# 4.2 Elementary properties of vector spaces

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We are going to prove several important, yet simple, properties of vector spaces. From now on, \(V\) will denote a vector space over \(\mathbb{F}\).

**Proposition 4.2.1.** *Every vector space has a unique additive identity.*

*Proof. * Suppose there are two additive identities \(0\) and \(0'\) Then

\[ 0'=0+0'=0, \]

where the first equality holds since \(0\) is an identity and the second equality holds since \(0'\) is an identity. Hence \(0=0'\), proving that the additive identity is unique.

**Proposition 4.2.2. ***Every *\(v \in V\) *has a unique additive inverse.*

*Proof. *Suppose \(w\) and \(w'\) are additive inverses of \(v\) so that \(v+w=0\) and \(v+w'=0\) Then

\[ w = w+0 = w+(v+w') = (w+v)+w' = 0+w' =w'. \]

Hence \(w=w'\), as desired.

Since the additive inverse of \(v\) is unique, as we have just shown, it will from now on be denoted by \(-v\) We also define \(w-v\) to mean \(w+(-v)\) We will, in fact, show in Proposition 4.2.5 below that \(-v=-1 v\)

**Proposition 4.2.3.** \(0v = 0\)* for all* \(v \in V\).

Note that the \(0\) on the left-hand side in Proposition 4.2.3 is a scalar, whereas the \(0\) on the right-hand side is a vector.

*Proof.* For \(v\in V\), we have by distributivity that

\[ 0v=(0+0)v=0v+0v. \]

Adding the additive inverse of \(0v\) to both sides, we obtain

\[ 0 = 0v-0v = (0v+0v)-0v=0v.\]

**Proposition 4.2.4. ** \(a0=0\) *for every *\( a \in \mathbb{F}\).

*Proof.* As in the proof of Proposition 4.2.3, if \(a\in \mathbb{F}\), then

\[ a0=a(0+0)=a0+a0. \]

Adding the additive inverse of \(a0\) to both sides, we obtain \(0=a0\), as desired.

**Proposition 4.2.5.** \((-1)v=-v\) *for every* \(v \in V\).

*Proof.* For \(v\in V\), we have

\[ v+(-1)v= 1v+(-1)v = (1+(-1))v=0v=0, \]

which shows that \((-1)v\) is the additive inverse \(-v\) of \(v\).

### Contributors

- Isaiah Lankham, Mathematics Department at UC Davis
- Bruno Nachtergaele, Mathematics Department at UC Davis
- Anne Schilling, Mathematics Department at UC Davis

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