From the Definition to a Proof Plan
A subset of a vector space may look like a smaller vector space, but resemblance is not enough: its vectors must remain inside the subset whenever the vector-space operations are performed. The subspace test turns that requirement into a compact proof plan. Instead of checking every vector-space axiom from the beginning, we use the fact that the operations and axioms are already available in the ambient space. The work is to show that the subset is nonempty and closed under addition and scalar multiplication.
Let \(\mathbb{F}\) be a field and let \(V\) be a vector space over \(\mathbb{F}\). A subset \(W\subseteq V\) is a subspace of \(V\) if \(W\), with the addition and scalar multiplication inherited from \(V\), is itself a vector space over \(\mathbb{F}\). For the polynomial and matrix examples below, the field is \(\mathbb{R}\). Closure means that performing an operation on elements of \(W\) produces another element of \(W\).
The test contains three requirements, not just two. Nonemptiness matters because a set with no elements cannot contain the zero vector or satisfy the vector-space axioms. In practice, nonemptiness is often shown by noting that \(\mathbf{0}\in W\). If the defining condition of \(W\) makes that fact immediate, it is still useful to state it explicitly before checking closure.
Why These Checks Are Enough
A subspace must contain the zero vector and additive inverses. These facts follow from the vector-space axioms for \(W\). In the other direction, if a nonempty subset is closed under the two operations, those operations always stay inside it; the other vector-space axioms are inherited from \(V\).
Proof. Since \(W\) is a vector space, it has a zero vector for its addition. The addition in \(W\) is inherited from \(V\), so this zero vector must be the zero vector \(\mathbf{0}\) of \(V\); hence \(\mathbf{0}\in W\). If \(\mathbf{u}\in W\), the vector-space axioms in \(W\) give an additive inverse \(\mathbf{w}\in W\) with \(\mathbf{u}+\mathbf{w}=\mathbf{0}\). The additive inverse in \(V\) is unique, so \(\mathbf{w}=-\mathbf{u}\). Thus \(-\mathbf{u}\in W\). \(\square\)
Proof. Suppose first that \(W\) is a subspace. It is a vector space, so it is nonempty. Its addition and scalar multiplication are the operations inherited from \(V\), and vector-space operations on elements of \(W\) remain in \(W\). Therefore \(W\) is closed under addition and scalar multiplication.
Conversely, suppose \(W\) is nonempty and has both closure properties. Choose \(\mathbf{w}\in W\). Scalar closure, applied with \(c=0\), gives \(0\mathbf{w}=\mathbf{0}\in W\). If \(\mathbf{u}\in W\), then scalar closure with \(c=-1\) gives \(-\mathbf{u}\in W\). Addition closure ensures that sums of elements of \(W\) remain in \(W\), and scalar closure ensures the same for scalar multiples. Thus the inherited operations are defined within \(W\), \(W\) has a zero vector and additive inverses, and all remaining vector-space axioms hold because they already hold in \(V\). Hence \(W\) is a subspace. \(\square\)
Sometimes it is convenient to combine the two closure checks. The following equivalent form is especially useful when a defining condition involves a linear combination of two arbitrary elements.
Proof. If \(W\) is a subspace, then \(a\mathbf{u}\in W\) and \(b\mathbf{v}\in W\) by scalar closure, so their sum \(a\mathbf{u}+b\mathbf{v}\) is in \(W\) by addition closure. For the converse, suppose \(W\) is nonempty and has the stated property. For any \(\mathbf{u},\mathbf{v}\in W\), choose \(a=b=1\); then \(\mathbf{u}+\mathbf{v}\in W\). For any \(\mathbf{u}\in W\), choose some \(\mathbf{v}\in W\), which is possible because \(W\) is nonempty, and set \(a=c\), \(b=0\). The property gives \(c\mathbf{u}+0\mathbf{v}=c\mathbf{u}\in W\) for every \(c\in\mathbb{F}\). The subspace test now proves that \(W\) is a subspace. \(\square\)
Worked Examples in Polynomial Spaces
Worked Example: Polynomials That Vanish at One
Consider \(W=\{p\in\mathbb{R}_2[x]:p(1)=0\}\), where \(\mathbb{R}_2[x]\) is the vector space of real polynomials of degree at most \(2\). Prove that \(W\) is a subspace.
The zero polynomial satisfies \(0(1)=0\), so \(W\) is nonempty. Take arbitrary \(p,q\in W\). By the definition of \(W\), \(p(1)=0\) and \(q(1)=0\). Evaluating their sum gives
Thus \(p+q\in W\). For any \(c\in\mathbb{R}\), scalar multiplication gives
so \(cp\in W\). The subspace test proves the claim.
Check with specific polynomials: Let \(p(x)=x-1\) and \(q(x)=x^2-1\). Their values at \(1\) are \(p(1)=1-1=0\) and \(q(1)=1-1=0\). Their sum is \(p(x)+q(x)=x^2+x-2\), and \((p+q)(1)=1+1-2=0\). Also, for \(c=-2\), \((cp)(x)=-2x+2\), so \((cp)(1)=-2+2=0\). Both computations agree with the general closure proof.
The important feature of this proof is not the particular number \(1\). It is that the defining condition is preserved by the operations. The proof starts with arbitrary elements \(p,q\in W\), uses exactly what membership tells us, and then verifies membership for the sum and scalar multiple. That order makes the argument apply to every element of the set rather than just to a few examples.
