One Closure Condition, Any Finite List
A subspace must be closed under addition and scalar multiplication. Those requirements concern one or two vectors at a time, but they also guarantee closure under any finite sequence of additions and scalar multiplications. The finite linear-combination test packages these requirements into one statement: choose any finite list of vectors in the subset and any real coefficients, and the resulting linear combination must remain in the subset.
Let \(V\) be a vector space over \(\mathbb{R}\), and let \(S\subseteq V\). A finite linear combination of vectors \(\mathbf{v}_1,\dots,\mathbf{v}_p\) is an expression \(a_1\mathbf{v}_1+\cdots+a_p\mathbf{v}_p\), where \(p\geq 1\) and \(a_1,\dots,a_p\in\mathbb{R}\). A subspace of \(V\) is a nonempty subset that is itself a vector space under the addition and scalar multiplication inherited from \(V\). In the test below, “nonempty” is a separate requirement; we do not include a combination of zero vectors.
The quantifiers matter. The condition is not about one particular list, or only about a list of two vectors. It must hold for every positive finite list from \(S\), with every choice of real coefficients. For instance, a one-term combination tests scalar multiplication, while a two-term combination can test addition. The proof explains why these cases together also give closure for a list of any finite length.
Why the Test Is Equivalent to the Subspace Conditions
First, finite combinations can be built by repeatedly applying addition and scalar multiplication. The following elementary fact makes that step explicit.
Proof. Let \(p\geq 1\), let \(\mathbf{v}_1,\dots,\mathbf{v}_p\in S\), and let \(a_1,\dots,a_p\in\mathbb{R}\). We use induction on \(p\). For \(p=1\), closure under scalar multiplication gives \(a_1\mathbf{v}_1\in S\). Assume the result holds for lists of length \(p\). For a list of length \(p+1\), the induction hypothesis gives \(a_1\mathbf{v}_1+\cdots+a_p\mathbf{v}_p\in S\), and scalar closure gives \(a_{p+1}\mathbf{v}_{p+1}\in S\). Addition closure then gives their sum in \(S\). Thus the claim holds for every positive integer \(p\). \(\square\)
Now prove both directions of the theorem. The forward direction uses the inherited operations of a subspace. The reverse direction uses the nonempty hypothesis to choose a vector, then uses particular finite combinations to recover the subspace conditions.
Proof. Suppose first that \(S\) is a subspace of \(V\). Its inherited operations are closed: if \(\mathbf{u},\mathbf{v}\in S\), then \(\mathbf{u}+\mathbf{v}\in S\), and if \(a\in\mathbb{R}\) and \(\mathbf{u}\in S\), then \(a\mathbf{u}\in S\). The lemma therefore shows that every finite linear combination of vectors in \(S\) belongs to \(S\).
Conversely, suppose \(S\) is nonempty and contains every finite linear combination of its vectors. Choose \(\mathbf{u}\in S\). Taking the one-term combination \(0\mathbf{u}\) shows that \(\mathbf{0}\in S\). For any \(\mathbf{u},\mathbf{v}\in S\), the two-term combination \(1\mathbf{u}+1\mathbf{v}\) belongs to \(S\), so \(S\) is closed under addition. For any \(a\in\mathbb{R}\) and \(\mathbf{u}\in S\), the one-term combination \(a\mathbf{u}\) belongs to \(S\), so \(S\) is closed under scalar multiplication. The vector-space axioms for \(S\) now follow from those in \(V\): the operations on \(S\) are the same operations, restricted to elements of \(S\). Hence \(S\) is a subspace of \(V\). \(\square\)
This proof also shows why nonemptiness must be stated. The closure condition concerns vectors already in \(S\); if there are no such vectors, it cannot supply a starting vector. Once there is a vector to use, the coefficient \(0\) produces the zero vector. There is no need to add zero as a separate requirement to the finite-combination condition.
Applying the Test to Vectors
Worked Example: The Span of Two Vectors
Let \(\mathbf{u}=\begin{bmatrix}1\\0\\1\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}0\\1\\1\end{bmatrix}\) in \(\mathbb{R}^3\), and define \(S=\{a\mathbf{u}+b\mathbf{v}:a,b\in\mathbb{R}\}\). Verify the finite-combination condition.
The set is nonempty because \(\mathbf{0}=0\mathbf{u}+0\mathbf{v}\in S\). Take any positive integer \(p\), any \(\mathbf{x}_1,\dots,\mathbf{x}_p\in S\), and any real numbers \(c_1,\dots,c_p\). By the definition of \(S\), each \(\mathbf{x}_j\) has the form \(\alpha_j\mathbf{u}+\beta_j\mathbf{v}\) for some \(\alpha_j,\beta_j\in\mathbb{R}\). Therefore, using distributivity,
The last expression is a real linear combination of \(\mathbf{u}\) and \(\mathbf{v}\), so it belongs to \(S\). The test proves that \(S\) is a subspace.
For a concrete check, take \(\mathbf{x}_1=\mathbf{u}+2\mathbf{v}=\begin{bmatrix}1\\2\\3\end{bmatrix}\), \(\mathbf{x}_2=3\mathbf{u}-\mathbf{v}=\begin{bmatrix}3\\-1\\2\end{bmatrix}\), and coefficients \(2\) and \(-1\). Then
The result is in \(S\), with coefficients \(-1\) and \(5\) on its defining vectors. The calculation checks this particular combination; the arbitrary-\(p\) argument above is what proves closure for every finite list.
