1.2 Vector Operations
1.2.1. Addition and subtraction
Numeric vectors. One adds and subtracts numeric vectors by adding and subtracting
corresponding components. In order to do this they have to have the same number of
components.
v1
w1
v1+w1
v1-w1
v
w
v
+w
v =  2 w =  2  v + w =  2 2
v - w = v2-w2.
v3
w3
v3+w3
v3-w3
For example
 2
 1
 2 + 1   3
 2 - 1   1
v = -4
w =  2
 v + w =  -4 + 2  = -2
v - w =  -4 - 2  = -6.
 3
-3
 3 + (-3)  0
 3 - (-3)  6
th
th
To summarize, the i component of (v+w) is the sum of the i component of v and the ith
component of w. This can be expressed symbolically by the formula (v+w)i = vi + wi for
i = 1, …, n where n is the number of components in v and w. Similarly, (v-w)i = vi - wi
for i = 1, …, n.
 0
A zero vector is a vector all of whose components are 0, e.g. v =  0.
 0
One forms the negative of a vector v by negating the components of v.
v1
- v1
v = v2  - v = - v2
v3
- v3
For example
 2
-2
v = -4  - v =  4 
 3
-3
So, (-v)i = -(vi) for i = 1, …, n.
Example 1.2.1. An electronics company makes two types of circuit boards for computers,
namely ethernet cards and sound cards. Last week the production was
20
20 ethernet cards
y =  30  =  30 sound cards 
This week the production is scheduled to be
25
25 ethernet cards
t =  35  =  35 sound cards 
The total production for the two week period is
20
25
55
55 ethernet cards
y + t =  30  +  35  =  65  =  65 sound cards 
The net change in production from last week to this week is
25
20
5
5 ethernet cards
t - y =  35  +  30  =  5  =  5 sound cards 
Geometric / physical vectors. We focus on the situation where the vectors are directed
line segments, since other physical quantities that are vectors can be represented by
directed line segments. We also assume that directed line segmens with the same length
and direction represent the same vector.
1.2 - 1

y
First consider addition. Suppose u = PQ

u + v = PT
and v = RS are directed line segments.
The sum u + v is defined as follows.

v = QT
W
Consider the directed line segments PW
S
u - v = QW

and QT starting at P and Q respectively
and having the same length and direction

T
v = PW
Q
v = RS

as RS. So v = PW = QT also. Then
PQTW is a parallelogram and the diagonal
P
u = PQ
R

x
PT of the parallelogram is the sum of u


and v, i.e. u + v = PT . Putting it another way, u + v = PT is the third side of the triangle,


two of whose sides are u = PQ and v = QT It is not hard to show that this geometric
definition of addition corresponds to the definition of addition of the numeric vectors
corresponding to u and v.
Subtraction of geometric vectors is also simple. In terms of the notation in the previous

paragraph, v – u = QU is the directed line segment from the tip of u to the tip of v. The



reason is because u + (v – u) = PQ + QU = PU = v.
For directed line segments, the zero vector is just a directed line segment which ends

where it starts, i.e. 0 = PP .
y
F1 + F2
If one has two forces F1 and F2 acting on an
object, the two force can be replaced by their
sum F1 + F2.
Functions. If f(x) and g(x) are two functions
then their sum and difference are defined in
the usual way, i.e.
(f + g)(x) = f(x) + g(x)
(f - g)(x) = f(x) - g(x)
F2
F1
For example, if f(x) = 4x2 – 2x + 3 and
g(x) = 2x2 + x - 5 then (f + g)(x) = 6x2 – x - 2 and (f - g)(x) = 2x2 – 3x + 8.
1.2 - 2
x
1.2.2. Multiplication of a vector by a number
Numeric vectors. One multiplies a numeric vector v by a number c by multiplying each
component by c. Their product is a vector and is often called a scalar multiple of v.
v1
cv1
v = v2  cv = cv2
v3
cv3
For example
 2
 (3)(2)  6 
v = -4
c=3
 cv = (3)(-4)) = -12
 5
 (3)(5)  15 
Thus (cv)i = c(vi) for i = 1, …, n.
Example 1.2.2. An electronics company makes two types of circuit boards for computers,
namely ethernet cards and sound cards. Last week the production was
25
25 ethernet cards
y =  35  =  35 sound cards 
Suppose this week the production of each component is to be cut by 20%. Then this weeks
prodcution will be
25
25
25
5
20
20 ethernet cards
t = y – (0.2)y =  35  - (0.2) 35  =  35  -  7  =  28  =  28 sound cards 
Geometric / physical vectors. Multiplication of
geometric / physical vectors by a number is defined as
y

