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Sophie Chrysostomou
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Mathematics

MATB24H3

Sophie Chrysostomou

Winter

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University of Toronto at Scarborough
Department of Computer & Mathematical Sciences
MAT B24S Fall 2011
MAT B24
Lectures 15.
1 The Gram-Schmidt Process
In this section we will develop a process that will give us a basis of a subspace
W of R with the property that any two vectors in this basis are orthogonal
and unit vectors.
DEFINITION: Let {v ,v ,...v }1be 2 set kf nonzero vectors in R . We n
say that this set is orthogonal if v ▯ i = 0jfor all i,j ∈ {1,2,...k} such that
i ▯= j
THEOREM: Let {v ,v ,...v } be an orthogonal set of vectors in R Then n
1 2 k
this set is independent.
proof. : Suppose that
0 = r v + r v + ... + r v
1 1 2 2 k k
then v ▯i0 = v ▯ ir v 1 r1v + 2..2+ r v ) Thksk
0 = r 1v i v )1+ r (2 ▯ i ) 2 ... + r (vi▯ vi) +i... + r (v k vi) k
2
or 0 = r i|v i| Since v ▯=i0 then ||v || =i ▯ 0 therefore r =i0. This is true for
all i, therefore r = r = ... = r = 0 and the vectors v ,v ,...v are linearly
1 2 k 1 2 k
independent.
n
If we have an orthogonal basis for a subspace W of R , then it is easy to
n
ﬁnd b W the projection of b on W for any vector b ∈ R using the following
theorem:
1 THEOREM: Projection Using an Orthogonal Basis. If {v1,v2,...k } be
an orthogonal basis for W in R , then for any b ∈ R the projection of b
on W is
b ▯ v1 b ▯ 2 b ▯ vk
b W = v 1 v2+ ... + v k
v1▯ v1 v2▯ v2 vk▯ vk
▯k ▯k
b ▯ i
Note that this is the sum vi= b vi
1 vi▯ vi 1
2 EXAMPLE: From our previous example: Let b = [22,11,11,33] ∈ R and 4
W = sp([1,0,2,−1],[−1,1,−2,2]).
We cannot use the theorem above since {[1,0,2,−1],[−1,1,−2,2]} is not an
orthogonal set. But W = sp([1,0,2,−1],[1,6,2,5]) as well since [1,6,2,5] ∈
W and the two vectors are linearly independent. Also, they are orthogonal
since [1,0,2,−1] ▯ [1,6,2,5] = 1 + 4 − 5 = 0
Thus
[22,11,11,33] ▯ [1,0,2,−1] [22,11,11,33] ▯ [1,6,2,5]
b W = [1,0,2,−1] +
[1,0,2,−1] ▯ [1,0,2,−1] [1,6,2,5] ▯ [1,6,2,6]
This gives
b = 11[1,0,2,−1] + 25[1,6,2,5] = [6,25,12,19]
W 6 6
as expected.
In this example, I did not explain how I found the vector [1,6,2,5]. In
n
general, given any basis for a subspace, W of R we want to be able to ﬁnd
a basis for W which is orthogonal so that we can apply the above theorem.
Further more, we want each basis vector to be a unit vector. We call such a
set an orthonormal basis for W
DEFINITION: Let W be a subspace of R . A basis {q ,q ,..1q 2 forkW
is orthonormal if it is orthogonal and each vector i is a unit vector for all
i = 1,2...,k.
Thus, the theorem in the previous page now gives the following:
THEOREM: The projection of b on W with orthonormal basis {q ,q ,...q }
1 2 k
is
▯k
b = (b ▯ q )q + (b ▯ q )q + ... + (b ▯ q )q = b
