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Lesson 4: Basis and Dimension

Estimated time: 35-45 minutes

Learning Objectives

Basis

Basis: A set of vectors B = {v_1, ..., v_k} is a basis for a subspace H if: (1) B is linearly independent, and (2) H = Span(B). A basis is a minimal spanning set -- no vector is redundant.

Example

The standard basis for R^3: {e_1, e_2, e_3} = {(1,0,0), (0,1,0), (0,0,1)}. Independent and spans all of R^3.

Finding Bases from Matrices

Basis for Col(A): The pivot columns of the original matrix A form a basis for the column space.

Basis for Nul(A): Row reduce A, express free variables as parameters, and the direction vectors from the parametric form constitute a basis for the null space.

Worked Example

A = [1 2 3; 2 4 7; 1 2 4]. RREF: [1 2 0; 0 0 1; 0 0 0]. Pivots in columns 1 and 3.

Basis for Col(A): Columns 1 and 3 of the original A: {(1,2,1), (3,7,4)}.

Basis for Nul(A): Free variable x_2 = t. From RREF: x_1 = -2t, x_3 = 0. Null space basis: {(-2, 1, 0)}.

Dimension

Dimension: The dimension of a subspace H, written dim(H), is the number of vectors in any basis for H. All bases for a given subspace have the same number of vectors.

dim(R^n) = n. dim({0}) = 0. dim of a line through origin = 1. dim of a plane through origin = 2.

Rank and Nullity

Rank: rank(A) = dim(Col(A)) = number of pivot columns.

Nullity: nullity(A) = dim(Nul(A)) = number of free variables.

Rank Theorem: For an m x n matrix A: rank(A) + nullity(A) = n (number of columns).

Example

From the example above: A is 3x3. Rank = 2 (two pivots). Nullity = 1 (one free variable). 2 + 1 = 3 = n. ✓

Check Your Understanding

1. What is dim(R^5)?

Answer: 5. The standard basis has 5 vectors.

2. A 4x6 matrix has rank 3. What is its nullity?

Answer: nullity = 6 - 3 = 3 (by the Rank Theorem).

3. Can {(1,2), (2,4)} be a basis for R^2?

Answer: No. (2,4) = 2(1,2), so they are linearly dependent. A basis must be independent.

4. An invertible 5x5 matrix has what rank?

Answer: Rank 5 (full rank). All 5 columns are pivot columns. Nullity = 0.

Key Takeaways