1/||v1||, v2/||v2||} = {[ }.
1
2[1,1,0],[0,0,−1]
LEMMA. If v is a unit vector and w any other vector, then
w = u+x where
x = component of w in the same direction as v = (wv)v
u = component of w orthogonal to v = w(wv)v.
v
w = u+x
T
u
x
w
v a unit vector
u = w - x
x = (w v)v
.
cos T = (wv)/||w||||v|| = (wv)/||w|| since ||v|| = 1.
Using trigonometry, cos T = ||x||/||w||.
W||x|| = ||w||cos T = ||w||(wv)/||w|| = wv.
Since x points in the direction of v and has length wv,
x = (wv)v.
Subtracting x from w gives u which is 7 to v.
CFind the component of w = [2, 3, 4] which is
perpendicular to the unit vector v = [0, 0, 1].
x = (wv)v = 4v = [0, 0, 4] (component in the same direction)
u = wx = [2, 3, 4][0, 0, 4] = [2, 3, 0].
Answer: [2, 3, 0] is the component of w 7 to v.
GRAM-SCHMIDT ORTHONORMALIZATION PROCESS. Any basis
{w1, w2, ..., wn} for a subspace can be converted into
an orthonormal basis {u1, u2, ..., un} as follows:
Let u1= w1/||w1||. Then for i = 1, 2, 3, ...
Let vi+1=
wi+1(wi+1u1)u1(wi+1u2)u2(wi+1u3)u3 ... (wi+1ui)ui
Simplify vi+1 and let ui+1 = vi+!/||vi+1||.
COrthonormalize the following basis for R3:
{w1, w2, w3} = {[0,0,3], [2,2,2], [0,1,0]}.
First simplify to (scalar multiples have the same direction)
{w1, w2, w3} = {[0,0,1], [1,1,1], [0,1,0]}
u1 = w1/||w1|| = [0,0,1]/||[0,0,1]|| = [0,0,1]/1 = [0,0,1]
v2 = w2 (w2 u1)u1
= [1, 1, 1] ([1, 1, 1][0, 0, 1])[0, 0, 1]
= [1, 1, 1] (1)[0, 0, 1]
= [1, 1, 1] [0, 0, 1] = [1, 1, 0]
u2 = v2/||v2|| = [1, 1, 0]/||[1, 1, 0]|| = [1, 1, 0]
1
2
v3 = w3 (w3u1)u1 (w3u2)u2
= [0, 1, 0]([0, 1, 0][0, 0, 1])[0, 0, 1]
([0,1,0]⋅ 1
2[1,1,0]) 1
2[1,1,0]
= [0, 1, 0] 0[0, 0, 1] (½)(1)[1, 1, 0] =
= [0, 1, 0] [½, ½, 0] = [-½, ½, 0]
Simplify to v3 = [-1, 1, 0].
u3 = v3/||v3|| = [-1, 1, 0]/||[-1, 1, 0]|| = [-1, 1, 0].
1
2
Hence the orthonormal basis is
{[0, 0, 1], [1, 1, 0], [-1, 1, 0]}
1
2
1
2
Since this can be a long error-prone process, check the
final basis vectors for orthogonality. Check:
[0, 0, 1] [ 1, 1, 0] = 0
1
2
[0,0,1] [-1, 1, 0] = 0
1
2
[1,1,0] [-1, 1,0] = 0
1
2
1
2
Applying the process to a dependent set will produce
some zero vectors. Just delete them.
THEOREM. In any inner product space, if U = {u1, u2, ...,
un} is an orthonormal basis, then for any vectors v, w
(v, w) = [v]U[w]U. That is, the inner product is just the
usual dot product of the coordinate vectors with
respect to the basis U.
PROOF. Omitted.
Hw 18 Answers
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2
D
[3,-8,-1]T
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F
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.
x
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