From Generators to a Subspace
A collection of vectors can generate a larger collection by taking every possible linear combination of them. The resulting set is called their span. It packages two ideas together: the generators themselves belong to the span, and adding or scaling vectors in the span never takes us outside it. This makes the span a natural way to describe the subspace built from a given set.
Let \(V\) be a vector space over a field \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\). A subspace of \(V\) is a subset that contains the zero vector and is closed under addition and scalar multiplication. For a set \(S\subseteq V\), a finite linear combination of elements of \(S\) is an expression \(c_1\mathbf{s}_1+\cdots+c_k\mathbf{s}_k\), where \(k\) is a nonnegative integer, each \(\mathbf{s}_i\in S\), and each \(c_i\in\mathbb{F}\). When \(k=0\), the expression is the empty sum, defined to be \(\mathbf{0}\). Even if \(S\) has infinitely many elements, each individual linear combination uses only finitely many of them.
The braces in this notation identify the generators as a set; the displayed vectors are often given as a list so that each one can be used in a linear combination. The span consists of the combinations, not just the original generators. For instance, it includes the zero vector by choosing all coefficients to be zero. If \(S=\{\mathbf{v}\}\), its span is \(\{c\mathbf{v}:c\in\mathbb{F}\}\), a line through the origin when \(\mathbf{v}\ne\mathbf{0}\), and \(\{\mathbf{0}\}\) when \(\mathbf{v}=\mathbf{0}\).
Why the Span Is the Smallest Such Subspace
Proof. First, the empty sum shows that \(\mathbf{0}\in\operatorname{span}S\). Let \(\mathbf{x},\mathbf{y}\in\operatorname{span}S\). By definition, there are finite expressions \(\mathbf{x}=\sum_{i=1}^{k}a_i\mathbf{s}_i\) and \(\mathbf{y}=\sum_{j=1}^{\ell}b_j\mathbf{t}_j\), with all \(\mathbf{s}_i,\mathbf{t}_j\in S\). Adding these expressions gives \(\mathbf{x}+\mathbf{y}=\sum_{i=1}^{k}a_i\mathbf{s}_i+\sum_{j=1}^{\ell}b_j\mathbf{t}_j\), again a finite linear combination of elements of \(S\). Hence \(\mathbf{x}+\mathbf{y}\in\operatorname{span}S\). For any \(c\in\mathbb{F}\), \(c\mathbf{x}=\sum_{i=1}^{k}(ca_i)\mathbf{s}_i\), so \(c\mathbf{x}\in\operatorname{span}S\). These facts show that the span is a subspace.
For each \(\mathbf{s}\in S\), the expression \(1\mathbf{s}\) shows that \(\mathbf{s}\in\operatorname{span}S\); if \(S=\varnothing\), this inclusion holds vacuously. Finally, let \(U\) be any subspace of \(V\) with \(S\subseteq U\). Every finite linear combination of elements of \(S\) belongs to \(U\), because \(U\) is closed under scalar multiplication and addition; the empty sum belongs to \(U\) because it contains \(\mathbf{0}\). Therefore \(\operatorname{span}S\subseteq U\), proving the smallest-subspace claim. \(\square\)
This result gives a way to describe a generated subspace without listing all of its vectors one by one: specify the generators and allow arbitrary coefficients. In familiar spaces, we can often simplify that description into coordinates, polynomial coefficients, or matrix entries.
Proof. There is at least one subspace containing \(S\), namely \(V\), so the intersection in the statement is defined. Since the theorem shows that \(\operatorname{span}S\) is itself one of these subspaces, the intersection is contained in \(\operatorname{span}S\). The theorem also shows that \(\operatorname{span}S\) is contained in every subspace in the intersection, so it is contained in their intersection. The two inclusions prove equality. \(\square\)
Spans in \(\mathbb{R}^n\)
For column vectors \(\mathbf{v}_1,\dots,\mathbf{v}_p\in\mathbb{R}^n\), their span consists of all vectors obtained by choosing real coefficients \(c_1,\dots,c_p\). If the vectors are placed as the columns of a matrix \(\mathbf{A}\), then \(\mathbf{A}\mathbf{c}=c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\), where \(\mathbf{c}=(c_1,\dots,c_p)^\top\). Thus the possible outputs of this matrix-vector product give a parameterized description of the span.
Worked Example: Two Generators in \(\mathbb{R}^3\)
Describe the span of \(\mathbf{v}_1=(1,0,1)^\top\) and \(\mathbf{v}_2=(0,1,1)^\top\) both as linear combinations and as a coordinate equation.
