Additive Inverse in Vector Space is Unique

Theorem

Let $\struct {\mathbf V, +, \circ}_F$ be a vector space over a field $F$, as defined by the vector space axioms.

Then for every $\mathbf v \in \mathbf V$, the additive inverse of $\mathbf v$ is unique:

$\forall \mathbf v \in \mathbf V: \exists! \paren {-\mathbf v} \in \mathbf V: \mathbf v + \paren {-\mathbf v} = \mathbf 0$

Proof

Proof of Existence

Follows from the vector space axioms.

$\Box$

Proof of Uniqueness

Let $\mathbf v$ have inverses $\mathbf x$ and $\mathbf y$.

Then:

\(\ds \mathbf v + \mathbf x\)

\(=\)

\(\ds \mathbf 0\)

\(\, \ds \land \, \)

\(\ds \mathbf v + \mathbf y\)

\(=\)

\(\ds \mathbf 0\)

\(\ds \leadsto \ \ \)

\(\ds \mathbf v + \mathbf x\)

\(=\)

\(\ds \mathbf v + \mathbf y\)

\(\ds \leadsto \ \ \)

\(\ds \mathbf x\)

\(=\)

\(\ds \mathbf y\)

Vectors are Left Cancellable

$\blacksquare$

Also see

• Zero Vector is Unique
• Vector Inverse is Negative Vector

Sources

• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Vector Spaces: $\S 32$. Definition of a Vector Space: Theorem $64 \ \text{(ii)}$
• 2015: Sheldon Axler: Linear Algebra Done Right (3rd ed.): p. 15