Why the Ambient Zero Vector Is a First Test
When deciding whether a subset is a subspace, begin by asking whether it contains the zero vector of the surrounding space. This check is often quicker than testing every closure condition: if the zero vector is missing, the subset cannot be a subspace. The test applies in any vector space, including spaces of matrices, polynomials, functions, and sequences. The zero vector depends on the ambient space: it is the zero column in \(\mathbb{R}^n\), the zero matrix in a matrix space, and the zero function in a function space.
Throughout this tutorial, spaces and scalars are over \(\mathbb{R}\). A subspace \(U\) of a vector space \(V\) is a nonempty subset of \(V\) closed under vector addition and scalar multiplication: if \(\mathbf{u},\mathbf{v}\in U\) and \(c\in\mathbb{R}\), then \(\mathbf{u}+\mathbf{v}\in U\) and \(c\mathbf{u}\in U\). Closure under linear combinations is an equivalent way to state these closure conditions. The ambient zero vector is written \(\mathbf{0}_V\), or simply \(\mathbf{0}\) when the space is clear.
Proof. Because \(U\) is nonempty, choose \(\mathbf{u}\in U\). Closure under scalar multiplication gives \(0\mathbf{u}\in U\). In a vector space, \(0\mathbf{u}=\mathbf{0}_V\): indeed, \((0+0)\mathbf{u}=0\mathbf{u}+0\mathbf{u}\), while \((0+0)\mathbf{u}=0\mathbf{u}\); cancellation in vector addition yields \(0\mathbf{u}=\mathbf{0}_V\). Hence \(\mathbf{0}_V\in U\). The final claim follows by contraposition: a set lacking the ambient zero vector cannot be a subspace. \(\square\)
This theorem gives a necessary condition, not a complete test. Necessary means that every subspace must pass the check. It does not mean that every subset containing zero is a subspace. After the zero check, a candidate still needs to satisfy the appropriate closure conditions.
Checking a Translated Set
A translated set is formed by adding a fixed vector to every element of a set. If \(W\) is a subspace of \(V\) and \(\mathbf{v}\in V\), the translate of \(W\) by \(\mathbf{v}\) is \( \mathbf{v}+W=\{\mathbf{v}+\mathbf{w}:\mathbf{w}\in W\}\). When \(\mathbf{v}=\mathbf{0}_V\), this is just \(W\). When the translation moves \(W\), the zero-vector test detects the change.
Proof. Suppose first that \(\mathbf{0}_V\in\mathbf{v}+W\). By the definition of the translate, there is some \(\mathbf{w}\in W\) such that \(\mathbf{0}_V=\mathbf{v}+\mathbf{w}\). Therefore \(\mathbf{w}=-\mathbf{v}\). Since \(W\) is closed under scalar multiplication, \(-\mathbf{w}\in W\), and \(-\mathbf{w}=\mathbf{v}\). Hence \(\mathbf{v}\in W\). Conversely, suppose \(\mathbf{v}\in W\). Then \(-\mathbf{v}\in W\), and \(\mathbf{v}+(-\mathbf{v})=\mathbf{0}_V\), so \(\mathbf{0}_V\in\mathbf{v}+W\).
It remains to prove the equality claim. If \(\mathbf{v}\in W\), then for every \(\mathbf{w}\in W\), closure under addition gives \(\mathbf{v}+\mathbf{w}\in W\), so \(\mathbf{v}+W\subseteq W\). For any \(\mathbf{w}\in W\), we also have \(\mathbf{w}-\mathbf{v}\in W\), because \(-\mathbf{v}\in W\) and \(W\) is closed under addition. Thus \(\mathbf{w}=\mathbf{v}+(\mathbf{w}-\mathbf{v})\in\mathbf{v}+W\), giving \(W\subseteq\mathbf{v}+W\). Therefore \(\mathbf{v}+W=W\). If a translate differs from \(W\), it follows that \(\mathbf{v}\notin W\); by the first part it omits \(\mathbf{0}_V\), so the zero-vector test rules it out as a subspace. \(\square\)
Worked Example: A Shifted Line in \(\mathbb{R}^2\)
Consider \(S=\{(t,2t+1):t\in\mathbb{R}\}\). This is the line obtained by translating the line \(W=\{(t,2t):t\in\mathbb{R}\}\) by \(\mathbf{v}=(0,1)\). Determine whether \(S\) is a subspace of \(\mathbb{R}^2\).
To check whether \(\mathbf{0}\in S\), we would need a single value of \(t\) that makes both coordinates zero. The first coordinate requires \(t=0\). Substituting that value into the second coordinate gives \(2(0)+1=1\), not \(0\). There is no such \(t\), so \((0,0)\notin S\). The zero-vector test proves that \(S\) is not a subspace.
This also agrees with the translation theorem. The translating vector \((0,1)\) is not in \(W\): an element of \(W\) with first coordinate \(0\) has \(t=0\), so its second coordinate is \(0\), not \(1\). Thus the translate is shifted away from the zero vector.
Check: The point at \(t=0\) is \((0,1)\), which is in \(S\), but it is not the zero vector. The set contains points and is nonempty; neither fact replaces the requirement to contain \((0,0)\).
Solution Sets with a Nonzero Right-Hand Side
The same check applies to sets defined by linear equations. Let \(\mathbf{A}\in\mathbb{R}^{m\times n}\) and \(\mathbf{b}\in\mathbb{R}^m\), and consider the solution set \(S=\{\mathbf{x}\in\mathbb{R}^n:\mathbf{A}\mathbf{x}=\mathbf{b}\}\). The zero vector can belong to this set only if substituting \(\mathbf{x}=\mathbf{0}\) satisfies the equation. Matrix multiplication gives \(\mathbf{A}\mathbf{0}=\mathbf{0}\), so the right-hand side determines the zero-vector check.
