CHAPTER 9.
VECTOR SPACE
Professor : Jorge M. Seminario
Date – Nov. 14th, 2011
Group 3 - Pratik Darvekar
Brian Zachary Harding
Min-Chi Hsieh
9.1 INTRODUCTION

A vector is a quantity which has both a magnitude and
a direction. Vectors arise naturally as physical
quantities.

Examples of vectors are displacement, velocity,
acceleration, force and electric field. Some physical
quantities cannot be added in the simple way
described for scalars.
[1]http://www.physchem.co.za/index.htm
9.1 INTRODUCTION
ex: Walk 4m in a northerly direction and then 3m in an
easterly direction, how far would you be from your starting
point?
 The answer is clearly NOT 7 m!
One could calculate the distance using the theorem of
Pythagoras, 𝑑 =
3𝑚 2 + 4𝑚 2 = 5𝑚
 You could have reached the same
final position by walking 5 m in
the direction 36.9° east of north.
 These quantities are called
displacements. Displacement is an
example of a vector quantity.

[1]http://www.physchem.co.za/index.htm
9.2 VECTORS;
GEOMETRICAL REPRESENTATION
9.2.1 Algebra of Vectors
F+G=G+F
 (F + G) + H = F + (G + H)
F+0=F
 α(F + G) = αF + αG
 (αβ)F = α(βF)
 (α + β)F = αF + βF

G
F
F
G
[1]Peter V. O’Neil, “Vectors and Vector Spaces” in Advanced
Engineering Mathematics 5th edition. Birmingham, AL: B. Stenquist,
2003, ch. 5.1, pp. 204-208.
9.2 VECTORS;
GEOMETRICAL REPRESENTATION
9.2.2 Norm of a vector

The norm, or magnitude, of a vector (a, b, c) is the
number ∥ 𝑎, 𝑏, 𝑐 ∥ defined by
Z
∥ (𝑎, 𝑏, 𝑐 ) ∥ =
𝑎2 + 𝑏 2 + 𝑐 2
18
ex : 𝑂𝑃 = (-1, 4, 1)
‖ 𝑂𝑃 ‖ =
−1
2
+ 4
2
+ 1
2
= 18
P
(-1, 4 ,1)
O
X
[1]Peter V. O’Neil, “Vectors and Vector Spaces” in Advanced
Engineering Mathematics 5th edition. Birmingham, AL: B. Stenquist,
2003, ch. 5.1, pp. 201-203.
Y
9.3 INTRODUCTION
PRODUCT
OF
ANGLE
AND
DOT
9.3.1 Dot Product

Let F = a1i + b1j + c1k , G = a2i + b2j + c2k
F· G = a1a2 + b1b2 + c1c2 = ∥ F ∥ ∥ G ∥ cosθ
[1]Peter V. O’Neil, “Vectors and Vector Spaces” in Advanced
Engineering Mathematics 5th edition. Birmingham, AL: B. Stenquist,
2003, ch. 5.2, pp. 209.
9.3 INTRODUCTION
PRODUCT
OF
ANGLE
AND
DOT
9.3.2 Properties of Dot product
F· G = G· F
(F + G)· H = F· H + G· H
α(F· G ) = (αF)· G = F· (αG)
F· F = ∥ F ∥ 2
F· F = 0 if and only if F = 0
‖𝛼𝐹 + 𝛽𝐺‖2 = 𝛼 2 ‖𝐹‖2 + 2𝛼𝛽𝐹 ⋅ 𝐺 + 𝛽2 ‖𝐺‖2
[1]Peter V. O’Neil, “Vectors and Vector Spaces” in Advanced
Engineering Mathematics 5th edition. Birmingham, AL: B. Stenquist,
2003, ch. 5.2, pp. 210-211.
9.3 INTRODUCTION
PRODUCT
OF
ANGLE
AND
DOT
9.3.3 Orthogonal Vectors

Vector F and G are orthogonal if and only if
F· G = 0
ex: F = -4i + j + 2k, G = 2i + 4k, H = 6i – j – 2k
F· G = (-4)(2) + (1)(0) + (2)(4) = 0
F· H = (-4)(6) + (1)(-1) + (2)(-2) = -29 ≠ 0
G· H = (2)(6) + (0)(-1) + (4)(-2) = 4 ≠ 0
only F & G are orthogonal, others are not!
[1]Peter V. O’Neil, “Vectors and Vector Spaces” in Advanced
Engineering Mathematics 5th edition. Birmingham, AL: B. Stenquist,
2003, ch. 5.2, pp. 213-215.
9.3 INTRODUCTION
PRODUCT
OF
ANGLE
AND
DOT
Application for finding plant equation
ex: Suppose we want the equation of a plant Π containing the
point (-6, 1, 1) and perpendicular to the vector N = -2i + 4j + k
First we suggest a point (x, y, z) is on Π, so the vector from (-6, 1, 1) to (x,
Z
y, z) must be orthogonal to N. this means that
(-6, 1, 1)
((x+6)i + (y-1)j + (z-1)k)· N = 0
((x+6)i + (y-1)j + (z-1)k)· (-2i + 4j + k) = 0
(x, y, z)
N
-2(x+6) + 4(y-1) + (z-1) = 0
-2x + 4y + z = 17
X
[1]Peter V. O’Neil, “Vectors and Vector Spaces” in Advanced
Engineering Mathematics 5th edition. Birmingham, AL: B. Stenquist,
2003, ch. 5.2, pp. 213-215.
Y
9.4&9.6 N-SPACE& GENERALIZED
VECTOR SPACE

