Hassani S. Mathematical Methods: For Students of Physics and Related Fields
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174 Planar and Spatial Vectors
Box 6.0.1. If we can find some set of real numbers, α
1
,α
2
,...,α
n
(not
all of which are zero), such that the sum in (6.1) is zero, we say that the
vectors are linearly dependent. If no such set of real numbers can be
found, then the vectors are called linearly independent.
6.1 Vectors in a Plane Revisited
Before elaborating further on the generalization of vectors and their spaces,
it is instructive to revisit the familiar vectors in a plane from a point of view
suitable for generalization. We first discuss the notion of linear independence
as applied to vectors in the plane.
The two vectors
ˆ
e
x
and
ˆ
e
y
(sometimes denoted as i and j) are linearly
independent because α
ˆ
e
x
+ β
ˆ
e
y
=0canbesatisfiedonlyifbothα and β are
zero. If one of them, say α, were different from zero, one could divide the
equation by α and get
ˆ
e
x
= −
β
α
ˆ
e
y
which is impossible because
ˆ
e
x
and
ˆ
e
y
cannot lie along the same line.
Example 6.1.1.
The arrows in the plane are not the only kinds of vectors dealt
with in physics. For instance, consider the set of all linear functions, or polynomials
of degree one (or less), i.e., functions of the form α
0
+ α
1
t where α
0
and α
1
are real
numbers and t is an arbitrary variable. Let us call this set P
1
[t], where P stands forpolynomials as
vectors? “polynomial,” 1 signifies the degree of these polynomials, and t is just the variable
used. We can add two such polynomials and get a third one of the same form. We
can multiply any such polynomial by a real number and get another polynomial. In
fact, P
1
[t] has all the properties of the vectors in a plane. We say that P
1
[t]and
the vectors in a plane are isomorphic which literally means they have the “same
shape.”
It is important to emphasize that two polynomials are equal if and only if all
their coefficients are equal. In particular, a polynomial is equal to zero only if it is so
for all values of t, i.e., only if its coefficients vanish. This immediately leads to the
fact that the two polynomials 1 and t are linearly independent because if α + βt =0
(for all values of t), then α = β = 0 (try t =0andt =1).
It is easy to show that any three vectors in the plane are linearly dependent.proof of the fact
that any three
vectors in the
plane are linearly
dependent
Figure 6.1 shows three arbitrary vectors drawn in a plane. From the tip of one
of the vectors (a
3
in the figure), a line is drawn parallel to one of the other two
vectors such that it meets the third vector (or its extension) at point D.The
vectors
−−→
OD and
−−→
DC are proportional to a
1
and a
2
, respectively, and their
sum is equal to a
3
.Sowecanwrite
a
3
=
−−→
OD +
−−→
DC = αa
1
+ βa
2
⇒ αa
1
+ βa
2
− a
3
= 0
and a
1
, a
2
,anda
3
are linearly dependent. Clearly we cannot do the same
with two arbitrary vectors. Thus
6.1 Vectors in a Plane Revisited 175
A
B
C
D
O
a
1
a
2
β a
2
a
3
α a
1
Figure 6.1: Any three vectors a
1
, a
2
,anda
3
in the plane are linearly dependent.
Box 6.1.1. The maximum number of linearly independent vectors in a
plane is two. Any vector in a plane can be written as a linear combination
of only two non-collinear (not lying along the same line) vectors.
We also say that any two non-collinear vectors span the plane.
Suppose that we can write a vector a as a linear combination of n vectors
a = α
1
a
1
+ α
2
a
2
+ ···+ α
n
a
n
.
We want to see under what conditions the coefficients are unique. Suppose
that we can also write
a = β
1
a
1
+ β
2
a
2
+ ···+ β
n
a
n
,
where the β’s are different from the α’s. Then, subtracting these two linear
combinations, we get
0 =(α
1
− β
1
)a
1
+(α
2
− β
2
)a
2
+ ···+(α
n
− β
n
)a
n
.
