satisfying axioms to 'prove' vector space
I need some clarity on the following please:
The question asked that it be shown that the following set is not a vector space:
'The set of all triples of real numbers with addition defined by
(x, y, z) + (u, v, w) = (z+w, y+v, x+u)'
I worked out that the following axiom holds:
(k+m)u = ku + ku
however, the answer works it out to not hold.
It goes as follows:
Set u = (u1, u2, u3), then
= (k+m)(u1, u2, u3)
= ((k+m)u1, (k+m)u2, (k+m)u3)
ku + mu
= (ku1, ku2, ku3) + (mu1, mu2, mu3)
= (ku3 + mu3, ku2 + mu2, ku1 + mu1)
since these are not equal, this axiom does not hold.'
Now, I don't understand why these two are not equal?
Can we not further simplify:
(ku3 + mu3, ku2 + mu2, ku1 + mu1)
= ((k+m)u3, (k+m)u2, (k+m)u1) ?
That way, does it not equal = ((k+m)u1, (k+m)u2, (k+m)u3)) ?
or is it not equal because the order of the triple is different?
I'm quite confused and any help would be much appreciated!
Have a great day all (Happy)