If n is a positive integer, an n-vector is an n-tuple (x1,
x2,…, xn), with each coordinate xj a real number.
The set of all n-vector is denoted Rn.
9.4.1 Algebra of Vectors - same as before
F+G=G+F
 (F + G) + H = F + (G + H)
F+0=F
 α(F + G) = αF + αG
 (αβ)F = α(βF)
 (α + β)F = αF + βF

[1]Peter V. O’Neil, “Vectors and Vector Spaces” in Advanced
Engineering Mathematics 5th edition. Birmingham, AL: B. Stenquist,
2003, ch. 5.4, pp. 222-223.
9.5 DOT PRODUCE, NORM
FOR N-SPACE
AND
ANGLE
9.5.1 Definition - Norm
The Norm of F = (x1, x2,…,xn)
∥ (x1, x2,…,xn ) ∥ =
x12 + x22 + ⋯ + xn2
9.5.2 Definition - Dot Product
We let F = (x1, x2,…,xn), G = (y1, y2,…,yn), we can get
F· G = x1y1 + x2y2 + … + xnyn
[1]Peter V. O’Neil, “Vectors and Vector Spaces” in Advanced
Engineering Mathematics 5th edition. Birmingham, AL: B. Stenquist,
2003, ch. 5.4, pp. 222-223.
9.5 DOT PRODUCE, NORM
FOR N-SPACE
AND
ANGLE
9.5.3 Properties of Dot product - same as
before
F· G = G· F
(F + G)· H = F· H + G· H
α(F· G ) = (αF)· G = F· (αG)
F· F = ∥ F ∥ 2
F· F = 0 if and only if F = 0
‖𝛼𝐹 + 𝛽𝐺‖2 = 𝛼 2 ‖𝐹‖2 + 2𝛼𝛽𝐹 ⋅ 𝐺 + 𝛽2 ‖𝐺‖2
[1]Peter V. O’Neil, “Vectors and Vector Spaces” in Advanced
Engineering Mathematics 5th edition. Birmingham, AL: B. Stenquist,
2003, ch. 5.4, pp. 222-223.
9.5 DOT PRODUCE, NORM
FOR N-SPACE
AND
ANGLE
9.5.4 Cauchy-Schwarz Inequality
Let F and G be vectors, then |𝐹 ⋅ 𝐺| ≤ ‖𝐹‖‖𝐺‖
Proof: from dot product we know cosθ =
𝐹⋅𝐺
‖𝐹‖‖𝐺‖
since −1 ≤ cos𝜃 ≤ 1
so
−1 ≤
𝐹⋅𝐺
‖𝐹‖‖𝐺‖
≤ 1
then −‖𝐹‖‖𝐺‖ ≤ 𝐹 ⋅ 𝐺 ≤ ‖𝐹‖‖𝐺‖
equivalent to the Cauchy-Schwarz inequality!
[1]Peter V. O’Neil, “Vectors and Vector Spaces” in Advanced
Engineering Mathematics 5th edition. Birmingham, AL: B. Stenquist,
2003, ch. 5.4, pp. 224.
9.7 SPAN
AND
SUBSPACE
9.7.1 Definition
If you have a vector or a group of vectors, the set of
all linear combinations of these vectors is called the
span.
Symbolically this is written as:
𝑢 = 𝛼1 𝑢1 + ⋯ + 𝛼𝑘 𝑢𝑘
where α1,… αk are scalars
[1]Greenberg, Michael D. "Vector Space: Span and Subspace."
Advanced Engineering Mathematics 2nd Edition. Upper Saddle River,
NJ: Prentice-Hall, 1998. ch. 9.7 pp. 439-441. Print.
9.7 SPAN
AND
SUBSPACE
9.7.2 Single Vector

