- #1

Delong66

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Suppose that u,v, w $$\in V$$. where u,v, w are vectors and $$\V$$ is a vector space

$$u + v \in V \tag{Closure under addition}$$

$$u + v = v + u \tag{Commutative property}$$

$$u + (v+w)=(u+v)+w \tag{Associative property}$$

V has a zero vector 0 such that for every $$\u \in \V$$, $$u+0=u$$. $$\tag{Additive identity}$$

For every $$u \in V$$, there is a vector in V denoted by −u such that u+(−u)=0. $$\tag{Additive inverse}$$

Now let's also assume that $$c,d \in \mathbb R$$

$$cu \in V \tag{Closure under scalar multiplication}$$

$$c(u+v)=cu+cv \tag{Distributive property}$$

$$(c+d)u=cu+du \tag{Distributive property}$$

$$c(du)=(cd)u \tag{Associative property}$$

$$1(u)=u \tag{Scalar identity}$$