An Axiom Audit
To decide whether a set with proposed operations is a real vector space, test the operations against the axioms rather than relying on how familiar the set looks. The key is to keep the set, addition, and scalar multiplication together as one candidate structure. If even one required condition fails, that structure is not a vector space. If an equation is in doubt, choose specific vectors and scalars and calculate both sides separately.
Let \(V\) be a set with proposed addition \(+\) and scalar multiplication, written \(a\mathbf{v}\) for \(a\in\mathbb{R}\) and \(\mathbf{v}\in V\). The operations must take their values in \(V\), and the following conditions must hold for all \(\mathbf{u},\mathbf{v},\mathbf{w}\in V\) and \(a,b\in\mathbb{R}\).
Closure is part of the audit: an operation that sends allowed inputs outside \(V\) is not an operation on \(V\) of the required kind. The zero vector and additive inverses are existence requirements: a candidate zero must work for every vector, and every vector must have an inverse in \(V\). The other conditions are equations required to hold for every allowed choice of inputs.
A useful order is to check closure first, then addition, then scalar multiplication and distributivity. This order does not change which axioms matter; it often makes a failure quicker to find. In particular, write down the proposed operations before assuming that familiar symbols behave in their usual way.
Confirm that adding two elements of \(V\), or multiplying an element of \(V\) by a real scalar, produces an element of \(V\).
Test associativity and commutativity, then find an element that is an additive identity for every vector and verify that every vector has an additive inverse.
Test compatibility with scalar multiplication by a product and the identity-scalar condition.
Compare the two sides of each law using the proposed operations, not assumed ordinary operations.
How a Failed Check Gives a Conclusion
Proof. Assume the proposed structure is a real vector space. By definition, each axiom holds for all elements of \(V\) and all real scalars. In particular, it must hold for any specific allowed inputs we choose. If a calculation exhibits allowed inputs for which the two sides of one required equation differ, that equation does not hold for all inputs, contradicting the assumption that the structure is a vector space. Hence the proposed structure is not a vector space. \(\square\)
This is a one-way shortcut: one failed equation settles the question, but a few successful test values do not prove that an equation holds for all inputs. Also distinguish universal equations from existence conditions. To disprove an addition identity, it is not enough to show that one proposed element fails; one must show that no element of \(V\) can serve as the identity.
Worked Example: A Set That Is Not Closed Under Addition
Let \(V=\{(t,t^2):t\in\mathbb{R}\}\subseteq\mathbb{R}^2\), with the usual vector addition and scalar multiplication. Test whether it is a real vector space.
Closure under addition is required. The vector \((1,1)\) belongs to \(V\), since it has the form \((t,t^2)\) with \(t=1\). Adding it to itself gives
For \((2,2)\) to belong to \(V\), its second coordinate would have to equal the square of its first coordinate. But \(2^2=4\ne2\), so \((2,2)\notin V\). Thus \(V\) is not closed under addition and is not a real vector space.
Check: the failure uses two vectors in \(V\), but their sum is outside \(V\). This directly violates the addition-closure requirement, so there is no need to test the remaining axioms.
Worked Example: A Modified Scalar Operation
Take \(V=\mathbb{R}^2\), use ordinary addition, and define scalar multiplication by \(c\odot(x,y)=(cx,y)\) for each real \(c\). Check the distributive law \((a+b)\odot\mathbf{v}=a\odot\mathbf{v}+b\odot\mathbf{v}\).
Choose \(a=b=1\) and \(\mathbf{v}=(0,1)\). The left side is
The right side is
Since \((0,1)\ne(0,2)\), this instance of distributivity fails. The proposed structure is not a real vector space. Both sides are in \(\mathbb{R}^2\), so this is not a closure failure: the specific failure is the distributive law for a sum of scalars.
Check the Operation That Is Actually Defined
Worked Example: Modified Addition with Ordinary Scalar Multiplication
On \(V=\mathbb{R}\), define \(x\oplus y=x+y+1\), but use ordinary scalar multiplication \(c\odot x=cx\). Check the distributive law \((a+b)\odot x=(a\odot x)\oplus(b\odot x)\).
Choose \(a=b=0\) and \(x=0\). The left side is
The right side is
The two sides differ, so this structure is not a real vector space. The calculation also shows why the proposed operations must be used literally: even though the scalar operation is ordinary multiplication, the symbol \(\oplus\) is not ordinary addition.
It is worth checking what the additive identity would have to be. If \(e\in\mathbb{R}\) is an identity for \(\oplus\), then \(x\oplus e=x\) for every \(x\), so \(x+e+1=x\), which requires \(e=-1\). This candidate does work on both sides: \(x\oplus(-1)=x\) and \((-1)\oplus x=x\). The structure nevertheless fails because the distributive law above fails. Passing one axiom does not compensate for failing another.