This is possible only if the vectors are linearly dependent. Therefore, if we
want the coefficients to be unique, the vectors have to be linearly independent.
In particular, we can have at most two such vectors in the plane. Thus,
choosing any two linearly independent vectors a
1
and a
2
in the plane, we can
expand any other vector uniquely as a linear combination of a
1
and a
2
.This
brings us to the notion of a basis.
basis defined
Box 6.1.2. Vectors that span the plane and are linearly independent are
called a basis for the plane.
The foregoing argument showed that any two non-collinear vectors form a
basis for the plane.
With the notion of a basis comes the concept of components of a vector.
Given a basis, there is a unique way in which a particular vector can be written
176 Planar and Spatial Vectors
in terms of the vectors in the basis. The unique coefficients of the basis vectors
are called the components of the particular vector in that basis. Another
components and
dimension
concept associated with the basis is dimension which is defined to be the
number of vectors in a basis. It follows that the plane has two dimensions.
Example 6.1.2.
The components of a
3
in the basis {a
1
, a
2
} of Figure 6.1 are
(α, β).
2
Given any basis {a
1
, a
2
} of the plane, it is readily seen that the components
of a
1
are (1, 0) and those of a
2
are (0, 1).
Example 6.1.3. The polynomials {1,t} form a basis for P
1
[t], because they are
linearly independent and they span P
1
[t]. Therefore P
1
[t] is a two-dimensional
vector space. The components of f = α
0
+ α
1
t are (α
0
,α
1
) in this basis. How do
we determine the components of f in another basis {a
1
, a
2
} with a
1
=1+t and
a
2
=1− t?Since{a
1
, a
2
} is a basis, we can write
f = x
1
a
1
+ x
2
a
2
= x
1
(1 + t)+x
2
(1 − t)=(x
1
+ x
2
)+(x
1
− x
2
)t
or
α
0
+ α
1
t =(x
1
+ x
2
)+(x
1
− x
2
)t ⇒ (α
0
− x
1
− x
2
) · 1+(α
1
− x
1
+ x
2
)t =0.
The linear independence of 1 and t now tells us that the coefficients of 1 and t should
vanish. This leads to two equations in two unknowns:
x
1
+ x
2
= α
0
,
x
1
− x
2
= α
1
.
The solution of these equations are easily found to be
x
1
=
1
2
(α
0
+ α
1
),x
2
=
1
2
(α
0
− α
1
).
Thus, the components of f are (
1
2
(α
0
+ α
1
),
1
2
(α
0
− α
1
)) in the new basis.
6.1.1 Transformation of Components
There are infinitely many bases in a plane, because there are infinitely many
pairs of vectors that are linearly independent. Therefore, there are infinitely
many sets of components for any given vector, and it is desirable to be able
to find a relation between any two such sets. Such a relation employs the
machinery of matrices.
Consider a vector a with components (α
1
,α
2
) in the basis {a
1
, a
2
} and
components (α
1
,α
2
) in the basis {a
1
, a
2
}.Wecanwrite
a = α
1
a
1
+ α
2
a
2
and a = α
1
a
1
+ α
2
a
2
. (6.2)
Since {a
1
, a
2
} form a basis, any vector, in particular, a
1
or a
2
, can be written
in terms of them:
a
1
= a
11
a
1
+ a
21
a
2
,
a
2
= a
12
a
1
+ a
22
a
2
, (6.3)
2
Since in this chapter we are dealing primarily with components (and not coordinates),
we shall use parentheses—instead of angle brackets—to list the components.
6.1 Vectors in a Plane Revisited 177
where (a
11
,a
21
)and(a
12
,a
22
) are, respectively, components of a
1
and a
2
in
the basis {a
1
, a
2
}. Combining Equations (6.2) and (6.3), we obtain
α
1
(a
11
a
1
+ a
21
a
2
)+α
2
(a
12
a
1
+ a
22
a
2
)=α
1
a
1
+ α
2
a
2
or
(α
1
− a
11
α
1
− a
12
α
2
)a
1
+(α
2
− a
21
α
1
− a
22
α
2
)a
2
=0.