In an application, it is often useful to work from how the set is described. If every element of \(S\) has a form that is preserved when finite linear combinations are taken, the test can establish subspace status without checking each vector-space axiom separately.
Worked Example: Diagonal Matrices
Let \(V=\mathbb{R}^{2\times2}\), the vector space of real \(2\times2\) matrices with entrywise addition and scalar multiplication, and let \(D\) be the set of diagonal matrices in \(V\). Show that \(D\) is a subspace using the finite-combination test.
A diagonal matrix has the form \(\begin{bmatrix}a&0\\0&b\end{bmatrix}\), with \(a,b\in\mathbb{R}\). The set is nonempty, since it contains \(\begin{bmatrix}0&0\\0&0\end{bmatrix}\). Let \(p\geq1\), take any matrices \(\mathbf{A}_1,\dots,\mathbf{A}_p\in D\), and choose real coefficients \(c_1,\dots,c_p\). Write \(\mathbf{A}_j=\begin{bmatrix}a_j&0\\0&b_j\end{bmatrix}\). Then
Both off-diagonal entries remain zero, so the sum is diagonal and belongs to \(D\). The finite-combination test proves that \(D\) is a subspace of \(V\).
For example, choose \(\mathbf{A}_1=\begin{bmatrix}1&0\\0&2\end{bmatrix}\), \(\mathbf{A}_2=\begin{bmatrix}-1&0\\0&3\end{bmatrix}\), and coefficients \(2\) and \(3\). Then
The result is diagonal, as the general calculation requires.
Finding a Combination That Fails
To prove that a nonempty subset is not a subspace, it is enough to find one finite list from the set and one choice of coefficients whose linear combination lies outside the set. This is the other practical strength of the test: failure needs only a single counterexample.
Worked Example: The Union of the Coordinate Axes
Consider \(S=\left\{\begin{bmatrix}x\\y\end{bmatrix}\in\mathbb{R}^2:x=0\text{ or }y=0\right\}\), the union of the two coordinate axes. This set is nonempty; for example, \(\begin{bmatrix}1\\0\end{bmatrix}\in S\). Decide whether it is a subspace.
Both \(\mathbf{e}_1=\begin{bmatrix}1\\0\end{bmatrix}\) and \(\mathbf{e}_2=\begin{bmatrix}0\\1\end{bmatrix}\) belong to \(S\). Use the two-term combination with coefficients \(1\) and \(1\):
The result is not in \(S\), because its first coordinate is nonzero and its second coordinate is also nonzero. Thus \(S\) fails the finite-combination condition and is not a subspace. Notice that each individual scalar multiple of a vector on either axis remains on that axis, but that alone does not establish closure under combinations of two different vectors.
Common Pitfall: Checking Only One Kind of Combination
A common mistake is to check addition and conclude that a set is closed under linear combinations. For a concrete counterexample, let \(Q=\left\{\begin{bmatrix}x\\y\end{bmatrix}\in\mathbb{R}^2:x\geq0,\ y\geq0\right\}\). Sums of vectors in \(Q\) remain in \(Q\), because sums of nonnegative coordinates are nonnegative. But \(\begin{bmatrix}1\\0\end{bmatrix}\in Q\), while the combination with coefficient \(-1\) is
So \(Q\) is not a subspace. The correction is to test arbitrary real coefficients, not just positive ones or sums with coefficient \(1\). The finite-combination condition includes negative and zero coefficients as well as positive ones.
The converse shortcut is also unsafe: scalar multiples alone do not guarantee closure under addition. In the union-of-axes example, both individual axes contain all scalar multiples of each of their vectors, but the sum of a nonzero vector on one axis and a nonzero vector on the other can lie outside the union. The single finite-combination test prevents both kinds of incomplete checking.
Why the Finite Form Is Useful
The finite-combination condition records exactly what it means for a set to be compatible with the linear structure of its ambient vector space. If a subset passes the test, then it contains zero, is closed under addition, and is closed under every real scalar multiple. If it fails, the particular combination that fails identifies precisely where that compatibility breaks.
When the set has a simple description, the most efficient proof usually has three parts: establish nonemptiness, take an arbitrary finite list from the set, and show that its arbitrary linear combination still has the required form. A proof for a single hand-picked list is only a check, not a closure argument. Conversely, a non-subspace claim needs only one verified failure, with the chosen vectors shown to be in the set and their combination shown to be outside it.
Check Your Understanding
For each question, identify whether the task calls for an arbitrary-combination proof or a specific counterexample.
- State the finite linear-combination test for a nonempty subset of a real vector space.
- Why is nonemptiness included as a separate hypothesis when the combination length satisfies \(p\geq1\)?
- Let \(S=\{(a,2a):a\in\mathbb{R}\}\subseteq\mathbb{R}^2\). Show that every finite linear combination of elements of \(S\) belongs to \(S\).
- Give a finite linear combination that shows \(\{(x,y)\in\mathbb{R}^2:xy=0\}\) is not a subspace.
- Explain why closure under addition alone does not establish the finite linear-combination condition.
