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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
- The span of vectors is every linear combination of them.
- Vectors are linearly independent if the only combination giving 0 uses all-zero weights.
- A basis is an independent spanning set; every basis of a space has the same size, its dimension.
- Column space: span of A's columns. Null space: all solutions of Ax = 0.
- Rank–nullity: rank A + dim Nul A = number of columns.
Formulas and rules
Common mistakes
- Using the reduced matrix's columns as the column-space basis. Use the original pivot columns.
- Forgetting a subspace must contain the zero vector.
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.
Are ⟨1, 2⟩ and ⟨2, 4⟩ linearly independent?
- Yes
- Can't tell
- Only in R³
- No
Show the answer and steps
Answer: D) No
- ⟨2, 4⟩ = 2⟨1, 2⟩.
- One is a multiple of the other, so they're dependent.
- Not independent.
Are ⟨1, 3⟩ and ⟨−2, −6⟩ linearly independent?
- Only in R³
- No
- Yes
- Can't tell
Show the answer and steps
Answer: B) No
- Check the 2 × 2 determinant: (1)(−6) − (3)(−2) = 0.
- Zero means one is a multiple of the other (dependent).
A is 4 × 6 with rank 3. Nullity?
- 1
- 3
- 4
- 2
Show the answer and steps
Answer: B) 3
- Rank–nullity: rank + nullity = number of columns.
- Nullity = 6 − 3 = 3.