The linear independence of a
1
and a
2
gives
α
1
= a
11
α
1
+ a
12
α
2
,
α
2
= a
21
α
1
+ a
22
α
2
. (6.4)
These equations can be written concisely as
3
α
1
α
2
=
a
11
a
12
a
21
a
22
α
1
α
2
or a
= Aa, (6.5)
wherewehaveintroducedthematrices
matrix and
column vector
a ≡
α
1
α
2
, a
≡
α
1
α
2
, A ≡
a
11
a
12
a
21
a
22
. (6.6)
The matrices a and a
are called column vectors or 2 × 1 matrices because
they each have two rows and one column. Similarly, A is called a 2×2matrix.
Let us now choose a third basis, {a
1
, a
2
},andwritea = α
1
a
1
+ α
2
a
2
.If
(a
11
,a
21
)and(a
12
,a
22
) are, respectively, the components of a
1
and a
2
in this
third basis, then
a
1
= a
11
a
1
+ a
21
a
2
,
a
2
= a
12
a
1
+ a
22
a
2
.
Substituting these in the second equation of (6.2) and equating the result to
a = α
1
a
1
+ α
2
a
2
yields
α
1
= a
11
α
1
+ a
12
α
2
,
α
2
= a
21
α
1
+ a
22
α
2
. (6.7)
We can write Equation (6.7) in matrix form:
α
1
α
2
=
a
11
a
12
a
21
a
22
α
1
α
2
or a
= A
a
, (6.8)
where
a
≡
α
1
α
2
, A
≡
a
11
a
12
a
21
a
22
, (6.9)
and a
is as defined before.
3
At this point, think of Equation (6.5) as a short-hand way of writing Equation (6.4).
Further significance of this notation will become clear after Box 6.1.3.
178 Planar and Spatial Vectors
We can also discover how a
and a are related by substituting (6.4) in
(6.7). This leads to the equation
two
transformations in
a row suggest the
rule of matrix
multiplication.
α
1
=(a
11
a
11
+ a
12
a
21
)α
1
+(a
11
a
12
+ a
12
a
22
)α
2
,
α
2
=(a
21
a
11
+ a
22
a
21
)α
1
+(a
21
a
12
+ a
22
a
22
)α
2
,
which, in matrix form, becomes
a
= A
a where A
≡
a
11
a
11
+ a
12
a
21
a
11
a
12
+ a
12
a
22
a
21
a
11
+ a
22
a
21
a
21
a
12
+ a
22
a
22
. (6.10)
On the other hand, the matrix equations (6.8) and (6.5) yield a
= A
(Aa),
which is consistent with Equation (6.10) only if matrix multiplication is
defined so that A
= A
A, i.e.,
a
11
a
12
a
21
a
22
a
11
a
12
a
21
a
22
=
a
11
a
11
+ a
12
a
21
a
11
a
12
+ a
12
a
22
a
21
a
11
+ a
22
a
21
a
21
a
12
+ a
22
a
22
. (6.11)
All discussions and all the equations obtained so far are based on fixing
a vector and looking at its components in different bases. However, there is
another, more physical, way of interpreting these equations. Consider (6.5).
active and passive
transformations
distinguished
Here the column vector on the RHS represents the components of a vec-
tor a in the basis {a
1
, a
2
}. Applying the matrix A to this column vector
yields a new column vector given on the LHS, which can be interpreted as
the components of a new vector a
in the same basis.So,inessencewehave
changed the vector a into a new vector a
via the transformation A. The first
interpretation mentioned above is called a passive transformation (a is
“passively” unchanged as basis vectors are altered); the second interpretation
is called active transformation (a is actively changed into a
). We shall
have occasion to employ both interpretations. However, the active transfor-
mation is more direct and we shall use that more often. The reader may
convince himself or herself that passive transformation in one “direction” is
completely equivalent to active transformation in the “opposite” direction.
A good example to keep in mind is the rotation of axes (passive rotation)
versus the rotation of a vector (active rotation) in the plane as shown in
Figure 6.2.
