Why Uniqueness Needs a Proof
A vector-space axiom guarantees that an additive identity exists and that every vector has an additive inverse. Existence alone does not establish that there is only one: in principle, a definition could allow several objects with the same stated property. The addition axioms rule this out. The proofs are short, but they illustrate an important method: assume two objects have the defining property, then use the axioms to show that the objects are equal.
Throughout, let \(V\) be a real vector space, with vector addition written \(+\). The arguments below use the additive identity, additive inverses, commutativity and associativity of vector addition. They do not need any scalar-multiplication axioms. The same arguments work for a vector space over any field.
We write the additive identity as \(\mathbf{0}\). Once it is known to be unique, this notation refers to one definite vector. The inverse of \(\mathbf{x}\) is commonly written \(-\mathbf{x}\); once uniqueness is established, that notation also refers to one definite vector. The scalar \(0\) and the vector \(\mathbf{0}\) are different kinds of objects, even though their notation is related.
The Additive Identity Is Unique
Proof. Suppose \(\mathbf{z}\) and \(\mathbf{w}\) are both additive identities in \(V\). Because \(\mathbf{z}\) is an identity, adding it to \(\mathbf{w}\) gives \(\mathbf{w}+\mathbf{z}=\mathbf{w}\). Because \(\mathbf{w}\) is an identity, adding it to \(\mathbf{z}\) gives \(\mathbf{w}+\mathbf{z}=\mathbf{z}\). The left sides are the same vector, so \(\mathbf{w}=\mathbf{z}\). Thus any two additive identities must be equal. The zero-vector axiom supplies at least one identity, so exactly one exists. \(\square\)
Notice that this proof uses the defining property for every vector, applied to the two proposed identities. It does not require cancellation, subtraction, or a coordinate description. In fact, the identity-uniqueness argument itself does not need commutativity: both identity equations have the same left side, \(\mathbf{w}+\mathbf{z}\).
Worked Example: Identifying the Zero Vector in \(\mathbb{R}^2\)
Use the usual addition in \(\mathbb{R}^2\) to identify the additive identity and confirm that no other vector can satisfy the identity condition.
For any \((u,v)\in\mathbb{R}^2\), coordinatewise addition gives
Thus \((0,0)\) is an additive identity. Now suppose \((a,b)\) is also an additive identity. Apply its defining condition to the vector \((0,0)\):
The left side is \((a,b)\), so \((a,b)=(0,0)\). The candidate satisfies the identity condition for every \((u,v)\), and the condition applied to one particular vector forces any other candidate to equal it.
Check: \((u,v)+(0,0)=(u,v)\) for arbitrary \(u,v\), and any proposed identity \((a,b)\) must satisfy \((0,0)+(a,b)=(a,b)=(0,0)\). Hence the identity is exactly \((0,0)\).
Each Vector Has One Additive Inverse
The inverse axiom says that for each \(\mathbf{x}\in V\), there is an additive inverse. The identity theorem ensures that the zero appearing in this condition is unambiguous. To prove uniqueness, suppose two vectors both add to \(\mathbf{x}\) to give that same zero. Commutativity and associativity let us rearrange the sum so that one proposed inverse can be compared directly with the other.
Proof. Let \(\mathbf{a}\) and \(\mathbf{b}\) be additive inverses of the same vector \(\mathbf{x}\), so \(\mathbf{x}+\mathbf{a}=\mathbf{0}\) and \(\mathbf{x}+\mathbf{b}=\mathbf{0}\). By the identity property, \(\mathbf{a}=\mathbf{a}+\mathbf{0}\). Substitute \(\mathbf{x}+\mathbf{b}\) for \(\mathbf{0}\), then use associativity, commutativity, and the inverse condition for \(\mathbf{a}\):
Therefore any two additive inverses of \(\mathbf{x}\) are equal. The inverse axiom guarantees that at least one exists, so each \(\mathbf{x}\) has exactly one additive inverse. \(\square\)
Each step has a specific reason: the first and last use the identity property, the second uses \(\mathbf{x}+\mathbf{b}=\mathbf{0}\), the third uses associativity, the fourth uses commutativity, and the fifth uses \(\mathbf{x}+\mathbf{a}=\mathbf{0}\). The proof does not assume that an inverse can be cancelled from an equation.
Worked Example: Finding the Inverse of a Vector in \(\mathbb{R}^2\)
Find the additive inverse of \(\mathbf{x}=(3,-2)\) under usual addition in \(\mathbb{R}^2\), then verify the answer and establish its uniqueness.
