Prove that in any vector space k × 0 = 0, where k is any real number and 0 is the zero vector. I know I have to use the vector axioms.. **Axiom****Signification**Associativity of addition**u** + (**v** + **w**) = (**u** + **v**) + **w**.Commutativity of addition**v** + **w** = **w** + **v**.Identity element of additionThere exists an element **0** ∈ *V*, called the *zero vector*, such that **v** + **0** = **v** for all **v** ∈ *V*.Inverse elements of additionFor all **v** ∈ V, there exists an element **w** ∈ *V*, called the *additive inverse* of **v**, such that **v** + **w** = **0**. The additive inverse is denoted −**v**.Distributivity of scalar multiplication with respect to vector addition *a*(**v** + **w**) = *a***v** + *a***w**.Distributivity of scalar multiplication with respect to field addition(*a* + *b*)**v** = *a***v** + *b***v**.Compatibility of scalar multiplication with field multiplication*a*(*b***v**) = (*ab*)**v** [nb 3]Identity element of scalar multiplication1**v** = **v**, where 1 denotes the multiplicative identity in *F*.
Do I use the distributivity of scalar multiplication with respect to field addition ??

(a+b)v= av+bv

let a=b=0

(0+0)v=0v+oV

?????

little confused since it's pretty much obvious!

Thanks