Why Altered Operations Need Careful Testing
A set alone is not a vector space. The operations assigned to it matter: addition must behave like addition of vectors, and scalar multiplication must interact with that addition in specific ways. If even one required property fails, the set with those operations is not a vector space, even if the operations look familiar or satisfy several other properties.
Here we work over the scalar field \(\mathbb{R}\). For a proposed vector space, there must be a set \(V\), an addition operation on pairs of elements of \(V\), and a scalar multiplication operation that takes a real number and an element of \(V\) to an element of \(V\). We write the operations as \(\oplus\) and \(\odot\) when they differ from the usual ones. For a set with these operations to be a real vector space, the following properties must hold for all \(\mathbf{u},\mathbf{v},\mathbf{w}\in V\) and \(a,b\in\mathbb{R}\):
- Addition is commutative and associative.
- There is an additive zero \(\mathbf{0}_V\in V\) such that \(\mathbf{v}\oplus\mathbf{0}_V=\mathbf{v}\) for every \(\mathbf{v}\).
- Every \(\mathbf{v}\) has an additive inverse \(-\mathbf{v}\in V\) such that \(\mathbf{v}\oplus(-\mathbf{v})=\mathbf{0}_V\).
- Scalar multiplication distributes over vector addition: \(a\odot(\mathbf{u}\oplus\mathbf{v})=(a\odot\mathbf{u})\oplus(a\odot\mathbf{v})\).
- Scalar addition distributes over scalar multiplication: \((a+b)\odot\mathbf{v}=(a\odot\mathbf{v})\oplus(b\odot\mathbf{v})\).
- Scalar multiplication is associative: \(a\odot(b\odot\mathbf{v})=(ab)\odot\mathbf{v}\).
- The scalar identity acts as an identity: \(1\odot\mathbf{v}=\mathbf{v}\).
Closure is also required: the proposed addition must return an element of \(V\), and scalar multiplication by any real number must return an element of \(V\). In the examples below the set is \(\mathbb{R}\), and the formulas always return real numbers, so closure holds. The issue is whether the returned values obey the required axioms.
A Failed Addition Axiom Is Enough
Worked Example: Subtraction Used as Addition
Take \(V=\mathbb{R}\), define \(x\oplus y=x-y\), and use the usual scalar multiplication \(a\odot x=ax\). Test whether these operations make \(V\) a real vector space.
Commutativity is required for vector addition. Evaluate the two orders using \(x=1\) and \(y=2\):
Since \(-1\ne1\), this operation is not commutative. That single failed axiom is enough to rule out a vector space. In fact, this operation does not even have a two-sided additive identity: \(x\oplus e=x\) would require \(x-e=x\), hence \(e=0\); but \(e\oplus x=x\) would then require \(0-x=x\) for every \(x\), which fails at \(x=1\).
Check: with \(e=0\), \(1\oplus e=1\), but \(e\oplus1=-1\ne1\). So \(0\) works only on the right, not on both sides.
This example illustrates a useful testing principle: look for a direct violation rather than trying to verify every axiom before deciding. A single pair of elements with unequal addition orders disproves commutativity. On the other hand, passing one test does not establish that the operations form a vector space. Each required axiom still needs justification before a positive conclusion is warranted.
Scalar Multiplication Must Respect Scalar Addition
Worked Example: Absolute Value in Scalar Multiplication
Keep the usual addition on \(V=\mathbb{R}\), but define \(a\odot x=|a|x\), where \(|a|\) is the usual absolute value. This operation returns a real number, and it passes some basic checks: \(1\odot x=|1|x=x\), and for all \(a,x,y\in\mathbb{R}\),
So the identity-scalar axiom and distributivity over vector addition hold. But the other distributive axiom requires \((a+b)\odot x=(a\odot x)+(b\odot x)\). Choose \(a=1\), \(b=-1\), and \(x=1\). The two sides are
They are unequal, so scalar addition does not distribute over scalar multiplication. Therefore these operations do not make \(\mathbb{R}\) a real vector space, even though the operation behaves correctly in the preceding checks.
Check: the failed equality is \(0\ne2\), obtained from permitted real scalars and a permitted vector. No further test is needed to establish that the vector space axioms fail.
When a formula is modified, it is important to check both distributive laws separately. One concerns adding vectors before multiplying by a scalar; the other concerns adding scalars before multiplying a vector. The absolute-value example passes the first calculation because multiplication by a fixed real number \(|a|\) distributes over ordinary addition. It fails the second because absolute value does not preserve sums of scalars in the required way.
