What a Counterexample Must Show
A claim that something is true for every subspace, every vector, or every linear map has a demanding logical form: one example that violates the claim is enough to disprove it. The useful question is not merely whether an object looks unusual, but which required condition it fails. For subspaces, test membership of the zero vector and closure under addition and scalar multiplication. For linear maps, test additivity and scalar homogeneity directly. Small examples often make the failed condition visible without a long calculation.
We work over \(\mathbb{R}\). A subset \(U\) of a real vector space \(V\) is a subspace if it is itself a real vector space using the addition and scalar multiplication inherited from \(V\). In practice, this means that it contains the zero vector and is closed under addition and scalar multiplication.
Proof. Suppose first that \(U\) is a subspace. Its zero vector is the same zero vector as in \(V\), so \(\mathbf{0}\in U\). Since the operations on \(U\) are inherited from \(V\), adding two elements of \(U\) gives an element of \(U\), and multiplying an element of \(U\) by any real scalar gives an element of \(U\). Conversely, suppose the three conditions hold. The addition and scalar multiplication on \(U\) are restrictions of those on \(V\), so all the vector-space axioms involving these operations hold on \(U\) because they hold in \(V\). The first condition supplies the zero vector in \(U\); closure under scalar multiplication gives \((-1)\mathbf{u}\in U\) for every \(\mathbf{u}\in U\), which is an additive inverse of \(\mathbf{u}\). Thus \(U\) is a vector space with the inherited operations, so it is a subspace. \(\square\)
To refute a subspace claim, it is enough to exhibit one failure of the test. Checking whether the zero vector belongs is often quickest. If it does belong, test closure using specific vectors in the proposed set; a failure for one pair or one scalar is decisive. Passing a few checks does not prove a set is a subspace, because the test requires the conditions to hold for every permitted vector and scalar.
Translated Lines and Nonhomogeneous Constraints
Worked Example: A Translated Line That Is Not a Subspace
Consider the line \(L\) in \(\mathbb{R}^2\) given by \(L=\left\{\begin{bmatrix}t\\1\end{bmatrix}:t\in\mathbb{R}\right\}\). Decide whether \(L\) is a subspace.
Every point of \(L\) has second coordinate \(1\), whereas the zero vector has second coordinate \(0\). Therefore \(\begin{bmatrix}0\\0\end{bmatrix}\notin L\), so the subspace test already shows that \(L\) is not a subspace.
Closure under addition also fails. The vector \(\mathbf{u}=\begin{bmatrix}0\\1\end{bmatrix}\) is in \(L\), and \(\mathbf{u}+\mathbf{u}=\begin{bmatrix}0\\2\end{bmatrix}\). Its second coordinate is \(2\), so it is not in \(L\). The calculation checks the failure directly; the missing zero vector was already enough to settle the question.
Check: The proposed counterexample uses a vector that really is in \(L\), since it has the required form with \(t=0\). Its sum is not in \(L\), since no value of \(t\) can change the second coordinate from \(2\) to \(1\).
The issue is not that the set is a line. A line through the origin can be a subspace; a translated line may not be. For example, the line described above is shifted away from the origin. A geometric label such as “line” does not replace the subspace test.
Worked Example: A Nonhomogeneous Constraint
Let \(S=\left\{\begin{bmatrix}x\\y\\z\end{bmatrix}\in\mathbb{R}^3:x-2y+z=4\right\}\). Test the claim that \(S\) is a subspace.
The zero vector does not satisfy the constraint: \(0-2(0)+0=0\ne4\). Hence \(\begin{bmatrix}0\\0\\0\end{bmatrix}\notin S\), so \(S\) is not a subspace.
There is also a direct addition failure. Choose \(\mathbf{u}=\begin{bmatrix}4\\0\\0\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}0\\-2\\0\end{bmatrix}\). Substitution gives \(4-2(0)+0=4\) for \(\mathbf{u}\), and \(0-2(-2)+0=4\) for \(\mathbf{v}\), so both belong to \(S\). But
Thus \(\mathbf{u}+\mathbf{v}\notin S\), which explicitly violates closure under addition.
Check: Both chosen vectors satisfy the original equation, and their sum does not. The nonzero right-hand side also excludes the zero vector, so either test independently disproves the subspace claim.
This example illustrates a useful way to inspect a constraint. An equation such as \(x-2y+z=0\) is homogeneous: substituting the zero vector gives a true equation. Replacing the right-hand side by \(4\) makes it nonhomogeneous, and the zero vector no longer satisfies it. The zero-vector check is a quick test, not a claim that every set described by an equation is a subspace exactly when its right-hand side is zero; the full subspace test remains the general criterion.
Testing a Formula-Defined Map
A function \(T:V\to W\) between real vector spaces is linear if it preserves addition and scalar multiplication: \(T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})\) for all \(\mathbf{u},\mathbf{v}\in V\), and \(T(c\mathbf{u})=cT(\mathbf{u})\) for all \(\mathbf{u}\in V\) and \(c\in\mathbb{R}\). In a counterexample, one failed equation is enough to disprove linearity.
Proof. If \(T\) is linear, preservation of addition and scalar multiplication are exactly the defining requirements, so both equations hold for every allowed input. Conversely, if both equations hold for every allowed input, then \(T\) preserves addition and scalar multiplication; by the definition of a linear map, \(T\) is linear. \(\square\)
Worked Example: A Map That Fails Additivity
Define \(T:\mathbb{R}^2\to\mathbb{R}^2\) by \(T\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}x+y\\xy\end{bmatrix}\). Test whether \(T\) is linear.
To test additivity, choose \(\mathbf{u}=\begin{bmatrix}1\\0\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}0\\1\end{bmatrix}\). First, \(\mathbf{u}+\mathbf{v}=\begin{bmatrix}1\\1\end{bmatrix}\), so
Separately,
Therefore \(T(\mathbf{u})+T(\mathbf{v})=\begin{bmatrix}2\\0\end{bmatrix}\), which is not equal to \(T(\mathbf{u}+\mathbf{v})=\begin{bmatrix}2\\1\end{bmatrix}\). Additivity fails, so \(T\) is not linear.
Check: The two input vectors sum to \(\begin{bmatrix}1\\1\end{bmatrix}\). Applying the stated formula to that sum gives second coordinate \(1\), while adding the separately computed outputs gives second coordinate \(0\). This difference is the required witness.
A useful feature of this example is that \(T\) does map the zero vector to the zero vector: \(T\left(\begin{bmatrix}0\\0\end{bmatrix}\right)=\begin{bmatrix}0\\0\end{bmatrix}\). That fact alone does not establish linearity. It checks only one input, while additivity and homogeneity impose conditions across many inputs.
Worked Example: A Map That Fails Scalar Homogeneity
Define \(G:\mathbb{R}^2\to\mathbb{R}^2\) by \(G\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}x\\|y|\end{bmatrix}\). Test scalar homogeneity using a negative scalar.
Take \(\mathbf{w}=\begin{bmatrix}0\\1\end{bmatrix}\) and \(c=-1\). Then \(c\mathbf{w}=\begin{bmatrix}0\\-1\end{bmatrix}\), and
On the other hand, \(G(\mathbf{w})=\begin{bmatrix}0\\|1|\end{bmatrix}=\begin{bmatrix}0\\1\end{bmatrix}\), so
Scalar homogeneity fails, and therefore \(G\) is not linear.
Check: The absolute value in the second coordinate gives \(|-1|=1\), while multiplying the output by \(-1\) gives second coordinate \(-1\). The unequal results verify the failure.
Common Mistakes and a Reliable Strategy
Mistake: treating every line as a subspace. A tempting but incorrect calculation is to take \(\begin{bmatrix}0\\1\end{bmatrix}\in L\) and claim that multiplying it by \(2\) stays in \(L\) because “a line is closed under scaling.” In fact, \(2\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}0\\2\end{bmatrix}\notin L\). The correction is to distinguish a line through the origin from a translated line and then test the actual set’s defining condition.
Mistake: checking only positive scalars. For the map \(G\), a positive scalar does not expose the absolute-value problem: with \(c=2\) and \(\mathbf{w}=\begin{bmatrix}0\\1\end{bmatrix}\), both \(G(c\mathbf{w})\) and \(cG(\mathbf{w})\) equal \(\begin{bmatrix}0\\2\end{bmatrix}\). But the required condition covers every real scalar, including \(-1\), which gives the mismatch above. When a formula contains an absolute value, a negative scalar is a natural test.
For a universal claim, plan the counterexample around the condition you want to break. Use an element of the proposed set when testing closure, and use a specific pair of inputs or a specific scalar when testing a map. Then write both sides of the relevant equation and compute them fully. A persuasive counterexample verifies that its inputs meet the hypotheses and that its outputs genuinely differ; intuition alone is not enough.
For a subspace claim, choose zero membership, addition, or scalar closure. For a linearity claim, choose additivity or scalar homogeneity.
Try zero coordinates, standard basis vectors, or the scalar \(-1\) when the formula makes those choices informative.
Substitute into the set definition or compute both sides of the required map identity. A mismatch disproves the claim; a match for one choice does not prove it.
Check Your Understanding
For each question, state the condition you are testing and verify the relevant calculation.
- Is \(\left\{\begin{bmatrix}t\\t+1\end{bmatrix}:t\in\mathbb{R}\right\}\) a subspace of \(\mathbb{R}^2\)? Test zero membership.
- Find two vectors in \(\left\{\begin{bmatrix}x\\y\end{bmatrix}:x+y=2\right\}\) whose sum is not in that set.
- For \(F\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}x\\x+y\end{bmatrix}\), test additivity using \(\begin{bmatrix}1\\0\end{bmatrix}\) and \(\begin{bmatrix}0\\1\end{bmatrix}\).
- For \(H(x)=|x|\) on \(\mathbb{R}\), test scalar homogeneity with \(x=1\) and \(c=-1\).
- Explain why one failed linearity identity is enough to conclude that a function is not linear.
