MTH2021 · Linear Algebra with Applications
Span, Linear Independence, Basis & Dimension
Week 4 assembles the core structural toolkit: linear combinations and span, linear (in)dependence, a basis as a minimal spanning set, and dimension. The examinable skill is testing whether vectors are independent or span a space (usually by row reduction), and extracting a basis. These ideas are drilled in Quiz 4 and sit inside Test 2 (Weeks 3–7); the dimension theorem and basis arguments recur throughout the proof-heavy final.
What this chapter covers
- 01Linear combination v = a₁w₁ + … + aᵣwᵣ; span(S) = all linear combinations = the smallest subspace containing S
- 02Linear independence: k₁v₁ + … + kᵣvᵣ = 0 ⇒ all kᵢ = 0; otherwise dependent
- 03A set is dependent ⇔ some vector is a linear combination of the others
- 04Basis = linearly independent + spanning; standard bases of ℝⁿ, Pₙ, ℝ^{2×2}
- 05In an n-vector basis space: more than n vectors ⇒ dependent; fewer than n ⇒ cannot span
- 06Dimension Theorem: every basis of a finite-dimensional V has the same size, dim(V)
- 07dim(ℝⁿ) = n, dim(Pₙ) = n+1, dim(ℝ^{2×2}) = 4; for a subspace W ⊆ V, dim(W) ≤ dim(V), equality ⇔ W = V
- 08Every spanning set contains a basis; every independent set extends to a basis
Testing independence and finding a basis and dimension
- +1Set up the dependence test as a matrix. Place the vectors as rows of A = [[1,2,3],[2,3,4],[3,5,7]] and row-reduce; the span's dimension is the number of nonzero rows in echelon form.
- +1Row-reduce: R2 → R2 − 2R1 = (0,−1,−2); R3 → R3 − 3R1 = (0,−1,−2). Then R3 → R3 − R2 = (0,0,0). Echelon form is [[1,2,3],[0,−1,−2],[0,0,0]] — two nonzero rows.
- +1Conclude dependence. There are only 2 pivots for 3 vectors, so the set is linearly DEPENDENT (rank 2 < 3). Equivalently det(A) = 0.
- +1Exhibit the dependence and a basis. Notice v₁ + v₂ = (3,5,7) = v₃, so v₃ is a linear combination of v₁, v₂. Hence {v₁, v₂} (independent, not parallel) is a basis of the span, and dim span{v₁,v₂,v₃} = 2 (a plane through the origin).
Key terms
- Linear combination
- A sum a₁w₁ + … + aᵣwᵣ of vectors with scalar coefficients aᵢ; the building block of span.
- Span
- span(S) is the set of all linear combinations of vectors in S; it is the smallest subspace containing S.
- Linear independence
- Vectors v₁, …, vᵣ are independent if k₁v₁ + … + kᵣvᵣ = 0 forces every kᵢ = 0; otherwise they are dependent and one is a combination of the others.
- Basis
- A linearly independent set that spans V; every vector of V then has a unique expansion in the basis.
- Dimension
- The common size of every basis of a finite-dimensional space V, written dim(V); dim(ℝⁿ) = n, dim(Pₙ) = n+1, dim(ℝ^{2×2}) = 4.
- Rank (of a set of vectors)
- The dimension of their span, equal to the number of pivots when the vectors are row-reduced; it caps how many of them can be independent.
Span, Linear Independence, Basis & Dimension FAQ
How do I test whether vectors are linearly independent?
Form a matrix from the vectors and row-reduce. If the number of pivots equals the number of vectors, they are independent; if there are fewer pivots, they are dependent. For n vectors in ℝⁿ you can shortcut with the determinant: nonzero ⇒ independent, zero ⇒ dependent.
What is the difference between a spanning set and a basis?
A spanning set reaches every vector but may carry redundancy; a basis is a spanning set with no redundancy — independent and minimal. Every spanning set contains a basis (throw out dependent vectors), and every independent set extends to a basis (add vectors until it spans).
Why does every basis of a space have the same number of vectors?
That is the Dimension Theorem: if one basis has n vectors, no independent set can have more than n and no spanning set fewer than n, so all bases have exactly n vectors. That invariant number is the dimension, and it is what makes dim a well-defined tool.
How do I find the dimension of a span?
Row-reduce the spanning vectors and count the pivots (nonzero rows) — that count is the dimension, and the original vectors corresponding to pivot positions form a basis. In the worked example three vectors gave two pivots, so the span is 2-dimensional.
Where does this show up in assessment?
Span, independence and basis are examined in Quiz 4 and inside Test 2 (Weeks 3–7). The final's handwritten section often asks for a short proof — for instance that removing a dependent vector leaves the span unchanged — so rehearse the two-inclusion argument alongside the computations.
Exam move
Make row reduction your single tool for independence, spanning and basis questions: pivots count the dimension, and pivot-position vectors form a basis. Practise both directions of the theory — trimming a spanning set to a basis and extending an independent set to one — and keep the dimension facts (n, n+1, 4 for ℝⁿ, Pₙ, ℝ^{2×2}) at your fingertips. This is examined in Quiz 4 and Test 2 (Weeks 3–7). When you are unsure whether a set is independent, ask Sia to walk the reduction and interpret the pivots.
Working through Span, Linear Independence, Basis & Dimension in MTH2021? Sia is AskSia’s AI Mathematics tutor — ask any MTH2021 Span, Linear Independence, Basis & Dimension question and get a clear, step-by-step explanation grounded in how MTH2021 is taught and assessed. Read this chapter free, then take your hardest questions to Sia.