Equation (6.11) defines the “product” of two matrices in a prescribed man-
ner. To find the entry in the first row and first column of the product, multiply
the entries of the first row of the first matrix by the corresponding entries of
the first column of the second matrix and add the terms thus obtained. To
find the entry in the first row and second column of the product, multiply
the entries of the first row of the first matrix by the corresponding entries of
the second column of the second matrix and add the terms. Other entries
are found similarly. This leads us to the following rule which applies to all
matrices, not just those that are 2 × 2:
6.1 Vectors in a Plane Revisited 179
O
a
'
(b)
x
a
O
(a)
y
x
y
a
O
'
(c)
x
y
y
'
x
'
Figure 6.2: (a) A vector a in a coordinate system Oxy can be (b) actively transformed
to a new vector a
in the same coordinate system, or (c) passively transformed to a
new coordinate system O
x
y
. Note that the relation of a
to Oxy is identical to the
relation of a to O
x
y
.
Box 6.1.3. (Matrix Multiplication Rule). To obtain the entry in the
ith row and jth column of the product of two matrices, multiply the entries
of the ith row of the matrix on the left by the corresponding entries of the
jth column of the matrix on the right and add the products thus obtained.
For this rule to make sense, the number of entries in a row of the matrix on for the product
rule to make
sense, number of
columns of the left
matrix must equal
number of rows of
the right matrix.
the left must equal the number of entries in a column of the matrix on the
right.
We identified a column vector as a 2 × 1 matrix. With this identification,
the RHS of Equation (6.5) can be interpreted as the product of two matrices,
a2× 2matrixanda2× 1 matrix, resulting in a 2 × 1 matrix, the column
vector on the LHS.
Matrices were obtained in a natural way in the discussion of basis changes,
matrices forming a
mathematical
structure with
operations other
than
multiplication
and the natural operation ensued was that of multiplication. Once a mathe-
matical entity is created in this manner, a full mathematical structure becomes
irresistibly enticing. For example, such operations as addition, subtraction,
division, inversion, etc., also demand our attention. We now consider such
operations.
First, we need to define the equality of matrices: Two matrices are equal
if they have the same number of rows and columns, and their corresponding
elements are equal. Addition of two matrices is defined if they have the same
number of rows and columns in which case the sum is defined to be the sum
of corresponding elements. A 2 × 2 matrix can be added to another 2 × 2
matrix, but a column vector cannot. Thus if
A =
a
11
a
12
a
21
a
22
and B =
b
11
b
12
b
21
b
22
,
180 Planar and Spatial Vectors
then
A + B =
a
11
+ b
11
a
12
+ b
12
a
21
+ b
21
a
22
+ b
22
.
From the definition of the sum and the product of matrices, it is clear that
addition is always commutative but product need not be:
A + B = B + A but AB = BA. (6.12)
Wecanturnthesetof2× 2 matrices into a vector space by defining the
2 × 2 matrices
form a vector
space
product of a number and a matrix as a new matrix whose elements are the old
elements times the number. The zero “vector” is simply the zero matrix—
the 2 × 2 matrix all of whose elements are zero. The reader may verify that
zero matrix
all the usual operations of vectors apply to this set.
4
If you multiply a matrix
by the number 0, you get the zero matrix.
Example 6.1.4.
Suppose
A =
1 −1
23
and B =
−10
12
.
Then
A + B =
1 − 1 −1+0
2+1 3+2
=
0 −1
35
= B + A
and
AB =
1 −1
23
−10
12
=
1 · (−1) + (−1) · 11·0+(−1) · 2
2 · (−1) + 3 · 12· 0+3· 2
=
−2 −2
16
,
while
BA =
−10
12
1 −1
23
=
(−1) · 1+0· 2(−1) ·(−1) + 0 ·3
1 · 1+2· 21· (−1) + 2 · 3
=
−11
55
.
Clearly, AB = BA.
