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Span



Definition
Geometric Interpretation
Spanning Sets
Testing Whether a Vector Is in a Span
Testing Whether a Set Spans Rⁿ
Span and Column Space
Redundancy and Reduction
Span in Abstract Vector Spaces
Summary: Diagnostic Questions about Span



Everything Reachable by Linear Combinations

The span of a set of vectors is the collection of all their linear combinations — every vector that can be built by adding scaled copies of the given vectors. It is always a subspace, and it is the smallest subspace containing the original set. Span is the other half of what makes a basis: the half that guarantees complete coverage.



Definition

Given vectors v1,v2,…,vk\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k in a vector space VV, their span is the set of all linear combinations:

Span{v1,…,vk}={c1v1+c2v2+⋯+ckvk:c1,…,ck∈R}\text{Span}\{\mathbf{v}_1, \dots, \mathbf{v}_k\} = \{c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k : c_1, \dots, c_k \in \mathbb{R}\}


Every choice of scalars c1,…,ckc_1, \dots, c_k produces one vector in the span. As the scalars range over all real numbers, the span sweeps out an entire subspace of VV.

The span is always a subspace — the smallest subspace of VV that contains all of v1,…,vk\mathbf{v}_1, \dots, \mathbf{v}_k. Adding two linear combinations of these vectors produces another linear combination, and scaling a linear combination by a scalar produces another, so both closure conditions hold automatically.

By convention, the span of the empty set is {0}\{\mathbf{0}\}. With no vectors to combine, the only reachable point is the zero vector (the combination with no terms).

Geometric Interpretation

In R2\mathbb{R}^2 and R3\mathbb{R}^3, the span of a set of vectors is a flat subspace whose shape depends on how many independent vectors the set contains.

The span of a single nonzero vector v\mathbf{v} is the line {tv:t∈R}\{t\mathbf{v} : t \in \mathbb{R}\} — a one-dimensional subspace passing through the origin in the direction of v\mathbf{v}.

The span of two non-parallel vectors is a plane through the origin — a two-dimensional subspace. If the two vectors happen to be parallel, one is a scalar multiple of the other, and the span collapses to a line.

The span of three vectors in R3\mathbb{R}^3 that do not all lie in a single plane is all of R3\mathbb{R}^3. If the three vectors are coplanar, the span is a plane. If two of the three are parallel, the span may be only a line.

The pattern is consistent: the dimension of the span equals the number of independent vectors in the set, regardless of how many total vectors are present. Redundant vectors — those already in the span of the others — add nothing to the reach.
Set in ℝⁿ Span dimension Geometric shape Note
1 nonzero vector 1 line through the origin basis for the line
2 parallel vectors 1 line through the origin one is a scalar multiple of the other — the pair is redundant
2 non-parallel vectors 2 plane through the origin basis for the plane
3 coplanar vectors in ℝ³ 2 plane through the origin the third vector lies in the span of two — redundant
3 non-coplanar vectors in ℝ³ 3 all of ℝ³ basis for ℝ³

Spanning Sets

A set of vectors spans a vector space VV if Span{v1,…,vk}=V\text{Span}\{\mathbf{v}_1, \dots, \mathbf{v}_k\} = V — every vector in VV can be written as a linear combination of the set. When this holds, the set is called a spanning set for VV.

In Rn\mathbb{R}^n, a spanning set must contain at least nn vectors. Fewer than nn vectors cannot reach every direction — their span is a proper subspace of dimension less than nn. A spanning set with exactly nn independent vectors is a basis. A spanning set with more than nn vectors contains redundancies that can be removed.

The standard basis {e1,…,en}\{\mathbf{e}_1, \dots, \mathbf{e}_n\} spans Rn\mathbb{R}^n with no redundancy. The set {e1,e2,e1+e2}\{\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_1 + \mathbf{e}_2\} also spans R2\mathbb{R}^2, but the third vector is redundant — it lies in the span of the first two. Removing it leaves a basis.

Testing Whether a Vector Is in a Span

The question "Is b\mathbf{b} in Span{v1,…,vk}\text{Span}\{\mathbf{v}_1, \dots, \mathbf{v}_k\}?" asks whether there exist scalars c1,…,ckc_1, \dots, c_k such that c1v1+⋯+ckvk=bc_1\mathbf{v}_1 + \cdots + c_k\mathbf{v}_k = \mathbf{b}. This is a linear system: arrange the vectors as columns of a matrix A=[v1  ⋯  vk]A = [\mathbf{v}_1 \; \cdots \; \mathbf{v}_k] and check whether Ac=bA\mathbf{c} = \mathbf{b} is consistent.

