mindstair

Home / Courses / Linear Algebra / Unit 4

Linear Algebra Unit 4: Vector spaces & subspaces

Vector spaces are the setting for all of linear algebra. Span, independence, basis and dimension describe their structure.

Covers: Span, linear independence, basis, dimension, null & column space, rank

Study guide

A vector space is a set of objects (vectors) closed under addition and scalar multiplication that satisfies the usual algebra rules. Examples include ℝⁿ, polynomials and matrices. A subspace is a subset that contains the zero vector and is closed under addition and scalar multiplication.

The span of a set of vectors is all their linear combinations. A set is linearly independent if the only combination equal to zero uses all zero weights; otherwise one vector is a combination of the others. To test vectors in ℝⁿ, put them as columns of a matrix and row reduce: they're independent exactly when every column has a pivot.

A basis is a linearly independent set that spans the space. Every basis of a space has the same number of vectors, called the dimension: ℝⁿ has dimension n, and polynomials of degree at most 2 have dimension 3. In a space of dimension n, any n independent vectors form a basis.

Two subspaces come from every matrix A. The null space Nul A is all solutions of Ax = 0; its basis comes from the parametric form of the solution, one vector per free variable. The column space Col A is the span of the columns; its basis is the pivot columns of the original A. The rank is dim Col A, the number of pivots. The rank-nullity theorem says rank A + dim Nul A = number of columns.

Every m×n matrix of rank r has four fundamental subspaces: the column space Col A (dimension r, in ℝᵐ), the null space Nul A (dimension n − r, in ℝⁿ), the row space Col Aᵀ (dimension r, in ℝⁿ) and the left null space Nul Aᵀ (dimension m − r, in ℝᵐ). A basis for the row space is the nonzero rows of any echelon form of A. The vectors b in Col A are exactly those satisfying the solvability conditions from row reducing [A | b]. Writing those conditions as rows of a matrix B gives Nul B = Col A.

Row operations don't change the null space or the row space, but they do change the column space. That's why a column space basis must use the pivot columns of the original A, not of its echelon form.

Key ideas

Formulas and rules

rank A + nullity A = n (columns)
Subspace test: contains 0, closed under + and scalar ×

Common mistakes

Words to know

Linearly independent
The only combination equal to 0 is the trivial one (all weights zero).
Basis
A linearly independent set that spans the space.
Rank–nullity theorem
rank A + dim Nul A = number of columns of A.
Basis for the column space
The pivot columns of the original matrix A.
Three subspace conditions
Contains 0, closed under addition, closed under scalar multiplication.
Vector space
A set closed under addition and scalar multiplication obeying the usual rules.
Subspace
Contains 0 and is closed under addition and scalar multiplication.
Span
All linear combinations of a set of vectors.
Linear combination
c₁v₁ + c₂v₂ + … + cₖvₖ
Dimension
The number of vectors in any basis.
Null space (Nul A)
All solutions of Ax = 0.
Column space (Col A)
The span of A's columns.
Rank
dim Col A, the number of pivots.

Sample practice questions

25 more questions like these, from level 1 to 5, are in the practice set.

Question 1 (level 1)

Are ⟨1, 2⟩ and ⟨2, 4⟩ linearly independent?

  1. Yes
  2. Can't tell
  3. Only in R³
  4. No
Show the answer and steps

Answer: D) No

  1. ⟨2, 4⟩ = 2⟨1, 2⟩.
  2. One is a multiple of the other, so they're dependent.
  3. Not independent.
Question 2 (level 1)

Are ⟨1, 3⟩ and ⟨−2, −6⟩ linearly independent?

  1. Only in R³
  2. No
  3. Yes
  4. Can't tell
Show the answer and steps

Answer: B) No

  1. Check the 2 × 2 determinant: (1)(−6) − (3)(−2) = 0.
  2. Zero means one is a multiple of the other (dependent).
Question 3 (level 3)

A is 4 × 6 with rank 3. Nullity?

  1. 1
  2. 3
  3. 4
  4. 2
Show the answer and steps

Answer: B) 3

  1. Rank–nullity: rank + nullity = number of columns.
  2. Nullity = 6 − 3 = 3.