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

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Published on 21 Apr 2013

Department

Mathematics

Course

MATB24H3

Professor

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 b∈Rnusing the following

theorem:

1

THEOREM: Projection Using an Orthogonal Basis. If {v1,v2, ...vk}be

an orthogonal basis for Win Rn, then for any b∈Rnthe 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

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

## 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.