Basis and Dimension - Vector Algebra, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

Basis

Definition : Let V be a vector space. A linearly independent spanning set for V is called a basis.

Equivalently, a subset S ⊂ V is a basis for V if any vector v ∈ V is uniquely represented as a linear combination

v = r₁v₁ + r₂v₂ + · · · + r_kv_k , where v₁, . . . , v_k are distinct vectors from S and r₁, . . . , r_k ∈ R.

Examples. 

• Standard basis for Rn :
  e₁ = (1, 0, 0, . . . , 0, 0),
  e₂ = (0, 1, 0, . . . , 0, 0),
  . . . ,
  e_n = (0, 0, 0, . . . , 0, 1).
• Matrices   form a basis for M_2,2(R).
• Polynomials 1, x , x², . . . , xⁿ⁻¹ form a basis for P_n = {a₀ + a₁x + · · · + a_n−1xⁿ⁻¹ : ai ∈ R}.
• The infinite set {1, x , x², . . . , xⁿ , . . . } is a basis for P , the space of all polynomials.

Bases for Rⁿ

Theorem : Every basis for the vector space Rⁿ consists of n vectors.
Theorem : For any vectors v₁, v₂, . . . , v_n ∈ Rⁿ the following conditions are equivalent:

(i) {v₁, v₂, . . . , v_n} is a basis for Rⁿ ;
(ii) {v₁, v₂, . . . , v_n} is a spanning set for Rⁿ ;
(iii) {v_1, v₂, . . . , v_n } is a linearly independent set.

Dimension

Theorem : Any vector space V has a basis. All bases for V are of the same cardinality.

Definition : The dimension of a vector space V , denoted dim V , is the cardinality of its bases.

Remark. By definition, two sets are of the same cardinality if there exists a one-to-one correspondence between their elements.

For a finite set, the cardinality is the number of its elements.

For an infinite set, the cardinality is a more sophisticated notion. For example, Z and R are infinite sets of different cardinalities while Z and Q are infinite sets of the same cardinality.

Examples.

• dim Rⁿ = n
• M₂,₂(R): the space of 2×2 matrices dim _M2,2(R) = 4
• M_m,n(R): the space of m×n matrices dim M_m,n(R) = mn
• P_n : polynomials of degree less than n dim P_n = n
• P : the space of all polynomials dim P = ∞
• {0}: the trivial vector space dim {0} = 0

Problem. Find the dimension of the plane x + 2z = 0 in R³.

The general solution of the equation x + 2z = 0 is

That is, (x , y , z ) = (−2s , t , s ) = t (0, 1, 0) + s (−2, 0, 1).
Hence the plane is the span of vectors v₁ = (0, 1, 0) and v₂ = (−2, 0, 1). These vectors are linearly independent as they are not parallel.
Thus {v₁, v₂} is a basis so that the dimension of the plane is 2.

How to find a basis?

Theorem Let S be a subset of a vector space V .

Then the following conditions are equivalent:

(i) S is a linearly independent spanning set for V , i.e., a basis;
(ii) S is a minimal spanning set for V ;
(iii) S is a maximal linearly independent subset of V .

“Minimal spanning set” means “remove any element from this set, and it is no longer a spanning set”.

“Maximal linearly independent subset” means “add any element of V to this set, and it will become linearly dependent”.

Theorem Let V be a vector space. Then

(i) any spanning set for V can be reduced to a minimal spanning set;
(ii) any linearly independent subset of V can be extended to a maximal linearly independent set.

Equivalently, any spanning set contains a basis, while any linearly independent set is contained in a basis.

Corollary A vector space is finite-dimensional if and only if it is spanned by a finite set.

How to find a basis?

Approach 1. : Get a spanning set for the vector space, then reduce this set to a basis.

Proposition : Let v₀, v₁, . . . , v_k be a spanning set for a vector space V . If v₀ is a linear combination of vectors v₁, . . . , v_k then v₁, . . . , v_k is also a spanning set for V .

Indeed, if v₀ = r₁v₁ + · · · + r_k v_k , then

t₀v₀ + t₁v₁ + · · · + t_k v_k =
= (t₀r₁ + t₁)v₁ + · · · + (t₀r_k + t_k)v_k .

How to find a basis?

Approach 2. : Build a maximal linearly independent set adding one vector at a time.

If the vector space V is trivial, it has the empty basis.

If V ≠ {0}, pick any vector v₁ ≠ 0.

If v₁ spans V , it is a basis. Otherwise pick any vector v₂ ∈ V that is not in the span of v₁.

If v₁ and v₂ span V , they constitute a basis.

Otherwise pick any vector v₃ ∈ V that is not in the span of v₁ and v₂.
And so on. . .

Problem. Find a basis for the vector space V spanned by vectors w₁ = (1, 1, 0), w₂ = (0, 1, 1), w₃ = (2, 3, 1), and w₄ = (1, 1, 1).

To pare this spanning set, we need to find a relation of the form r₁w₁+r₂w₂+r₃w₃+r₄w₄ = 0, where r_i ∈ R are not all equal to zero. Equivalently,

To solve this system of linear equations for r₁, r₂, r₃, r₄, we apply row reduction.

 (reduced row echelon form)

General solution: (r₁, r₂, r₃, r₄)=(−2t , −t , t , 0), t ∈ R.

Particular solution: (r₁, r₂, r₃, r₄) = (2, 1, −1, 0).

Problem. Find a basis for the vector space V spanned by vectors w₁ = (1, 1, 0), w₂ = (0, 1, 1), w₃ = (2, 3, 1), and w₄ = (1, 1, 1).

We have obtained that 2w₁ + w₂ − w₃ = 0.

Hence any of vectors w₁, w₂, w₃ can be dropped.

For instance, V = Span(w₁, w₂, w₄).

Let us check whether vectors w₁, w₂, w₄ are linearly independent:

They are!!! It follows that V = R³ and {w₁, w₂, w₄} is a basis for V.

Vectors v₁ = (0, 1, 0) and v₂ = (−2, 0, 1) are linearly independent.

Problem. Extend the set {v1, v₂} to a basis for R³.
Our task is to find a vector v³ that is not a linear combination of v₁ and v₂.

Then {v₁, v₂, v₃} will be a basis for R³.

Hint 1. v₁ and v₂ span the plane x + 2z = 0.

The vector v₃ = (1, 1, 1) does not lie in the plane x + 2z = 0, hence it is not a linear combination of v₁ and v₂. Thus {v₁, v₂, v₃} is a basis for R³.

Vectors v₁ = (0, 1, 0) and v₂ = (−2, 0, 1) are linearly independent.

Problem. Extend the set {v₁, v₂} to a basis for R³.

Our task is to find a vector v₃ that is not a linear combination of v₁ and v₂.

Hint 2. At least one of vectors e₁ = (1, 0, 0), e₂ = (0, 1, 0), and e₃ = (0, 0, 1) is a desired one.

Let us check that {v₁, v₂, e₁} and {v₁, v₂, e₃} are two bases for R³: