Dimension of a Vector Space

The content in today's blog is taken from Hans Schnedier and George Phillip Barker's Matrices and Linear Algebra.

Definition 1: Finite-dimensional space

A vector space V is called finite-dimensional if and only if there is a finite family of vectors spanning V.

[For background: see Definition 2 here for vector space; see Definition 4 here for family of vectors; see Definition 3 here for spanning a vector space.]

Theorem 1:

Let V be a nonzero finite-dimensional vector space. Then:
(1) There is a finite basis for V.
(2) All bases for V have the same number of elements.

Proof:

(1) Let V be a nonzero finite-dimensional vector space.

[see Definition 2 here for vector space; see Definition 1 above for finite-dimensional]

(2) Then, there exists a finite family (x¹, ..., xⁿ) that spans V.

[See Definition 1 above; see Definition 4 here for family of vectors; see Definition 3 here for spanning a vector space.]

(3) Then there is a finite basis for V.

[See Corollary 1.1, here; see Definition 2 here for basis]

(4) Let X = the family of vectors (x¹, ..., x^m)

(5) Let Y = the family of vectors (y¹, ..., yⁿ)

(6) Let X,Y be two bases for V.

(7) X is linearly independent [see Definition 2, here]

(8) Y spans V. [see Definition 2, here]

(9) Using Theorem 2 here, we can conclude that m ≤ n.

(10) Since Y is linearly independent and X spans V, we can also conclude that n ≤ m.

(11) Hence, we have shown that m = n.

(12) This proves that all bases for V have the same number of elements.

QED

From Theorem 1, since all bases have the same number of elements, it makes sense to define a notation for this number of elements.

Definition 2: Dimension

Let V ≠ {0} be a finite-dimensional vector space and let X be a basis for V. Then |X| = n is called the dimension of V. We write dim V = n. If V = {0}, we define dim V = 0.

Lemma 2:

Let V be a nonzero finite-dimensional vector space.

Then, if the family of vectors (x¹, ..., x^m) is linearly independent, then there exists the family of vectors (y¹, ..., yⁿ) such that:

(x¹, ..., x^m, y¹, ..., yⁿ) is a basis.

Proof:

(1) Since V is a nonzero finite-dimensional space, there exists a family of vectors Y = (y¹, ..., y^p) such that Y is the basis for V. [see Theorem 1 above]

(2) Since X = (x¹, ..., x^m) is linearly independent, we can apply Theorem 1 here to show that there exists a family of vectors (x¹, ..., x^m, y¹, ..., yⁿ) which is a basis.

QED

Lemma 3:

Let n = dim V where n is greater than 0.

(1) Then every linearly independent family in V has at most n members.
(2) Every family spanning V has at least n members.

Proof:

(1) Let the family of vectors (x¹, ..., x^t) be a linearly independent family of vectors.

(2) By Lemma 2 above, it can be completed to a basis (x¹, ..., x^t, y¹, ..., y^r).

(3) By Theorem 1 above, t + r = n since t + r and n are both bases and all bases have the same number of elements.

(4) Hence, t ≤ n.

(5) Let [[ y¹, ..., y^u ]] = V.

(6) By a previous result (see Corollary 1.1, here), we can reduce (y¹, ..., y^u) to a basis (y¹, ..., y^s)

(7) So that s ≤ u since [[y¹, ..., y^s]] ⊆ [[y¹, ..., y^u]]


(8) Using Theorem 1 above, s = n which means that n ≤ u.

QED

Lemma 4:

Let W be a subspace of a finite-dimensional vector space V.

Then dim W ≤ dim V

Proof:

(1) Let n = dim V

(2) Let W be a nonzero subspace of V. [See Definition 3 here for definition of subspace]

(3) Since W is nonzero, we know that there exists x such that x ≠ 0 and x ∈ W.

(4) Let (x¹, x², ..., x^r) be a linearly independent family of elements of W. [See Definition 1 here for linearly independent]

(5) Then, we know that r ≤ n by Lemma 3 above.

(6) Let m be the largest integer for which there is a linearly independent family (x¹, ..., x^m) with each xⁱ ∈ W.

(7) Using Lemma 3 above, we can see that m ≤ n.

(8) I will now show that (x¹, ..., x^m) is a basis for W.

(9) Since W is nonzero, there exists y such that y ∈ W.

(10) Let Y = the family of vectors (x¹, x², ..., x^m, y).

(11) By the choice of m, this family is linearly dependent since we assumed that m is the largest family of linearly independent vectors.

(12) Using Lemma 4 here, we know that some element of Y is a linear combination of the preceding elements.

(13) But x^j is not a linear combination of xⁱ for i is less than j from step #4 since (x¹, x², ..., x^m) is linearly independent. [See Corollary 3.1, here]

(14) Hence, y is a linear combination of (x¹, ..., x^m).

(15) Thus, (x¹, ..., x^m) spans W since this is true for all y and y can be any element of W. [See step #9 above]

(16) Since (x¹, ..., x^m) is linearly independent and spans W this shows that it is a basis. [See Definition 2 here for definition of basis]

QED

References

• Hans Schneider, George Philip Barker, Matrices and Linear Algebra, 1989.