This document discusses subspaces, spanning sets, and bases in vector spaces. Some key points: - A subspace is a subset of a vector space that is closed under vector addition and scalar multiplication. - The span of a set S in a vector space V is the smallest subspace of V containing S. - A set S is a basis for V if every vector in V can be uniquely written as a linear combination of vectors in S. - Examples are provided to illustrate subspaces, spans, and finding the coordinate representation of vectors with respect to given bases.

1. NPTEL – Physics – Mathematical Physics - 1
Lecture 8
Basis & Dimension
Subspaces
A vector space V is called a subspace of 𝑅𝑛 if 𝑉 ∈ 𝑅𝑛 and the linear operations
applicable to 𝑅𝑛 (as defined earlier) are also applicable to V.
Thus a subset S of a vector space V is a subspace of V if and only if S is non-empty and is
closed under linear operations, that is
𝑥, 𝑦 ∈ 𝑆 implies 𝑥 + 𝑦 ∈ 𝑆
𝑥 ∈ 𝑆 implies 𝛼𝑥 ∈ 𝑆 for all 𝛼 ∈ 𝑅
An example of such a subspace can be given as follows. If P is a set of all polynomials
and if 𝑃𝑛 is a set of polynomials with degree less than n, then 𝑃𝑛 is a subspace of P.
Spanning Set
Let S be a subset of a vector space V the span of the set S, denoted by span (S), is the
smallest subspace of V that contains S. In other words, span (S) is a subspace of
V also for any subspace 𝑊 ⊂ 𝑉, one should have 𝑆 ⊂ 𝑊 which implies
𝑆𝑝𝑎𝑛(𝑆) ⊂ 𝑊. Example of the span can be given as follows. Let S be a subset of the
vector space V. If 𝑆 ={𝑉⃗⃗1, 𝑉⃗⃗2, …….𝑉⃗⃗𝑛}, then span (S) is the set of all linear
combinations
𝛼1𝑣⃗1 + 𝛼2𝑣⃗2 + … … . . 𝛼𝑛𝑣⃗𝑛 where all 𝛼𝑖 ∈ 𝑅
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As a second example, consider a subspace of 𝑀2×2(𝑅). The span of ( ) and ( )
1 0 0 0
0 0 0 1
consists of all matrices of the form,
𝑎 ( ) + 𝑏 ( ) = ( )
1 0 0 0
0 0 0 1 0 𝑏
𝑎 0
That is the subspace of diagonal matrices.
Now if span(S) = V, then S is called as the spanning set of V. So, as in the
former example, the spanning set for 𝑀2×2(𝑅) as
(𝛾 𝛿) = 𝛼 ( ) + 𝛽 ( ) + 𝛾 ( ) + 𝛿 ( )
𝛼 𝛽 1 0 0 1 0 0 0 0
0 0 0 0 1 0 0 1