Choose arbitrary real coefficients \(a\) and \(b\). Multiplying each vector by its coefficient and adding gives
Therefore the span is \(\{(a,b,a+b)^\top:a,b\in\mathbb{R}\}\). Writing the coordinates as \(x,y,z\), each such vector satisfies \(z=x+y\). Conversely, if \((x,y,z)^\top\) satisfies \(z=x+y\), then
So the span is exactly the plane described by \(z=x+y\). Check: for any \(a,b\), the coordinates of the computed vector satisfy \((a+b)=a+b\), as required. The converse calculation expresses every vector on that plane using the two generators.
Spans in Polynomial Spaces
The same definition applies in a vector space whose elements are polynomials. Let \(\mathcal{P}_2(\mathbb{R})\) denote the real polynomials of degree at most two, including the zero polynomial. A polynomial in this space has the form \(c_0+c_1x+c_2x^2\). To find the span of polynomial generators, multiply each generator by an arbitrary scalar and collect like powers of \(x\).
Worked Example: A Span in \(\mathcal{P}_2(\mathbb{R})\)
Describe \(\operatorname{span}\{1+x,\;x+x^2\}\) by its polynomial coefficients.
Let \(a,b\in\mathbb{R}\). Expanding the linear combination gives
Thus every polynomial in the span has constant coefficient \(a\), coefficient of \(x\) equal to \(a+b\), and coefficient of \(x^2\) equal to \(b\). In other words, the span can be written as
To check the description in both directions, the expansion shows that any linear combination has \(c_1=c_0+c_2\). Conversely, if a polynomial has that coefficient relation, set \(a=c_0\) and \(b=c_2\). Then \(a+b=c_0+c_2=c_1\), so the polynomial equals \(a(1+x)+b(x+x^2)\). Check: substituting these coefficients into the expansion recovers \(c_0+c_1x+c_2x^2\) exactly.
Spans in Matrix Spaces
Matrices of a fixed size form a vector space: addition is entrywise, and scalar multiplication multiplies every entry by the scalar. A span in a matrix space is therefore computed just as in \(\mathbb{R}^n\), except that the generated vectors are matrices. Combining entries in the same position makes the resulting structure visible.
Worked Example: A Span in \(\mathbb{R}^{2\times2}\)
Describe the span of
For arbitrary \(a,b\in\mathbb{R}\), compute each scaled matrix and add corresponding entries:
Consequently, the span consists of all matrices of the form \(\begin{bmatrix}a&b\\b&-a\end{bmatrix}\). These are precisely the symmetric \(2\times2\) real matrices whose diagonal entries sum to zero. Indeed, the displayed form is symmetric and its diagonal entries have sum \(a+(-a)=0\). Conversely, any symmetric matrix with diagonal sum zero has the form \(\begin{bmatrix}r&s\\s&-r\end{bmatrix}\), which is \(r\mathbf{A}+s\mathbf{B}\). Check: substituting \(a=r\) and \(b=s\) in the computed combination reproduces every such matrix.
A Common Misreading
Pitfall: treating the span as a union of scalar multiples. For example, take \(\mathbf{u}=(1,0)^\top\) and \(\mathbf{v}=(0,1)^\top\) in \(\mathbb{R}^2\). A mistaken description would include only vectors of the form \(a\mathbf{u}\) or \(b\mathbf{v}\), and would therefore leave out \((1,1)^\top\). But the definition allows a linear combination using both generators: \(1\mathbf{u}+1\mathbf{v}=(1,0)^\top+(0,1)^\top=(1,1)^\top\). The correct span is \(\{(a,b)^\top:a,b\in\mathbb{R}\}\), because every such vector equals \(a\mathbf{u}+b\mathbf{v}\). A span allows the generators to be added together, not merely scaled one at a time.
The smallest-subspace property is useful whenever a problem gives a set of vectors, polynomials, or matrices and asks for the subspace they generate. The defining formula supplies a direct description, while the theorem explains why no smaller subspace can contain all the generators: every subspace containing them must also contain all their finite linear combinations.
Check Your Understanding
Use the definition of span and the smallest-subspace property to justify your answers.
- What is \(\operatorname{span}\{\mathbf{0}\}\) in \(\mathbb{R}^3\)? Explain using arbitrary scalar multiples.
- Describe the span of \((1,1,0)^\top\) and \((0,1,1)^\top\) using two real parameters.
- For \(a,b\in\mathbb{R}\), expand \(a(1-x)+b x^2\) and state the resulting coefficient relation among constant, linear, and quadratic terms.
- Write a general element of the span of \(\begin{bmatrix}1&1\\0&0\end{bmatrix}\) and \(\begin{bmatrix}0&0\\1&1\end{bmatrix}\).
- Why must every subspace containing a set \(S\) also contain \(\operatorname{span}S\)?