The 2 ×2matrix
1 ≡
10
01
is called the 2 × 2 identity matrix or unit matrix, and has the property
identity matrix or
unit matrix
that when it multiplies any other matrix (on the right or on the left), the
latter does not get affected. The unit matrix is used to define the inverse of
amatrixA as a matrix B that multiplies A on either side and gives the unit
inverse of a matrix
matrix. The inversion of a matrix is a much more complicated process than
that of ordinary numbers, and we shall discuss it in greater length later. At
this point, suffice it to say that, contrary to numbers, not all nonzero matrices
have an inverse. For example, the reader can easily verify that the nonzero
matrix
10
00
!
cannot have an inverse.
4
Note that the extra operation of multiplication of a matrix by another matrix is not
part of the requirement for the set to be a vector space.
6.1 Vectors in a Plane Revisited 181
We have introduced 2 × 2andcolumn(or2× 1) matrices. To complete
the picture, we also introduce a row vector,ora1× 2matrix. Theruleof
row vector
matrix multiplication allows the multiplication of a 2×2 matrix and a column
vector, as long as the latter is to the right of the former: You cannot multiply
a2× 1 matrix situated to the left of a 2 × 2 matrix. Similarly, you cannot
multiply two 2 × 1 matrices. However, the product of a row vector (a 1 × 2
matrix) and a column vector (a 2 ×1 matrix) is defined—as long as the latter
is to the right of the former—and the result is a 1 × 1 matrix, i.e., a number.
This is because we have only one row to the left of a single column. What
about the product of a row vector and a 2 ×2 matrix? As long as the matrix
is to the right of the row vector, the product is defined and the result is a row
vector.
Example 6.1.5.
With A and B as defined in Example 6.1.4 and
x =
1
−1
, y =
−12
!
,
we have
Ax =
1 −1
23
1
−1
=
2
−1
,
yB =
−12
!
−10
12
=
34
!
,
yx =
−12
!
1
−1
=(−3) = −3,
yAx =
−12
!
1 −1
23
1
−1
=
−12
!
2
−1
= −
4,
yBAx =
34
!
2
−1
=2
=
−12
!
−11
55
=BA
1
−1
=
−12
!
−2
0
=2,
yABx =
−12
!
−2 −2
16
1
−1
=
414
!
1
−1
= −10.
In the manipulations above, we have used the associativity of matrix multiplication
and multiplied matrices in different orders without, of course, commuting them.
Products such as Ay, By, yy,andxx are not defined; therefore, we have not considered
them here.
There is a new operation on matrices which does not exist for ordinary
numbers. This is called transposition and is defined as follows:
transpose of a
matrix
Box 6.1.4. The transpose of a matrix is a new matrix whose rows are
the columns of the old matrix and whose columns are the rows of the old
matrix. The transpose of A is denoted by A
t
or
8
A.
182 Planar and Spatial Vectors
Therefore
A =
a
11
a
12
a
21
a
22
⇒ A
t
=
8
A =
a
11
a
21
a
12
a
22
.
If A
t
= A,wesaythatA is symmetric.symmetric matrix
Example 6.1.6. With A, B, x,andy as defined in Example 6.1.5, we have
A
t
=
12
−13
,
8
B =
−11
02
, x
t
=
1 −1
!
,
8
y =
−1
2
.
Note that although xx and yy are not defined, all the combinations
8
xx, y
8
y,
8
yy,and
x
8
x are defined: In the first two cases one gets a number, and in the last two cases a
2 × 2matrix.
It should be clear from the definition of the transpose thatproperties of
transposition
(A + B)
t
= A
t
+ B
t
, (AB)
t
= B
t
A
t
, (A
t
)
t
= A. (6.13)
Of the three relations, the middle one is the least obvious, but the reader
can verify it directly by choosing appropriate general matrices and carrying
through the multiplications on both sides of the relation.
6.1.2 Inner Product
From our discussion of Chapter 1, we know that if a and b are vectors in the
plane having components (a
x
,a
y
)and(b
x
,b
y
)alongthex-andy-axes, then
their dot product is
a ·b = a
x
b
x
+ a
y
b
y
. (6.14)
We want to generalize this dot product so that it applies to arbitrary bases.