When this definition is applied to a single vector, the
span is a line extending infinitely from the vector

ex: The span of the vector u1=(4,7) is the set of all
scalar multiples of (4,7).
2u1=(8,14)
0u1=(0,0)
-5u1=(-20,-35)
[1]Greenberg, Michael D. "Vector Space: Span and Subspace."
Advanced Engineering Mathematics 2nd Edition. Upper Saddle River,
NJ: Prentice-Hall, 1998. ch. 9.7 pp. 439-441. Print.
9.7 SPAN
AND
SUBSPACE
9.7.3 Two Vectors

Similar to the span of a single vector, the span of
two vectors is all linear combinations of both
vectors.
ex: u1=(4,7), u2=(8,14).
The span of these two vectors is the same as the
span of just u1 because u2 is a scalar multiple of u1.
[1]Greenberg, Michael D. "Vector Space: Span and Subspace."
Advanced Engineering Mathematics 2nd Edition. Upper Saddle River,
NJ: Prentice-Hall, 1998. ch. 9.7 pp. 439-441. Print.
9.7 SPAN
AND
SUBSPACE
9.7.4 Subspace
Up to this point, all of the vectors have been in R2.
That is to say that the spans of all of the vectors
have been subspaces of R2.
 The governing definition of subspace is:
If a subset T of a vector space S is itself a vector
space, then T is a subspace of S.
 Using this definition a subspace is either all or part
of a vector space, meaning R2 is a subspace of R2.

[1]Greenberg, Michael D. "Vector Space: Span and Subspace."
Advanced Engineering Mathematics 2nd Edition. Upper Saddle River,
NJ: Prentice-Hall, 1998. ch. 9.7 pp. 439-441. Print.
9.7 SPAN
AND
SUBSPACE
ex : If the span of u1=(2,3) u2=(2,4) all or part of R2.
To determine the span of (u1, u2) let v=(v1, v2) be any
vector in R2 and try to express v=(α1u1+α2u2)
(v1, v2)=α1(2,3)+α2(2,4)
v1=2α1+2α2 v2=3α1+4α2
Now, use Gauss elimination:
α1= 2v1-v2
This system is consistent, meaning that you can
solve α1 for every v
9.7 SPAN
AND
SUBSPACE
ex : In R3 do a similar exercise u1 = (2, 1, 2), u2 = (-2,
1, 1) and v = (α1u1+α2u2)
Gauss elimination:
2α1-2α2=v1
α1+α2=v2 2α1+α2=v3
0=v3-(1/4)v1-(3/2) v2
This means that the span is constrained by this
equation. Any two can be chosen arbitrarily and the
last is set. The solution is a plane and is not all of R3,
rather it is a subspace
9.8: LINEAR DEPENDENCE
9.8.1 Governing definition

A finite set of vectors is LI if and only if there
exist scalar multipliers such that
0=α1u1+α2u2…αjuj
is only true if all scalar multipliers are zero
Independent
Dependent
[1]Greenberg, Michael D. "Vector Space: Span and Subspace."
Advanced Engineering Mathematics 2nd Edition. Upper Saddle River,
NJ: Prentice-Hall, 1998. ch. 9.7 pp. 444-446. Print.
9.8: LINEAR DEPENDENCE
ex : Consider the 2-tuples, u1=(2, 1), u2=(-2, 1)
Now, 2α1-2α2=0 α1+α2=0
Gauss elimination gives:
α1-α2=0
α1=0
This gives only the trivial solution α1=α2=0
This fits the governing definition so both of these
are LI
9.8: LINEAR DEPENDENCE
9.8.2 Some important theorems
concerning linear independence



