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# 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

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