Zero Scalar Multiplication Must Give the Additive Zero
Proof. Assume the vector space axioms hold. Since \(0=0+0\) in \(\mathbb{R}\), distributivity over scalar addition gives \(0\odot\mathbf{v}=(0+0)\odot\mathbf{v}=(0\odot\mathbf{v})\oplus(0\odot\mathbf{v})\). Let \(\mathbf{w}=0\odot\mathbf{v}\). Then \(\mathbf{w}=\mathbf{w}\oplus\mathbf{w}\). Add the additive inverse of \(\mathbf{w}\) to both sides, using associativity and the identity property: \((-\mathbf{w})\oplus\mathbf{w}=(-\mathbf{w})\oplus(\mathbf{w}\oplus\mathbf{w})\) gives \(\mathbf{0}_V=\mathbf{w}\). Thus \(0\odot\mathbf{v}=\mathbf{0}_V\). \(\square\)
This consequence gives another fast diagnostic. If a proposed scalar multiplication sends a vector to anything other than the additive zero when the scalar is \(0\), then at least one vector space axiom must fail. The statement does not replace the axiom checks: it is a consequence of them, not a separate permission to ignore the distributive laws.
Translating Addition Changes Which Axioms Fit
Proof. If \(c=0\), then \(x\oplus y=x+y\) and \(a\odot x=ax\) are the usual operations on \(\mathbb{R}\), which satisfy the real vector space axioms. Conversely, suppose \(c\ne0\). Test distributivity over vector addition with scalar \(a=0\) and vectors \(x=y=0\). The left side is \(0\odot(0\oplus0)=0\odot c=0\). The right side is \((0\odot0)\oplus(0\odot0)=0\oplus0=c\). Since \(c\ne0\), the two sides differ, so a vector space axiom fails. Therefore the operations define a vector space exactly when \(c=0\). \(\square\)
Worked Example: Addition Shifted by One
Set \(c=1\). Then \(x\oplus y=x+y+1\), while \(a\odot x=ax\). First, the proposed addition is commutative because \(x+y+1=y+x+1\). It is also associative: for any \(x,y,z\in\mathbb{R}\),
The additive identity for this operation is \(-1\), since \(x\oplus(-1)=x-1+1=x\) and \((-1)\oplus x=-1+x+1=x\). An inverse for \(x\) must add to this identity: solving \(x\oplus y=-1\) gives \(x+y+1=-1\), so \(y=-x-2\). Substitution confirms \(x\oplus(-x-2)=x-x-2+1=-1\).
Despite these successful addition checks, scalar distributivity fails. With \(a=0\) and \(x=y=0\),
Thus \(0\ne1\), and the operations do not define a vector space. The calculation also agrees with the theorem: the shift \(c=1\) is nonzero.
Common Mistake: Finding One Good Property and Stopping
A frequent error is to find a plausible additive identity and conclude that the altered operations work. For the shifted addition \(x\oplus y=x+y+1\), it is wrong to claim that \(0\) is the additive identity because “adding zero changes nothing.” Under the proposed rule, the actual calculation is \(x\oplus0=x+0+1=x+1\), which is not \(x\) in general. The correction is to solve the identity equation: \(x\oplus e=x\) requires \(x+e+1=x\), so \(e=-1\); checking the other order gives \((-1)\oplus x=-1+x+1=x\).
Finding the correct identity still does not prove that the structure is a vector space. In this case, the zero-scalar distributivity calculation gives \(0\ne1\), so a different axiom fails. The reliable approach is to keep the operations visible in every calculation, test the full list of requirements when proving a positive result, and stop as soon as one exact counterexample disproves a requirement.
Write down the proposed addition and scalar multiplication separately; do not silently replace either with its usual formula.
Test commutativity, associativity, the existence of one two-sided identity, and additive inverses using the proposed addition.
Use \(a\odot(\mathbf{u}\oplus\mathbf{v})\) versus \((a\odot\mathbf{u})\oplus(a\odot\mathbf{v})\), and \((a+b)\odot\mathbf{v}\) versus \((a\odot\mathbf{v})\oplus(b\odot\mathbf{v})\).
Check scalar associativity and the action of scalar \(1\). If a proposed operation fails any required test, give the specific values and calculations.
These examples show why changed operations cannot be judged by their names or appearance. Subtraction fails commutativity; absolute value in scalar multiplication fails scalar distributivity; and shifting addition can preserve familiar addition properties while breaking compatibility with scalar multiplication. A precise diagnosis always identifies the particular axiom and exhibits the values for which it fails.
Check Your Understanding
For each question, identify the axiom being tested and show enough calculation to justify the conclusion.
- On \(\mathbb{R}\), define \(x\oplus y=x+y-2\) and use ordinary scalar multiplication. What is the additive identity for \(\oplus\)? Do these operations define a vector space?
- With usual addition on \(\mathbb{R}\), define \(a\odot x=a^2x\). Test the scalar identity axiom and both distributive laws.
- Suppose a proposed vector space has additive zero \(\mathbf{0}_V\). Explain why \(0\odot\mathbf{v}\) must equal \(\mathbf{0}_V\).
- For \(x\oplus y=x-y\), give a pair that disproves commutativity.
- In the shifted-addition theorem, use \(a=0\) to compare the two sides of distributivity for arbitrary \(x,y\) when \(c\ne0\).