Valid Operations Can Look Unfamiliar
Modified formulas do not automatically make a vector space invalid. One way to verify an unfamiliar structure is to relate its operations to those of a known vector space by a bijection. The following result states precisely why this method works.
Proof. Because \(\phi^{-1}\) maps elements of \(W\) into \(V\), both proposed operations are closed in \(V\). For any \(\mathbf{x},\mathbf{y},\mathbf{z}\in V\), applying \(\phi\) to \((\mathbf{x}\oplus\mathbf{y})\oplus\mathbf{z}\) gives \((\phi(\mathbf{x})+\phi(\mathbf{y}))+\phi(\mathbf{z})\), while applying it to \(\mathbf{x}\oplus(\mathbf{y}\oplus\mathbf{z})\) gives \(\phi(\mathbf{x})+(\phi(\mathbf{y})+\phi(\mathbf{z}))\). These are equal by associativity in \(W\); since \(\phi\) is injective, the two elements of \(V\) are equal. Commutativity follows in the same way from \(\phi(\mathbf{x})+\phi(\mathbf{y})=\phi(\mathbf{y})+\phi(\mathbf{x})\) in \(W\). The element \(\phi^{-1}(\mathbf{0}_W)\) is an additive identity, because its image under \(\phi\) is \(\mathbf{0}_W\). The element \(\phi^{-1}(-\phi(\mathbf{x}))\) is an additive inverse of \(\mathbf{x}\), because the sum of their images is \(\mathbf{0}_W\). For scalar multiplication, applying \(\phi\) to \(a\odot(b\odot\mathbf{x})\) gives \(a(b\phi(\mathbf{x}))=(ab)\phi(\mathbf{x})\), the image of \((ab)\odot\mathbf{x}\); applying it to \(1\odot\mathbf{x}\) gives \(1\phi(\mathbf{x})=\phi(\mathbf{x})\). Finally, the images of the two sides of \(a\odot(\mathbf{x}\oplus\mathbf{y})=a\odot\mathbf{x}\oplus a\odot\mathbf{y}\) are equal by scalar distributivity in \(W\), and the images of the two sides of \((a+b)\odot\mathbf{x}=a\odot\mathbf{x}\oplus b\odot\mathbf{x}\) are equal by the other distributive law in \(W\). Injectivity of \(\phi\) makes each corresponding pair equal in \(V\). Every required axiom holds, so \(V\) is a real vector space. \(\square\)
Worked Example: A Valid Structure on the Real Numbers
On \(V=\mathbb{R}\), define \(x\oplus y=x+y-1\) and \(a\odot x=ax+1-a\). These operations differ from ordinary addition and scalar multiplication. Set \(\phi(x)=x-1\). This is a bijection from \(\mathbb{R}\) to \(\mathbb{R}\), with inverse \(\phi^{-1}(w)=w+1\).
For addition, the transported operation from the theorem is
For scalar multiplication, it is
Thus both operations are exactly the operations in the theorem, so they satisfy every real vector-space axiom. For a direct check of the additive identity, the theorem gives \(\phi^{-1}(0)=1\), and indeed \(x\oplus1=x+1-1=x\). The inverse of \(x\) under \(\oplus\) is \(2-x\), since \(x\oplus(2-x)=x+2-x-1=1\). The unusual formulas are valid because they transport the familiar vector-space operations through a bijection.
Common Pitfalls
Do not assume that the number \(0\) is the additive identity. For the operation \(x\oplus y=x+y+1\), a mistaken check might claim that \(0\) is the identity because \(x+0=x\). That calculation uses ordinary addition, not \(\oplus\). In fact, \(2\oplus0=2+0+1=3\ne2\). The correct candidate is \(-1\), and \(2\oplus(-1)=2-1+1=2\). Even after finding the correct identity, the remaining axioms still need to be checked.
Do not declare success after a few successful substitutions. A vector-space axiom must hold for every allowed input. A useful counterexample disproves a universal equation; a handful of examples where the equation works cannot establish it. For a zero or inverse, check the defining condition for all vectors, not just one.
In practice, the most informative conclusion names the failed condition: “not closed under addition,” “the scalar-sum distributive law fails at these inputs,” or “no element can be an additive identity.” Naming the failure makes the calculation verifiable and prevents one unusual operation from being mistaken for another.
Check Your Understanding
For each candidate, identify the relevant axiom and justify your conclusion with a calculation or argument.
- With ordinary operations, is \(V=\{(x,y)\in\mathbb{R}^2:y=2x\}\) closed under addition? Give a general calculation.
- On \(\mathbb{R}\), define \(a\odot x=a^2x\) and use ordinary addition. Test \((a+b)\odot x=a\odot x+b\odot x\) at \(a=b=x=1\).
- For \(x\oplus y=x+y-1\), find the additive identity and verify it on both sides of the operation.
- Explain why finding one input for which an axiom equation fails proves that a candidate is not a vector space.
- In the transport theorem, which property of \(\phi\) lets equality of images imply equality of elements of \(V\)?