W 1 1 2 2 k k qi
1
3 HOW TO FIND SUCH AN ORTHONORMAL BASIS OF W:
THEOREM: The Gram-Schmidt Theorem. Let W be a subspace of R n
with basis a 1a 2...,a k Let
W = sp(a ,a ,...,a )
j 1 2 j
for j = 1,2,...,k. Then there is an orthonormal basis {q ,q ,1..q2} fok W
such that W =jsp(q ,q 1..2q ) j
proof. To prove this theorem I will use the following two theorems done
earlier:
n n
THEOREM A : If W is a subspace of R and b ∈ R , then there are
⊥
unique vectors b W ∈ W and b W ⊥ ∈ W such that
b = b + b ⊥ or that b = b − b .
W W W⊥ W
n
THEOREM B: If {v ,v ,...v1} 2e an krthogonal basis for W in R , then
n
for any b ∈ R the projection of b on W is
b·v1 b·2 b·k
b W = b W = v1·1 v1+ v2·v2v2+ ... + vk·vkv k
We start oﬀ by ﬁnding an orthogonal basis {v ,v ,...v } of W. Then we
1 2 k
v i
can ﬁnd {q ,1 ,2..q }kby letting q =i for i = 1,2,...,k
||vi||
(i) We let v1= a a1d W = s1(v ) = s1(a ). 1
n
(ii) To get v2: By theorem A applied on the vector a ∈ R 2nd the
⊥
subspace W we1get unique vectors a2W1 ∈ W 1 and a2W ⊥ ∈ W 1 such
1
that: a2W ⊥ = a 2 a 2W 1.
1
a2▯ v1
Let v2= a 2 ⊥ . By theorem B we get that a 2W = v1. Thus:
W1 1 v1▯ v1
a ▯ v
v = a = a − a = a − 2 1 v
2 W 1 2 W 1 2 v ▯ v 1
1 1
Finally, v ∈ W ⊥ so v ▯ v = 0.
2 1 2 1
Since v1= a ,1then v ∈ 2p(a ,a 1 an2 {v ,v } 1s a2 orthogonal set by
construction. Now let W = sp(v ,v ) = sp(a ,a ).
2 1 2 1 2
n
(iii) To get v3: By theorem A applied on the vector a ∈ R a3d the
4 ⊥
subspace W we2get unique vectors a 3W2 ∈ W a2d a 3 ⊥ ∈ W 2 such that:
W2
a3 ⊥ = a 3 a 3W . Let v 3 a 3 ⊥
W 2 2 W 2
By theorem B , since v ,v is an orthogonal basis for W we get that
1 2 2
v 1 a 3 v2▯ a3
a3W 2= v1+ v2
v 1 v 1 v 2 v 2
▯ ▯
v1▯ a 3 v 2 a 3
v3= a 3W ⊥ = a 3 a 3W2 = a 3 v 1 v 2
2 v1▯ v 1 v2▯ v 2
⊥
Finally, v3∈ W 2 and since v 1v ∈2W then2 v ▯ v =30, v1▯ v = 0.3 2
Also, v ∈ sp(a ,v ,v ) = sp(v ,v ,v ). Now let W = sp(v ,v ,v ) =
3 3 1 2 1 2 3 3 1 2 3
sp(a 1a 2a )3 Then, {v ,v 1v }2is 3n orthogonal basis for W by const3uc-
tion.
(iv) Continuing the process, if we have W i−1 = sp(v ,v1,.2.,v i−1) =
sp(a 1a 2...,a i−1) such that {v ,1 ,2..,v i−1} is an orthogonal basis for W i−1,
constructed as above, then we can get v by litting
▯ ▯
v 1 a i v 2 a i vi−1 ▯ ai
v i a iW⊥ = a ia W i−1 = a i v 1 v 2 ... + vi−1
i−1 v1▯ v 1 v 2 v 2 v i−1v i−1
And W = ip(v ,...1 ) = ip(a ,a ,.1.,2 ) . i
Eventually, we get the orthogonal basis {v ,v ,.1.,v2} of Wksuch that
W = sp(v ,..v ) = sp(a ,...,a ).
i 1 i 1 i
vi
By letting q =i for i = 1,2,...,k we get an orthonormal set of vectors
||vi||
{q 1q 2...,q k such that W = sp(v ,...1 ) = kp(q ,q ,..1,q2) withkthe prop-
erty that W =isp(v ,..1v ) =isp(q ,q ,1..,2 ) i
EXAMPLE: Let {a 1 [1,1,1,0],a = [021,1,0],a = [0,0,130]} be a
n
basis for the subspace W of R . Find an orthogonal basis for

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