Prove a Corollary: u ⋅ (av) = a(u ⋅ v)

[Mampac]

For some vectors u and v in Rn and a scalar a belonging to R, prove that
u ⋅ (av) = a(u ⋅ v)

WITHOUT using the definition of the dot product!
I know this is some very simple and basic stuff, but I see no way of proceeding after I consider that av has coordinates [ax₁, ax₂, ... ax_n]. What do I do after that if I can't open up the ⋅s using the dot product?

 

[lev888]

For some vectors u and v in Rn and a scalar a belonging to R, prove that
u ⋅ (av) = a(u ⋅ v)

WITHOUT using the definition of the dot product!
I know this is some very simple and basic stuff, but I see no way of proceeding after I consider that av has coordinates [ax₁, ax₂, ... ax_n]. What do I do after that if I can't open up the ⋅s using the dot product?

Please post a photo of the question. This doesn't sound right. Proving something about a concept without using its definition doesn't make sense to me.

 

[Mampac]

Please post a photo of the question. This doesn't sound right. Proving something about a concept without using its definition doesn't make sense to me.

here are the propositions and the corollary:

I managed to prove the distributivity, but can't see any way I can use it in Corollary 1.7

 

[HallsofIvy]

So "not from the definition" means from the properties proven before, in 1.6.

From 1.6.1, \(\displaystyle u\cdot (av)= (av)\cdot u\).
From 1.6.2, \(\displaystyle (av)\cdot u= a(v\cdot u)\)
From 1.6.1 again, \(\displaystyle a(v\cdot u)= a(u\cdot v)\).

 

[Mampac]

So "not from the definition" means from the properties proven before, in 1.6.

From 1.6.1, \(\displaystyle u\cdot (av)= (av)\cdot u\).
From 1.6.2, \(\displaystyle (av)\cdot u= a(v\cdot u)\)
From 1.6.1 again, \(\displaystyle a(v\cdot u)= a(u\cdot v)\).

oh my... it never occurred to me
i thought i had to use only the point i proved
thank you so much