Question

Question about vector space:

set of all triples of real numbers with standard vector addition but with scalar multiplication defined by
k(x,y,z)=(k^2,k^y,k^z)
axiom 8 doesn't hold but I don't see how
(axiom 8 is (k+m)u=ku+km) which seems like it holds to me.

Answer 1

consider a vector (1,1,1)
for axiom8 to hold, you want k(1,1,1) = (k, k, k) right ?

Answer 2

well yeah, i just used u=(u1,u2,u3)
so therefore, LHS would give:

(k+m)u --> k^2u1,k^2u2,k^2u3 + m^2u1,k^2u2,k^2u3
RHS would give the samething no?