Worked Examples in Matrix Spaces
Worked Example: Symmetric Two-by-Two Matrices
Let \(W=\{\mathbf{A}\in\mathbb{R}^{2\times2}:\mathbf{A}^{\top}=\mathbf{A}\}\), the set of symmetric real \(2\times2\) matrices. Prove that \(W\) is a subspace of \(\mathbb{R}^{2\times2}\).
The zero matrix \(\mathbf{O}\) satisfies \(\mathbf{O}^{\top}=\mathbf{O}\), so \(W\) is nonempty. Take \(\mathbf{A},\mathbf{B}\in W\). Then \(\mathbf{A}^{\top}=\mathbf{A}\) and \(\mathbf{B}^{\top}=\mathbf{B}\). Using the transpose rules,
Hence \(\mathbf{A}+\mathbf{B}\in W\). For \(c\in\mathbb{R}\),
so \(c\mathbf{A}\in W\). The subspace test gives the result.
Check with specific matrices: Take
Both are symmetric because their off-diagonal entries agree. Their sum is
which is symmetric. Scaling the first matrix by \(-2\) gives
which is also symmetric. These calculations check the operations in this instance; the argument with arbitrary \(\mathbf{A}\) and \(\mathbf{B}\) establishes closure for the whole set.
Worked Example: A Polynomial Condition That Fails
Consider \(S=\{p\in\mathbb{R}_2[x]:p(0)=1\}\). This set is nonempty because the constant polynomial \(p(x)=1\) belongs to it: \(p(0)=1\). But it is not a subspace. Indeed, scalar multiplication by \(0\) gives the zero polynomial, and its value at \(0\) is \(0\), not \(1\). More explicitly, \(0p(x)=0\), so \((0p)(0)=0\not=1\), and \(0p\notin S\).
There is also a direct addition failure: for \(p(x)=1\) and \(q(x)=1\), both lie in \(S\), but \((p+q)(0)=2\), so \(p+q\notin S\). Either failed closure check is enough to rule out a subspace. The defining equation fixes a nonzero value, so the set cannot contain the zero polynomial.
Worked Example: Singular Matrices Are Not a Subspace
In \(\mathbb{R}^{2\times2}\), let \(S\) be the set of singular matrices, meaning matrices that are not invertible. The matrices
are singular: each has a zero row, so neither can have an inverse. Their sum is
The identity matrix is invertible, since \(\mathbf{I}_2\mathbf{I}_2=\mathbf{I}_2\). Thus \(\mathbf{A},\mathbf{B}\in S\) but \(\mathbf{A}+\mathbf{B}\notin S\). Closure under addition fails, so the singular matrices do not form a subspace. This example shows why a condition on each matrix separately must still be tested against the operations; a collection of individually special matrices need not be closed.
Common Pitfall: Checking Addition but Not Scalars
Closure under addition alone does not pass the subspace test. For a concrete counterexample, let \(C\subseteq\mathbb{R}_2[x]\) consist of polynomials whose coefficients are all nonnegative. The zero polynomial is in \(C\). If \(p,q\in C\), each coefficient of \(p+q\) is the sum of two nonnegative numbers, so \(p+q\in C\). However, \(p(x)=1\) belongs to \(C\), while \((-1)p(x)=-1\) has a negative constant coefficient and is not in \(C\). Therefore \(C\) is not a subspace.
A flawed proof might say, “The set contains zero and is closed under addition, so it is a subspace.” The correction is to check scalar multiplication as a separate condition, including negative and zero scalars. In this example, multiplication by \(-1\) exposes the failure immediately. More generally, checking only a few positive scalars does not establish closure under every scalar in the field.
A Reliable Proof Routine
For a subset described by a condition, a closure proof is most dependable when it follows the quantifiers in the test. First identify the ambient vector space and the scalar field. Then establish nonemptiness, often by checking the zero vector or zero matrix. Next take arbitrary elements of the proposed subset, state what their membership means, and use that information to check the defining condition after addition. Finally, take an arbitrary scalar and an arbitrary element of the subset and check the condition after scalar multiplication.
Name the ambient vector space and the field; specify the subset being tested.
Verify that a particular element, often the zero vector, satisfies the membership condition.
Take arbitrary elements of the subset and prove their sum still satisfies its defining condition.
Take an arbitrary field scalar and an arbitrary subset element, and verify that their product remains in the subset.
This routine matters because it turns a definition into a proof that can be checked line by line. A successful example establishes the universal closure statements for arbitrary elements; a single counterexample to any required condition disproves the claim. Keeping those two tasks distinct—proving every required case or finding one failure—makes subspace arguments precise and efficient.
Check Your Understanding
For each proposed subset, identify what must be checked before concluding that it is a subspace.
- Let \(W=\{p\in\mathbb{R}_3[x]:p(2)=0\}\). How would you prove nonemptiness and closure under addition and scalar multiplication?
- For a subset of \(\mathbb{R}^{2\times2}\), why does showing that the zero matrix belongs to it not by itself prove that it is a subspace?
- Suppose \(W\) is nonempty and \(a\mathbf{u}+b\mathbf{v}\in W\) for every \(\mathbf{u},\mathbf{v}\in W\) and \(a,b\in\mathbb{R}\). Explain how to obtain addition closure and scalar closure.
- Give a specific pair of singular \(2\times2\) matrices whose sum is invertible, and use it to identify the failed subspace condition.
- Why is it necessary to check negative scalars when testing a subset of a real vector space for scalar closure?