Row reduce the augmented matrix [A∣b][A \mid \mathbf{b}]. If a row of the form [0  0  ⋯  0∣d][0 \; 0 \; \cdots \; 0 \mid d] with d≠0d \neq 0 appears, the system is inconsistent and b\mathbf{b} is not in the span.

Example: In the Span


Is b=(5,3,7)\mathbf{b} = (5, 3, 7) in Span{(1,0,1),(0,1,2)}\text{Span}\{(1, 0, 1), (0, 1, 2)\}? Form and reduce:

(105013127)→R3−R1(105013022)→R3−2R2(10501300−4)\begin{pmatrix} 1 & 0 & 5 \\ 0 & 1 & 3 \\ 1 & 2 & 7 \end{pmatrix} \xrightarrow{R_3 - R_1} \begin{pmatrix} 1 & 0 & 5 \\ 0 & 1 & 3 \\ 0 & 2 & 2 \end{pmatrix} \xrightarrow{R_3 - 2R_2} \begin{pmatrix} 1 & 0 & 5 \\ 0 & 1 & 3 \\ 0 & 0 & -4 \end{pmatrix}


The last row reads 0=−40 = -4, a contradiction. So b\mathbf{b} is not in the span.

Example: Not in the Span Versus In the Span


Changing the target to b=(5,3,11)\mathbf{b} = (5, 3, 11) and repeating:

(1050131211)→R3−R1(105013026)→R3−2R2(105013000)\begin{pmatrix} 1 & 0 & 5 \\ 0 & 1 & 3 \\ 1 & 2 & 11 \end{pmatrix} \xrightarrow{R_3 - R_1} \begin{pmatrix} 1 & 0 & 5 \\ 0 & 1 & 3 \\ 0 & 2 & 6 \end{pmatrix} \xrightarrow{R_3 - 2R_2} \begin{pmatrix} 1 & 0 & 5 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \end{pmatrix}


No contradiction. The solution is c1=5c_1 = 5, c2=3c_2 = 3, so (5,3,11)=5(1,0,1)+3(0,1,2)(5, 3, 11) = 5(1, 0, 1) + 3(0, 1, 2).

Testing Whether a Set Spans Rⁿ

The question "Does {v1,…,vk}\{\mathbf{v}_1, \dots, \mathbf{v}_k\} span Rn\mathbb{R}^n?" asks whether Ac=bA\mathbf{c} = \mathbf{b} is consistent for every possible right-hand side b∈Rn\mathbf{b} \in \mathbb{R}^n, where AA is the matrix with the vectors as columns.

Row reduce AA. The columns span Rn\mathbb{R}^n if and only if every row of the echelon form contains a pivot. If any row is entirely zero, then for some choices of b\mathbf{b} the augmented system will produce a [0  ⋯  0∣d][0 \; \cdots \; 0 \mid d] contradiction, meaning those b\mathbf{b} are unreachable.

For the square case — nn vectors in Rn\mathbb{R}^n — spanning reduces to a determinant test: the columns span Rn\mathbb{R}^n if and only if det⁡(A)≠0\det(A) \neq 0. This is equivalent to independence when the count matches the dimension, which is why the spanning and independence conditions coincide for sets of exactly nn vectors in Rn\mathbb{R}^n.

Fewer than nn vectors can never span Rn\mathbb{R}^n. An n×kn \times k matrix with k<nk < n has at most kk pivots, so at least n−kn - k rows of the echelon form will be zero. More than nn vectors can span Rn\mathbb{R}^n, but only if the set contains at least nn independent vectors among them — the extras are redundant.

Span and Column Space

The column space of a matrix AA is exactly the span of its columns:

Col(A)=Span{a1,a2,…,an}\text{Col}(A) = \text{Span}\{\mathbf{a}_1, \mathbf{a}_2, \dots, \mathbf{a}_n\}


This identity connects the abstract concept of span to the concrete question of system solvability. The system Ax=bA\mathbf{x} = \mathbf{b} has a solution if and only if b\mathbf{b} lies in the column space of AA — that is, if and only if b\mathbf{b} is in the span of the columns.

The rank of AA equals the dimension of the column space, which equals the number of independent columns. The pivot columns of the echelon form identify which original columns form a basis for the column space. The non-pivot columns are redundant — they lie in the span of the pivot columns.

This also explains why row reduction is the universal computational tool for span questions. Every question about span — membership, spanning, independence — translates into a question about the column space of some matrix, and row reduction answers all of them.

