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4.2: Elementary properties of vector spaces

Last updated

Mar 5, 2021
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- 4.1: Deﬁnition of vector spaces
- 4.3: Subspaces

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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 $v$ .

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