# MATB24H3 Lecture Notes - Unit Vector, Dot Product, Row And Column Spaces

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School
UTSC
Department
Mathematics
Course
MATB24H3 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
Wof Rnwith the property that any two vectors in this basis are orthogonal
and unit vectors.
DEFINITION: Let {v1,v2, ...vk}be a set of nonzero vectors in Rn. We
say that this set is orthogonal if vi·vj= 0 for all i, j ∈ {1,2, ...k}such that
i6=j
THEOREM: Let {v1,v2, ...vk}be an orthogonal set of vectors in RnThen
this set is independent.
proof. : Suppose that
0=r1v1+r2v2+... +rkvk
then vi·0=vi·(r1v1+r2v2+... +rkvk) Thus,
0 = r1(vi·v1) + r2(vi·v2) + ... +ri(vi·vi) + ... +rk(vi·vk)
or 0 = ri||vi||2Since vi6=0then ||vi|| 6= 0 therefore ri= 0.This is true for
all i, therefore r1=r2=... =rk= 0 and the vectors v1,v2, ...vkare linearly
independent.
If we have an orthogonal basis for a subspace Wof Rn, then it is easy to
ﬁnd bWthe projection of bon Wfor any vector bRnusing the following
theorem:
1
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Unlock all 10 pages and 3 million more documents. THEOREM: Projection Using an Orthogonal Basis. If {v1,v2, ...vk}be
an orthogonal basis for Win Rn, then for any bRnthe projection of b
on Wis
bW=b·v1
v1·v1
v1+b·v2
v2·v2
v2+... +b·vk
vk·vk
vk
Note that this is the sum
k
X
1
b·vi
vi·vi
vi=
k
X
1
bvi
2
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Unlock all 10 pages and 3 million more documents. EXAMPLE: From our previous example: Let b= [22,11,11,33] R4and
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]
Wand the two vectors are linearly independent. Also, they are orthogonal
since [1,0,2,1] ·[1,6,2,5] = 1 + 4 5 = 0
Thus
bW=[22,11,11,33] ·[1,0,2,1]
[1,0,2,1] ·[1,0,2,1] [1,0,2,1] + [22,11,11,33] ·[1,6,2,5]
[1,6,2,5] ·[1,6,2,6]
This gives
bW=11
6[1,0,2,1] + 25
6[1,6,2,5] = [6,25,12,19]
as expected.
In this example, I did not explain how I found the vector [1,6,2,5]. In
general, given any basis for a subspace, Wof Rnwe want to be able to ﬁnd
a basis for Wwhich 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 Wbe a subspace of Rn. A basis {q1,q2, ...qk}for W
is orthonormal if it is orthogonal and each vector qiis a unit vector for all
i= 1,2..., k.
Thus, the theorem in the previous page now gives the following:
THEOREM: The projection of bon Wwith orthonormal basis {q1,q2, ...qk}
is
bW= (b·q1)q1+ (b·q2)q2+... + (b·qk)qk=
k
X
1
bqi
3
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## Document Summary

In this section we will develop a process that will give us a basis of a subspace. W of rn with the property that any two vectors in this basis are orthogonal and unit vectors. Definition: let {v1, v2, vk} be a set of nonzero vectors in rn. We say that this set is orthogonal if vi vj = 0 for all i, j {1, 2, k} such that i 6= j. Theorem: let {v1, v2, vk} be an orthogonal set of vectors in rn then this set is independent. proof. 0 = r1v1 + r2v2 + + rkvk then vi 0 = vi (r1v1 + r2v2 + + rkvk) thus, 0 = r1(vi v1) + r2(vi v2) + + ri(vi vi) + + rk(vi vk) or 0 = ri||vi||2 since vi 6= 0 then ||vi|| 6= 0 therefore ri = 0.