Proof. By definition, \(\mathbf{0}\) belongs to the solution set exactly when \(\mathbf{A}\mathbf{0}=\mathbf{b}\). Each entry of \(\mathbf{A}\mathbf{0}\) is a sum of products of matrix entries with zero coordinates, so every entry is zero; hence \(\mathbf{A}\mathbf{0}=\mathbf{0}\). Therefore \(\mathbf{0}\) is a solution exactly when \(\mathbf{b}=\mathbf{0}\). If \(\mathbf{b}\ne\mathbf{0}\), the solution set omits the ambient zero vector, and the zero-vector test shows it is not a subspace. \(\square\)
Worked Example: A System with a Nonzero Right-Hand Side
Consider the solution set in \(\mathbb{R}^2\) defined by \(\begin{bmatrix}1&2\\3&-1\end{bmatrix}\mathbf{x}=\begin{bmatrix}4\\2\end{bmatrix}\). Decide whether this set can be a subspace.
The right-hand side \(\begin{bmatrix}4\\2\end{bmatrix}\) is nonzero. Substituting the ambient zero vector gives
Thus the zero vector is not a solution, and the solution set is not a subspace. To confirm that the system does have solutions, take \(\mathbf{x}=\begin{bmatrix}2\\1\end{bmatrix}\). Multiplying gives
This trial vector does not solve the second equation, so it is not a valid solution. Instead, take \(\mathbf{x}=\begin{bmatrix}8/7\\10/7\end{bmatrix}\). Then
This confirms that the solution set is nonempty, while the zero-vector test still rules it out as a subspace. Nonemptiness and subspace status are different questions.
When \(\mathbf{b}=\mathbf{0}\), the zero vector passes this particular test. That fact alone does not prove that a candidate set is a subspace: it only means the zero-vector check has not rejected it. Closure must still be established before concluding that a set is a subspace.
Passing the Test Is Not Enough
A subset can contain the zero vector and still fail a closure condition. For instance, in \(\mathbb{R}^2\) consider the union of the two coordinate axes. It contains \((0,0)\), but adding one point from each axis can produce a point outside the union.
Worked Example: A Set That Contains Zero but Is Not a Subspace
Let \(U=\{(x,0):x\in\mathbb{R}\}\cup\{(0,y):y\in\mathbb{R}\}\). Both axes contain \((0,0)\), so \((0,0)\in U\). Now choose \(\mathbf{u}=(1,0)\) and \(\mathbf{v}=(0,1)\). Each vector belongs to \(U\), but
The point \((1,1)\) lies on neither coordinate axis: its first and second coordinates are both nonzero. Therefore \((1,1)\notin U\), so \(U\) is not closed under addition and is not a subspace. Here the zero-vector test passes, but the addition test rejects the set.
A Reliable Order for the Check
The zero-vector test is most useful when treated as the first step in a decision, not as a replacement for the full subspace test. For a proposed subset \(S\) of an ambient vector space \(V\), identify the ambient zero vector and test membership directly. If membership fails, stop: \(S\) is not a subspace. If membership succeeds, continue by checking closure under the operations required by the subspace definition.
For sets given by a parameterization, set the whole vector equal to zero and solve for the parameter values. Every coordinate must vanish for the same choice of parameters. For a translated set \(\mathbf{v}+W\), ask whether \(\mathbf{v}\in W\). For a solution set \(\mathbf{A}\mathbf{x}=\mathbf{b}\), substitute \(\mathbf{x}=\mathbf{0}\), which immediately requires \(\mathbf{b}=\mathbf{0}\). These are all applications of one question: does the ambient zero vector actually belong to the proposed set?
Common Mistake: Treating Nonempty as Containing Zero
A frequent error is to argue that a solution set is a subspace because it has at least one solution. Nonempty is part of the subspace definition, but it does not imply that the zero vector is in the set. For the system in the worked example, \(\begin{bmatrix}8/7\\10/7\end{bmatrix}\) is a solution, so the set is nonempty. However, substituting \(\begin{bmatrix}0\\0\end{bmatrix}\) gives \(\begin{bmatrix}0\\0\end{bmatrix}\), not the required right-hand side \(\begin{bmatrix}4\\2\end{bmatrix}\). The set is therefore not a subspace.
A related error is to set one parameter to zero and assume the resulting point is the zero vector. In the translated line \(S=\{(t,2t+1):t\in\mathbb{R}\}\), choosing \(t=0\) gives \((0,1)\), not \((0,0)\). The correction is to check every coordinate at once: the first coordinate forces \(t=0\), while the second would then equal \(1\). Since the requirements cannot both be met, the zero vector is absent.
Check Your Understanding
For each candidate, first apply the zero-vector test. If it passes, decide whether another subspace condition is needed.
- Can \(\{(x,y)\in\mathbb{R}^2:x+y=5\}\) be a subspace? Explain using the ambient zero vector.
- For \(W=\{(t,3t):t\in\mathbb{R}\}\), determine whether \((2,1)+W\) contains \((0,0)\).
- What condition on \(\mathbf{b}\) is necessary for the solution set of \(\mathbf{A}\mathbf{x}=\mathbf{b}\) to be a subspace of \(\mathbb{R}^n\)?
- Give two vectors in the union of the coordinate axes whose sum is not in that union.
- Explain why containing the zero vector is necessary but not sufficient for being a subspace.