Redundancy and Reduction

A spanning set may contain more vectors than necessary. A vector vj\mathbf{v}_j in a spanning set is redundant if removing it does not shrink the span:

Span{v1,…,vk}=Span{v1,…,vj−1,vj+1,…,vk}\text{Span}\{\mathbf{v}_1, \dots, \mathbf{v}_k\} = \text{Span}\{\mathbf{v}_1, \dots, \mathbf{v}_{j-1}, \mathbf{v}_{j+1}, \dots, \mathbf{v}_k\}


This happens exactly when vj\mathbf{v}_j is already a linear combination of the other vectors — it contributes no new directions.

Removing all redundant vectors from a spanning set produces a basis: a minimal spanning set with no waste. The process can be carried out systematically via row reduction. Arrange the vectors as columns, reduce, and identify the pivot columns. The original vectors corresponding to pivot positions form a basis for the span. The non-pivot columns are the redundant ones.

For example, if four vectors in R3\mathbb{R}^3 are given and row reduction reveals pivots in columns 11, 22, and 44, then the original first, second, and fourth vectors form a basis for the span. The third vector is a combination of the first two.

This reduction process is always possible in finite-dimensional spaces: every spanning set can be trimmed to a basis, and every independent set can be extended to one. The basis sits at the exact boundary between "too few vectors to span" and "too many vectors to be independent."

Span in Abstract Vector Spaces

The definition of span — the set of all linear combinations — applies in any vector space, not just Rn\mathbb{R}^n.

In the polynomial space P2\mathcal{P}_2, the span of {1,x,x2}\{1, x, x^2\} is all of P2\mathcal{P}_2, since every polynomial a+bx+cx2a + bx + cx^2 of degree at most 22 is a linear combination of these three. The set is independent and has the right count (3=dim⁡(P2)3 = \dim(\mathcal{P}_2)), so it is the standard basis.

A subtler example is Span{1,sin⁡2x,cos⁡2x}\text{Span}\{1, \sin^2 x, \cos^2 x\} in the space of continuous functions. This set has three elements, but the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 provides a dependence relation: 1⋅1+(−1)sin⁡2x+(−1)cos⁡2x=01 \cdot 1 + (-1)\sin^2 x + (-1)\cos^2 x = 0. The span is therefore two-dimensional, not three-dimensional. Any two of the three functions form a basis for the span — the third is redundant.

In abstract spaces, the column-matrix approach is unavailable. Testing whether a specific vector lies in a span, or whether a set spans the whole space, requires working directly from the definition: write the target as a combination and check whether the resulting equation has a solution. The algebraic structure of the particular space — polynomial coefficients, function identities, matrix entries — determines how this check proceeds.
Vector space Spanning set Result Reason
𝒫₂ (polynomials of degree ≤ 2) {1, x, x²} spans all of 𝒫₂; dimension 3 three independent vectors — no relations among them; standard basis
C(ℝ) (continuous real-valued functions) {1, sin²x, cos²x} spans a 2-dimensional subspace the identity sin²x + cos²x = 1 forces 1 to be a combination of the others

Summary: Diagnostic Questions about Span

Every span question — whether a given vector is reachable, whether a set covers Rn\mathbb{R}^n, which vectors are redundant, whether a set is a basis — reduces to building a matrix from the given vectors, row-reducing it, and inspecting the result. What changes between questions is which part of the reduced form to look at: the augmented column for membership, the rows for spanning, the columns for redundancy, both for a basis. The table below collects these questions side by side, showing what to set up, what to look for in the reduced form, and what a positive answer means in each case.
Question Setup Reduce and check A "yes" means
Is b in Span{v₁, …, vₖ}? augmented matrix [A | b] with the vᵢ as columns of A no contradiction row [0 ⋯ 0 | d ≠ 0] appears b is a linear combination of the vᵢ; the solution gives the coefficients
Does {v₁, …, vₖ} span ℝⁿ? n × k matrix A with the vᵢ as columns every row of the echelon form contains a pivot every vector in ℝⁿ is in the span (no unreachable directions)
Which vᵢ are redundant? n × k matrix A with the vᵢ as columns identify non-pivot columns of the echelon form the non-pivot vᵢ lie in the span of the pivot ones; removing them leaves a basis
Is {v₁, …, vₙ} a basis of ℝⁿ? square n × n matrix A with the vᵢ as columns (k = n) every row and every column has a pivot (equivalently, det A ≠ 0) the set spans ℝⁿ with no redundancy — minimal spanning set