follows. Suppose u = PQ and c is a number. Then cu
has the same direction as u if c  0 and the opposite
direction of u if c < 0. The length of cu is |c| times the
length of u. It is not hard to show that this geometric
definition multiplication of a vector by a number c
corresponds to the definition of multiplication of the
numeric vector corresponding to u by the number c.
R
Q
u = PQ
P
cu = PR
x
Functions. If f(x) is a function and c is a number then their product is defined in the
usual way, i.e. (cf)(x) = c(f(x)). For example, if f(x) = 4x2 – 2x + 3 and c = 3
then (3f)(x) = 12x2 – 3x + 9.
1.2.3. Linear combinations (or superpositions)
A linear combination (or superposition) of vectors is just a sum of scalar multiples of the
vectors. More specifically, if u1, u2, …, un are vectors and c1, c2, …, cn are numbers, then,
then the vector v = c1u1 + c2u2 + … + cnun is a linear combination (or superposition) of
u1, u2, …, un. Linear combinations arise frequently in problems involving vectors.
1.2 - 3
Example 1.2.3. Suppose
2
1
u1 =  1  and u2 =  2  and
c1 = 2 and c2 = 3. Then the
corresponding superposition is
v = c1u1 + c2u2
2
1
= 2  1  + 3  2 
4
3
7
=  2  +  6  =  8 
y
7
v = 2u1 + 3u2 = 8
()
8
7
3
3u2 = 6
()
6
5
4
3
u2 =
(12)
1
4
2u1 = 2
()
u1 =
(21)
x
1
2
3
4
5
6
7
1.2.4. Algebraic
properties of the vector operations
Many of the familiar algebaic properties of numbers have analogues for the vector
operations. The first ten of these properties are regarded as special. Suppose V is a
collection of objects, called vectors. Then V is called a vector space if the following ten
properties hold for all u, v and w in V and all numbers c and d.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
u + v is in V.
u+v = v+u
(u + v) + w = u + (v + w)
u+0 = u
u + (-u) = 0
cu is in V
c(u + v) = cv + cu
(c + d)u = cu + du
c(du) = (cd)u
1u = u
(commutative property)
(associative property)
(0 is the zero vector)
(distributive property)
(another distributive property)
(another associative property)
Most of these properties are easy to prove for numeric vectors or functions.
1.2 - 4
For example, to prove property 7 (the second distributive property (c + d)u = cu + du if
u
u = u12 is a vector with two components we argue as follows
u
(c + d)u
cu + du
cu
du
u
u
(c + d)u = (c + d)u12 = (c + d)u21 = cu12 + du12 = cu12 + du12 = c u12 + d u12 = cu + du.
Similarly, to prove this property 7 if u(x) is a function defined for all real x we argue as
follows ((c + d)u)(x) = (c + d)(u(x)) = cu(x) + du(x) = (cu)(x) + (du)(x) = (cu + du)(x).
The reason the above ten properties are regarded as special is because many other
familiar properties can be deduced from them. In a general vector space the operation of
subtraction is defined by v – u = v + (-u). It is not hard to see that this holds for numeric
vectors, directed line segments and functions.
Proposition 1. The following hold for all u, v and w in V and all numbers c and d.
(11)
c0 = 0
(0 is the zero vector)
(12)
0u = 0
(0 on the left is the number 0 and 0 on the right is
the zero vector)
(13)
(-1)u = - u
(14)
u + (v – u) = v
(v – u) is what you add to u to get v
(15)
- (u + v) = (- u) + (- v)
Proof. As an illustration we prove (12) and leave the rest as exercises.
0u = (0 + 0)u
by (4) for numbers
= 0u + 0u
by (8)
Add the negative of 0u to both sides:
0u + (- (0u)) = [0u + 0u] + (- (0u))
= 0u + [0u + (- (0u))]
by (3)
0 = 0u + 0
by (5)
0 = 0u
by (4)
//
The operation of transposing for numeric vectors also has some useful properties.
Proposition 2. If u and v are numeric vectors with the same number of components then
(16)
(17)
(18)
(19)
(20)
(uT)T = u
(u + v)T = uT + vT
(u - v)T = uT - vT
(cu)T = cuT
(-u)T = -(uT)
u
v
Proof. As an illustration we prove (17) if u = u12 and v = v12 are column vectors with
u
v
two components and leave the rest for exercises. One has (u + v)T = (u12 + v12)T =
u1 + v1T = (u1 + v1, u2 + v2) = (u1, u2) + (v1, v2) = u1T + v1T = uT + vT. //
u2 + v2
u2
v2
1.2 - 5
One implication of these algebraic properties is that often one can work with vectors as a
whole instead of the individual components. For example, suppose one wants to find a
3
1
3
1
2
2
1
vector u such that 3u + 5 = 5u + 7. Then 2u = 5 - 7 = -2. So u = ½ -2 = -1.
1.2 - 6