This generalization is called the inner product.
Recall that any two non-collinear vectors {a
1
, a
2
} in the plane form a
basis and any vector can be written as a linear combination of them with
the unique coefficients being the components of the vector in the basis. In
particular, the components of a
1
are (1, 0) and those of a
2
are (0, 1). If we were
to define the dot product in terms of components, we would have to modify
Equation (6.14) because that equation would give zero for a
1
·a
2
which wouldequation (6.14)
will not work for
arbitrary bases!
be inconsistent with (1.1). How should we modify (6.14)? Since we want to
deal with components, a natural setting would be the language of matrices.
If a and b are the column vectors
a
x
a
y
!
and
b
x
b
y
!
, respectively, then we can
rewrite Equation (6.14) as
inner (dot)
product in terms
of row and column
vectors
a ·b = a
t
b =
a
x
a
y
!
b
x
b
y
= a
x
b
x
+ a
y
b
y
. (6.15)
It is this matrix relation that we want to generalize so that the result is the
true dot product of vectors no matter what basis we choose in which to express
our vectors.
6.1 Vectors in a Plane Revisited 183
Besides the failure of Equation (6.15) for general bases, the demand for
generalization stems from another source: There are other kinds of “vectors”
that are not just arrows in the plane. For instance, the polynomials P
1
[t]of
degree one that we introduced in Example 6.1.1 are such vectors. How do we
define inner products for these vectors? We cannot use Equation (1.1) because
neither the length of a polynomial nor the angle between two polynomials is
defined. In fact, both the length and the angle are defined only after an
inner product has been introduced. Furthermore, there is no guarantee that
Equation (6.15) will make sense.
Let’s see how far we can go using the general properties of the inner prod-
uct discussed at the beginning of Section 1.1.1. Write a and b as a linear
combination of the basis vectors {a
1
, a
2
}:
a = α
1
a
1
+ α
2
a
2
, b = β
1
a
1
+ β
2
a
2
Take the dot-product of these vectors and write it in terms of the dot-products
of the basis vectors:
a ·b =
α
1
a
1
+ α
2
a
2
!
·
β
1
a
1
+ β
2
a
2
!
= α
1
β
1
a
1
· a
1
+ α
1
β
2
a
1
· a
2
+ α
2
β
1
a
2
·a
1
+ α
2
β
2
a
2
· a
2
Define a matrix with elements
g
11
= a
1
· a
1
,g
12
= a
1
· a
2
= a
2
· a
1
= g
21
,g
22
= a
2
·a
2
Then, representing a and b as column vectors a =
α
1
α
2
!
and b =
β
1
β
2
!
,the
dot product can be generalized to
a symmetric
matrix G is needed
to generalize the
inner product.
a ·b = a
t
Gb =
α
1
α
2
!
g
11
g
12
g
21
g
22
β
1
β
2
, (6.16)
where G is a symmetric matrix.
Example 6.1.7.
In this example, we shall define an inner product for the vectors inner (dot)
product of two
polynomials
in P
1
[t] that happens to be useful in physical applications. The idea is to find a rule
that takes two “vectors” in P
1
[t] and gives a real number. Since the vectors in P
1
[t]
are functions (albeit a very special kind), one natural way of getting numbers out
of functions is by integrating them. It turns out that this is indeed the most useful
way of defining the inner product for such polynomials. So, let (a, b)beaninterval
on the real line and let f = α
0
+ α
1
t and g = β
0
+ β
1
t be two “vectors” in P
1
[t].
We define
f · g ≡
#
b
a
f(t)g(t) dt. (6.17)
One can show that Equation (6.17) exhibits all the properties expected of an inner
product (as outlined in Section 1.1.1). For instance, f · f is always positive because
the integrand [f(t)]
2
is always positive. Furthermore, f ·g = g ·f, and, as the reader
may check,
f · (g + h)=f ·g + f · h.