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NPTEL – Physics – Mathematical Physics - 1
Example 1.
Let 𝑉⃗⃗1 = (1,2,0) and 𝑉⃗⃗2 = (3,1,1) and 𝑊⃗⃗ = (4, −7,3). Find whether 𝑊⃗⃗ belongs
to span (𝑉⃗⃗1, 𝑉⃗⃗2).
Solution
Here we have to check 𝛼1, 𝛼2 ∈ 𝑅 such that 𝑊⃗⃗ = 𝛼1𝑉⃗⃗1 + 𝛼2𝑉⃗⃗2. Thus we get a
system of linear equations,
4 = 𝛼1 + 3𝛼2
−7 = 2𝛼1 + 𝛼2 3 =
𝛼2
𝛼1 = −5 and 𝛼2 =
3
Hence
Hence 𝑊⃗⃗ = −5𝑉⃗⃗1 + 3𝑉⃗⃗2 ∈ Span(𝑉⃗⃗1, 𝑉⃗⃗2)
A set S = {𝑢1, 𝑢2 … … … . . 𝑢𝑛} of vectors is a basis of v if every 𝑣 ∈ 𝑉 can be
written uniquely as a linear combination of the basis vectors. Vector spaces
containing finite number of elements is said to have a finite basis.
Suppose a vector space does not have a finite basis. Then v is said to be of infinite
dimensions and has a special name: Hilbert space.
Consider the following n vectors in 𝐾𝑛,
𝑒1 = (1,0 … … . .0), 𝑒2 = (0,1, … … … 0), … … … … 𝑒𝑛 = (0,0 … … … .1)
These vectors are linearly independent. Further, any vector
𝑣 = (𝑎1, 𝑎2, … … … … . . 𝑎𝑛) in 𝐾𝑛 can be written as a linear combination of the
above vectors.
𝑣⃗ = 𝑎1𝑒1 + 𝑎2𝑒2 + ⋯ … 𝑎𝑛𝑒𝑛 8.1
Accordingly, the vectors form a basis of 𝐾𝑛 called the usual or standard basis of 𝐾𝑛
Example 2. Consider a vector space 𝑅2(𝑡) of polynomial of degree ≤ 2. The
polynomials 𝑟1 = 𝑡 + 1, 𝑟2 = 𝑡 − 1 and 𝑟3 = (𝑡 − 1)2 form a basic S of 𝑅2(𝑡). The
coordinate vector v of [V], v = 2𝑡2 − 5𝑡 + 9 relative to S is obtained as follows-
2𝑡2 − 5𝑡 + 9 = 𝑥(𝑡 + 1) + 𝑦(𝑡 − 1) + 𝑧(𝑡 − 1)2
Equating coefficients of equal powers of t,

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NPTEL – Physics – Mathematical Physics - 1
𝑧 = 2, 𝑥 + 𝑦 – 2𝑧 = −5, 𝑥 – 𝑦 + 𝑧 = 9
⇒ 𝑥 = 3, 𝑦 = −4, 𝑧 = 2.
𝑣 = 3𝑟1 − 4𝑟2 + 2𝑟3
Example 3.
Consider real space 𝑅3. The following vectors form a basis sof 𝑅3.
𝑢1 = (1, −1,0), 𝑢2 = (1,1,0) and 𝑢3 = (0,1,1). The coordinates of
𝑣 = (5, 3, 4) relative to the basis S is obtained as follows,
5 1 1 0
[3] = 𝑥 [−1] + 𝑦 [1] + 𝑧 [1]
4 0 0 1
Solving 𝑥 = 3, 𝑦 = 2 and 𝑧 = 4 then
𝑣 = 3𝑢1 + 2𝑢2 + 4𝑢3.
Example 4.
Consider the vector space 𝑀 = 𝑀2×2 consisting of all 2 × 2 matrices and consider the
following four matrices in M.
𝐸 = [ ] , 𝐸
11 12
1 0 0 1
0 0 0 0 1 0 0 1
= [ ] , 𝐸 = [ ] , 𝐸 = [ ]
21 22
0 0 0 0
Any matrix A in M can be written as a linear combination of the 4 matrices.
𝐴 = [
7 8
] = 5𝐸11 − 6𝐸12 + 7𝐸21 + 8𝐸22
Accordingly the 4 matrices 𝐸11, 𝐸12, 𝐸21, 𝐸22 span M. Example 5.
Express 𝑣 = (2, −5, 3) in 𝑅3 as a linear combination of the vectors
𝑢1 = (1, −3,2), 𝑢2 = (2, −4, −1) and 𝑢3 = (1, −5,7)
5 − 6
2 1 2 1
[−5] = 𝑥 [−3] + 𝑦 [−4] + 𝑧 [−5]
3 2 −1 7
It can be checked that the system is inconsistent and has no solution, so v can
not be written as a linear combination of 𝑢1, 𝑢2 and 𝑢3.