Write a candidate as \((a,b)\). The inverse condition requires
Equality of the coordinates gives \(3+a=0\) and \(-2+b=0\). Thus \(a=-3\) and \(b=2\), so the candidate inverse is \((-3,2)\).
Check: Adding the candidate to the original vector gives
It is an inverse because the sum is the zero vector. If another vector \((c,d)\) were an inverse, it would also satisfy \((3,-2)+(c,d)=(0,0)\). The inverse-uniqueness theorem then gives \((c,d)=(-3,2)\). The calculation finds an inverse; the theorem shows that no different inverse can also work.
Worked Example: An Inverse Under Entrywise Matrix Addition
Consider the matrix \(\mathbf{A}=\begin{bmatrix}1&-2\\3&0\end{bmatrix}\), with addition performed entry by entry. Identify the additive identity and an additive inverse of \(\mathbf{A}\).
The all-zero matrix \(\mathbf{O}=\begin{bmatrix}0&0\\0&0\end{bmatrix}\) is an identity for entrywise addition: for any matrix \(\begin{bmatrix}r&s\\t&u\end{bmatrix}\), adding \(\mathbf{O}\) leaves each entry unchanged. Negate each entry of \(\mathbf{A}\) to obtain the candidate inverse:
Adding entry by entry verifies the inverse condition:
Check: The sum is \(\mathbf{O}\), so the proposed matrix is an inverse. If a matrix \(\mathbf{B}\) also satisfied \(\mathbf{A}+\mathbf{B}=\mathbf{O}\), the inverse-uniqueness theorem applied to entrywise addition would imply \(\mathbf{B}=-\mathbf{A}\). Likewise, the identity-uniqueness theorem says that no other matrix can act as an additive identity.
A Common Mistake: Cancelling Before It Is Justified
Suppose \(\mathbf{x}+\mathbf{a}=\mathbf{0}\) and \(\mathbf{x}+\mathbf{b}=\mathbf{0}\). A tempting argument is: “Subtract \(\mathbf{x}\) from both sides of both equations, so \(\mathbf{a}=\mathbf{b}\).” This may sound reasonable, but it uses an unproved cancellation or subtraction rule. In a development of vector-space properties, it is important to derive the needed fact from the axioms already available rather than assume it.
The correction is to use the identity, inverse, commutativity and associativity axioms directly. Starting from \(\mathbf{a}=\mathbf{a}+\mathbf{0}\), replace \(\mathbf{0}\) by \(\mathbf{x}+\mathbf{b}\), regroup, and reorder:
Every equality now follows from a stated axiom or from one of the two assumed inverse equations. The argument proves the required equality without relying on cancellation.
What the Uniqueness Results Establish
The existence axioms and the uniqueness proofs do different jobs. Existence says there is an identity and that each vector has an inverse. Uniqueness says that once such an object is found, there is no competing choice. This distinction is useful whenever a definition specifies an object by a property: first show that an object with the property exists, then show that any two objects with the property must be equal.
In a real vector space, the unique identity is denoted by \(\mathbf{0}\), and the unique inverse of \(\mathbf{x}\) is denoted by \(-\mathbf{x}\). The symbols are justified by the proofs, not merely by a convenient choice of names. The additive axioms also ensure that an inverse can be used on either side: if \(\mathbf{x}+\mathbf{y}=\mathbf{0}\), commutativity gives \(\mathbf{y}+\mathbf{x}=\mathbf{0}\). The results here concern uniqueness; other consequences of the vector-space axioms can be derived separately.
Check Your Understanding
For each question, identify the axiom or defining property that justifies each step of your argument.
- If \(\mathbf{z}\) and \(\mathbf{w}\) are additive identities, write the two equations that have the same left side and conclude that \(\mathbf{z}=\mathbf{w}\).
- In \(\mathbb{R}^2\), find and check the additive inverse of \((-4,1)\).
- Suppose \(\mathbf{x}+\mathbf{a}=\mathbf{0}\) and \(\mathbf{x}+\mathbf{b}=\mathbf{0}\). Give a proof that \(\mathbf{a}=\mathbf{b}\) using the identity, commutativity and associativity axioms.
- Why does finding one additive inverse not, by itself, establish that it is the only inverse?
- In the matrix example, verify entry by entry that adding \(\mathbf{O}\) leaves \(\begin{bmatrix}r&s\\t&u\end{bmatrix}\) unchanged.
