Finite-dimensional vector spaces are vector spaces over real or complex fields, which are spanned by a finite number of vectors in the basis of a vector space. Let V(F) be a vector space over field F (F could be a field of real numbers or complex numbers). Then a subset S of V(F) is said to be the basis of V(F) if

• The elements of S are linearly independent.
• Each vector in V(F) is a linear combination of elements of S; that is, S spans V(F).

Dimension of a vector space could be considered the number of degrees of freedom of a particle in space. For example, a particle moving in a three-dimensional space has three degrees of freedom. We can express the position of that particle at any instant by a point

P(x, y, z) ∈ R³. If V(F) = R³, then the dimension of vector space V(F) will be 3.

Let us discuss more about finite-dimensional vector spaces, their properties and some significant results.

Definition of Finite Dimensional Vector Spaces

Let V(F) be a vector space over field F (where F = R or C) is said to be a finite-dimensional vector space or finitely generated vector space, if the subset S of V, which spans V(F), has a finite number of elements. That is, if S = { 𝛼₁, 𝛼₂, 𝛼₃, …, 𝛼_n} is finite and linearly independent. Every x ∈ V is such that

x = a₁𝛼₁ + a₂𝛼₂ + … + a_n𝛼_n where a_i’s are scalars in F. Then V(F) is called finite-dimensional vector space.

Dimension of a Finite Dimensional Vector Space

The dimension of a finite-dimensional vector space V(F) is the number of elements in the basis of V. That is, the number of elements in the linearly independent subset S of V, which spans V. It is denoted by dim V. If the number of elements in S is n then dim V = n.

A vector space whose dimension is not finite is known as infinite-dimensional vector space. For example, The vector space F[x] of all polynomials over a field F is an infinite-dimensional vector space.

Properties of Finite Dimensional Vector Spaces

Following are some important results related to finite-dimensional vector spaces.

• The Existence Theorem: A linearly independent subset S of vectors of a finite-dimensional vector space V always exists, which forms the basis of V.
• The Dimension Theorem: If V is a finite-dimensional vector space over real or complex field F, then any two bases of V have the same number of elements.
• The Extension Theorem: If V is a finite-dimensional vector space over real or complex field F, then any linearly independent subset of vectors of V either forms a basis of V or can be extended to form a basis of V.
• Let V is a finite-dimensional vector space over a real or complex field F such that dim V=n, then each set which (n + 1) elements of more is a linearly dependent subset of V.
• Let V is a finite-dimensional vector space over a real or complex field F such that dim V=n, then any subset of V containing n linearly independent vectors of V forms a basis of V.
• Dimension of Subspace: The dimension of a subspace of a finite-dimensional vector space cannot exceed the dimension of the vector space.

If V(F) is a finite-dimensional vector over real or complex fields, and W be any subspace of V. Then dim W ≤ dim V = n. Moreover, V = W if and only if dim V = dim W.

Solved Examples on Finite Dimensional Vector Spaces

Example 1:

Prove that the vectors 𝛼₁ = (1, 0, –1), 𝛼₂ = (1, 2, 1) and 𝛼₃ = (0, –3, 2) form a basis of V₃(R).

Solution:

Let S = {𝛼₁, 𝛼₂, 𝛼₃}, if S is a basis of V₃(R) then we have to prove that S is linearly independent and S spans V₃(R), that is L(S) = V₃(R) where L(S) contains all the linear combinations of elements of S.

(i) S is linearly independent

a₁𝛼₁ + a₂𝛼₂ + a₃𝛼_3,= 0 where a₁, a₂, a₃ are scalars in R

⇒ a₁ (1, 0, –1) + a₂ (1, 2, 1) + a₃ (0, –3, 2)_,= 0

⇒ a₁ + a₂ + 0a₃ = 0,

0a₁ + 2a₂ – 3a₃ = 0,

–a₁ + a₂ + 2a₃ = 0

are the system of linear equations. The coefficient matrix of the above equation is

Here, |A| = 1(4 – 3) –1( – 3 – 0) + 0 ( 0 + 2) = –2 ≠ 0

Rank of A = 3 = number of unknown constants

Hence, a₁ = a₂ = a₃ = 0

Consequently, 𝛼₁, 𝛼₂, 𝛼₃ are linearly independent

⇒ S is linearly independent.

(ii) L(S) = V₃(R)

Since S ⊆ V₃(R) ⇒ L(S) ⊆ V₃(R), we just need to prove that V₃(R) ⊆ L(S)

Let (x, y, z) be arbitrary in V₃(R) where x, y, z ∈ R

Let (x, y, z) = a₁𝛼₁ + a₂𝛼₂ + a₃𝛼₃ where a₁, a₂, a₃ are scalars in R

⇒ (x, y, z) = a₁ (1, 0, –1) + a₂ (1, 2, 1) + a₃ (0, –3, 2)

⇒ (x, y, z) = (a₁ + a₂, 2a₂ – 3a₃, –a₁ + a₂ + 2a₃)

⇒ x = a₁ + a₂; y = 2a₂ – 3a₃; z = –a₁ + a₂ + 2a₃ ……..(1)

We get,

a₁ = (7x – 2y – 3z)/10; a₂ = (3x + 2y + 3z)/10; a₃ = (x – y + z)/10

Clearly, a₁, a₂, a₃ ∈ R satisfies the system of equations in (1). Hence, every element of V₃(R) can be expressed as a linear combination of vectors 𝛼₁, 𝛼₂, 𝛼₃

Therefore, L(S) = V₃(R)

⇒ S is a basis of V₃(R)

Since, the number of elements in S is finite; hence V₃(R) is finite-dimensional vector space.

Example 2:

Show that the vectors (1, 2), (3, 4) form a basis of R².

Solution:

Let S = { (1, 2), (3, 4)}. We show that S is linearly independent.

Let a and b be any scalars in R.

Then, a (1, 2) + b (3, 4) = 0

⇒ a + 3b = 0

2a + 4b = 0

Solving both the equations we get a = b = 0.

Thus S is linearly independent.

By theorem, “Let V is a finite-dimensional vector space over real or complex field F such that dim V=n, then any subset of V containing n linearly independent vectors of V forms a basis of V.”

Since dim R² = 2, then S is a basis of R².