1) The document discusses vector spaces and linear combinations. It defines a vector space as a set of objects called vectors that can be added together and multiplied by scalars, following certain properties like closure and the existence of additive inverses. 2) Examples of vector spaces include Rn (n-dimensional coordinate vectors), matrices, and sets of functions. A linear combination is an expression of a vector as a sum of other vectors multiplied by scalars. 3) The key characteristics of a vector space are that it is closed under vector addition and scalar multiplication, has an additive identity element (the zero vector), and vectors have additive inverses. Any set satisfying these properties forms a vector space.

1. NPTEL – Physics – Mathematical Physics - 1
Module 2 Vector spaces Lecture 7
In various branches of Physics and Mathematics, one restricts to a set of numbers
or physical quantities, where it is meaningful and interesting to talk about
‘linear combinations’ of these objects in the set. Simple examples of such
combinations have already been seen for vectors in three dimensional space, and in
particular with linear combination of such vectors.
As we shall see later that a linear vector space is defined formally through a
number of properties. The word ‘space’ suggests something geometrical, as does
the word ‘vector’ to most readers. As we get into the chapter on linear vector
spaces, the reader will comprehend and come to terms with the fact that much of the
terminology has geometrical connotations.
Linear Vector spaces
Consider a vector 𝑟⃗ = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂. The vector starts from origin (0, 0, 0) and ends at a
point (𝑥, 𝑦, 𝑧). Thus there is a one-to-one correspondence between the vectors and the
variables (𝑥, 𝑦, 𝑧). The collection of all such points or all such vectors makes up the three
dimensional space, 𝑟, 𝑣, 𝑒 often called as 𝑅3 (𝑟 ∶ 𝑟𝑒𝑎𝑙) or 𝑉3 (𝑣: 𝑣𝑒𝑐𝑡𝑜𝑟) or 𝐸3
(E: Euclidiean; after the Greek Mathematician named Euclid). Similarly a set of two
dimensional vector will constitute a 𝑅2 space. Thus an n-dimensional vector is defined on
a 𝑅𝑛 space. For example, 4-vectors of special relativity are ordered sets of four numbers
or variables and we call space-time is 4-dimensional.
Before going ahead with the discussion on details of the vector space, it is necessary to
define the ‘dimension’ of the vector space and to understand the dimension of the space.
It is also necessary to understand the linear independence of vectors. Formally, the linear
independence is defined as in the following. For a vector space V with a subset S, if there
exists a set of vectors 𝛼1, 𝛼2, … … … . 𝛼𝑛 in S with some coefficients 𝐶1, 𝐶2, … … . 𝐶𝑛
(at least one of them not being zero), such that
𝐶1𝛼1 + 𝐶2𝛼2 + … … . . 𝐶𝑛 𝛼𝑛 = 0
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(7.1)
If any of the 𝛼𝑖 has a relation with any of the 𝛼𝑗 (𝑖 ≠ 𝑗), then the linear independence is no
longer there.
Suppose we have two linearly independent vectors. They define a plane. All linear
combination of these two vectors lie in the plane. Since all the vectors making up this

2. NPTEL – Physics – Mathematical Physics - 1
plane 𝑅2 (𝑜𝑟 𝑉2) is also a part of a 3-dimensional space 𝑅3 (𝑜𝑟 𝑉3). Thus 𝑅2 is called as
the subspace of 𝑅3.
Either the original vectors in 𝑅3 or the independent ones in 𝑅2 span the space 𝑅2. Thus
the definition of spanning is as follows - a set of vectors spans a space if all the vectors
in space can be written as linear combination of the spanning set. A set of linearly
independent vectors which span a vector space is called a basis.
The dimension of a vector space is equal to the number of basis vectors. It does not
matter how one chooses the basis vectors for a given vector space corresponding to a
particular problem, the number of basis vectors will always be the same. This number
denotes the dimension of the space. In 3-dimensions, one frequently uses the unit vectors
𝑖̂ , 𝑗̂, 𝑘̂, to describe a particular coordinate system called as Cartesian coordinate system.
Vector space: properties
At the outset, let us define the vector fields. Let us define a function
𝐹(𝑥, 𝑦) = (𝑦, −𝑥) in a two dimensional space (𝑅2). We can calculate the values of the
function at a set of discrete points as,
𝐹(1,0) = (0, −1)
𝐹(0,1) = (1,0)
𝐹(1,1) = (1, −1)
𝐹(1,2) = (2, −1)
𝐹(2,1) = (1, −2) etc.
Also we can plot the value of 𝐹(𝑥, 𝑦) as a vector with tail anchored at some
arbitrary
(𝑥, 𝑦) such as given below,
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NPTEL – Physics – Mathematical Physics - 1
Thus by putting each of the above vectors defined by 𝐹(𝑥, 𝑦) = (𝑦, −𝑥) and many more
of them we can see the structure of the vector field, which appears as arrow
heads rotating in a clockwise sense. This is the vector field corresponding to 𝐹(𝑥, 𝑦).
The definition of a vector space 𝑉 , whose elements are called vectors involves an
arbitrary field K whose elements are all scalar. Thus to make notations clear,
Vector space – 𝑉 Elements of 𝑉 − 𝑢1, 𝑣, 𝑤 Arbitrary field –𝐾
Elements of 𝐾 – 𝑎, 𝑏, 𝑐
The vector space has the following characteristics,
(i)
(ii)
(iii)
Vector Addition: for any 𝑢, 𝑣 ∈ 𝑉, 𝑢 + 𝑣 also ∈ 𝑉.
Scalar Multiplication: 𝑢 ∈ 𝑉, 𝑘 ∈ 𝐾, a product 𝑘𝑢 ∈ 𝑉
There is a null in V, denoted by 0 and called zero vector, such that for any
𝑢 ∈ 𝑉,
𝑢 + 0 = 0 + 𝑢 = 𝑢
For 𝑢 ∈ 𝑉, there is 𝑎 – 𝑢, such that
𝑢 + (−𝑢) = 0
(iv)
(v) (𝑎 + 𝑏)𝑢 = 𝑎𝑢 + 𝑏𝑢 for two scalars a and b
(vi) (𝑎𝑏)𝑢 = 𝑎(𝑏𝑢) for two scalars a and b
From the above discussion, we wish to say that any set of objects obeying the
same properties form a vector space.
More formally one can define a vector space V is a collection of
objects
|1 >, |2 > … … … . |𝑣 > … … . . |𝑤 > … … … ., called vectors, which there exists (with |
>
: 𝑘𝑒𝑡)
(A) A definite rule for forming the vector sum, that is |𝑣 > + |𝑤 >.
A definite rule for multiplication by scalars 𝛼, 𝛽, … … … denoted by 𝛼|𝑣 > with the
following features:

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NPTEL – Physics – Mathematical Physics - 1
(i) The result of these operations is another element in the same space.
This property is called closure.
|𝑣 > + |𝑤 >∈ 𝑉.
Scalar multiplication is distributive :
𝛼(|𝑣 > +|𝑤 >) = 𝛼|𝑣 > +𝛼|𝑤 > Scalar multiplication is
distributive : (𝛼 + 𝛽)|𝑣 > = 𝛼|𝑣 > +𝛽|𝑣 >
Scalar multiplication is associative :
𝛼(𝛽|𝑣 >) = 𝛼𝛽|𝑣 >
Addition is commutative :
|𝑣 > +|𝑤 > =|𝑤 > +|𝑣 >
Addition is associative:
|𝑤 > +(|𝑣 > +|𝑧 >) = (|𝑤 + |𝑣 >) + |𝑧 >
There exists a null vector |0 > obeying
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
|𝑣 > +|0 > = |𝑣 >
(viii) For every vector |𝑣 > there exists an inverse |– 𝑣 >, such that if
added with the origin it yields zero,
|𝑣 > +|−𝑣 >= 0. For example, a set of arrows with only positive
z-components do not form a vector space as there is no inverse.
Consider the set of all 2× 2 matrices. We know how to add them
and multiply them by scalars. They obey associative, distributive
and closure requirements. A null matrix exists with all the elements
as zero. Thus we have a genuine vector space
As another example, consider all functions 𝑓(𝑥) defined in an
interval −𝑎 ≤ 𝑥 ≤ 𝑎. A scalar multiplication by 𝛼 simply yields 𝛼𝑓
(𝑥). The sum of two functions f and g has the value 𝑓(𝑥) + 𝑔(𝑥)
at the point x. the null function is zero everywhere and
additive inverse of 𝑓(𝑥) 𝑖𝑠 – 𝑓(𝑥).
Examples of vector spaces (usual notation in literature)
(a)𝑅𝑛: 𝑛 − dimensional coordinate vectors.
(b)𝑀𝑚,𝑛(𝑅): 𝑚 × 𝑛 matrices with real elements.
(c){0}: null vector space with all ‘0’ elements.

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(d)F(R): the set of all functions.
(e)C(R): the set of all continuous functions.
(f)P: set of all polynomial functions,
𝑃(𝑥) = 𝑎0 + 𝑎1𝑥 + … … … . 𝑎𝑛𝑥𝑛
It is important to realize that the elements of a vector space are characterized by
two operations: one in which two elements are added together, and another in
which the elements are multiplied by scalars.
Linear Combination
Let V be a vector space. A vector v in V is a linear combination of
vectors
𝑢1, 𝑢2 … … … 𝑢𝑚; if there exist scalars 𝑎1𝑎2 … …. such that
𝑣 = 𝑎1𝑢1 + 𝑎2𝑢2 + … … … 𝑎𝑚𝑢𝑚
Example of Linear Combination
Suppose we want to express 𝑣 = (3, 7, −4) in 𝑅3 as a linear combination of the
vectors
𝑢1 = (1,2,3) ; 𝑢2 = (2,3,7); 𝑢3 = (3,5,6)
We seek scalars 𝑥, 𝑦, 𝑧 such that,
𝑣 = 𝑥𝑢1 + 𝑦𝑢2 + 𝑧𝑢3
3 1 2 3
[ 7 ] = 𝑥 [2] + 𝑦 [3] + 𝑧 [5]
−4 3 7 6
𝑥 + 2𝑦 + 3𝑧 = 3 2𝑥 + 3𝑦 + 5𝑧 = 7
3𝑥 + 7𝑦 + 6𝑧 = −4
Solving 𝑥 = 2, 𝑦 = − 4 and 𝑧 = 3
𝑣 = 2𝑢1 − 4𝑢2 + 3𝑢3