If a set of 2 vector is LD, then one vector must be
a scalar multiple of the other
Every finite set of orthogonal vectors is LI
A set containing the zero vector is LD
[1]Greenberg, Michael D. "Vector Space: Span and Subspace."
Advanced Engineering Mathematics 2nd Edition. Upper Saddle River,
NJ: Prentice-Hall, 1998. ch. 9.7 pp. 446. Print.
9.8: LINEAR DEPENDENCE
ex: u1=(2, 1, 2), u2=(1, -1, 0) and u3=(-2, -1, 1)
2α1+α2-2α3=0
α1-α2-α3=0
2α1+α3=0
Gauss elimination gives:
2α1+α2-2α3=0 -3α2=0 3α3=0
All α are equal to 0,
so the set is LI
Independent Dependent
9.9 BASES, EXPANSIONS, DIMENSION
9.9.1 Definition : Basis
The finite set of vectors {e1,e2,…,ek } in a vector space
S is a basis for S if each vector u in S can be can be
expressed uniquely in the formu=∝1e1+…+∝ 𝑘 ek=
𝑘
𝑗=1
∝ 𝑗𝑒𝑗
[1]Greenberg, Michael D. "Vector Space: Span and Subspace."
Advanced Engineering Mathematics 2nd Edition. Upper Saddle River,
NJ: Prentice-Hall, 1998. ch. 9.9 pp. 470-484. Print.
9.9 BASES, EXPANSIONS, DIMENSION
9.9.2 Test for basis
The finite set {e1,e2,…,ek } in a vector space S is a
basis for S if and only if it spans S and is Linearly
Independent.
[1]Greenberg, Michael D. "Vector Space: Span and Subspace."
Advanced Engineering Mathematics 2nd Edition. Upper Saddle River,
NJ: Prentice-Hall, 1998. ch. 9.9 pp. 470-484. Print.
9.9 BASES, EXPANSIONS, DIMENSION
ex : Show whether the vectors (3, 2) and (-1, -5) is a basis for
R2
(3, 2) and (-1, -5) is a basis for R2 if
α1 (3, 2)+α2 (-1, -5) = (u1,u2) admits a unique solution for α1,
α2 for any values of u1,u2.
3 α1 – α2= u1
2 α1 - 5α2= u2, solving we get,α1 = 15u139−3u2
α = 2u1−3u2
2
13
Thus we get unique α1 , α2 for every u1,u2. So it’s a basis for R2
9.9 BASES, EXPANSIONS, DIMENSION
9.9.3 Definition: dimension
If the greatest number of LI vectors that can be found
in a vector space S is k, where 1≤k≤∞, then S is kdimensional.
dim S = k

If a vector space S admits a basis consisting of k
vectors, then S is k-dimensional.
[1]Greenberg, Michael D. "Vector Space: Span and Subspace."
Advanced Engineering Mathematics 2nd Edition. Upper Saddle River,
NJ: Prentice-Hall, 1998. ch. 9.9 pp. 470-484. Print.
9.9 BASES, EXPANSIONS, DIMENSION
9.9.4 Orthogonal bases

For non-orthogonal bases, the expansion process of a
given vector can be quite laborious. For example, if we
seek to expand a given vector u in R8(8 dimensional
space), then there will be 8 base vectors(𝑒𝑗) and 8
expansion coefficients(∝ 𝑗), and these will be found by
solving a system of 8 equations in 8 unknowns(∝ 𝑗).
[1]Greenberg, Michael D. "Vector Space: Span and Subspace."
Advanced Engineering Mathematics 2nd Edition. Upper Saddle River,
NJ: Prentice-Hall, 1998. ch. 9.9 pp. 470-484. Print.
9.9 BASES, EXPANSIONS, DIMENSION
9.9.4 Orthogonal bases
 Hence orthogonal bases are preferred, in whichei.ej = 0 if i≠j.
if {e1,e2,…,ek } are orthogonal bases, the expansion
of any vector u is simplyu = e𝐮..ee1 e1+…+ e𝐮..ee𝑘 ek .
1 1
k 𝑘
[1]Greenberg, Michael D. "Vector Space: Span and Subspace."
Advanced Engineering Mathematics 2nd Edition. Upper Saddle River,
NJ: Prentice-Hall, 1998. ch. 9.9 pp. 470-484. Print.
9.9 BASES, EXPANSIONS, DIMENSION
ex : Expand u = (4, 3, -3, 6) in terms of the orthogonal
base vectors e1=(1, 0, 2, 0), e2=(0, 1, 0, 0), e3=(-2, 0,
1, -1) and e4=(-2, 0, 1, -1) of R4.
Compute 𝐮.e1= -2, e1.e1= 5, and so on,
u=
u=
𝐮.e1
e1.e1
−𝟐
𝟓
e1+…+
e1 + 3 e2 +
𝐮.e4
e4.e4
𝟏𝟗
𝟑𝟎
e3 +
e4 ,
−𝟏𝟕
𝟔
e4
9.10 BEST

APPROXIMATION
If the vectors {e1,e2,…,en } are orthonormal, but
fall short of being a basis for S (i.e., N < dim S),
and we still wish to expand a vector u in S which
does not fall within the span {e1,e2,…,en }, in such
a case the best approximation of u in terms of
.
{e1,e2,…,en } is given by- u = 𝑁
𝑗=1 u ej ej
[1]Greenberg, Michael D. "Vector Space: Span and Subspace."
Advanced Engineering Mathematics 2nd Edition. Upper Saddle River,
NJ: Prentice-Hall, 1998. ch. 9.10 pp. 470-484. Print.
9.10 BEST
APPROXIMATION
ex : Let S be R2, N=1, e1=
1
13
(12,5), and u= (1,1).
Find the best approximation u≈c1e1.
The best approximation of u in terms of {e1,e2,…,en }
is given by
𝑵
.
𝒋=𝟏 𝐮 ej ej
12 5
17
𝐮. e1= + =
13 13
13
u=
Hence u =
17
13
e1