Thread: satisfying axioms to 'prove' vector space

hello there!
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)u
= (k+m)(u1, u2, u3)
= ((k+m)u1, (k+m)u2, (k+m)u3)

and
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

Originally Posted by mathchickie

hello there!
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

Well, that's not what you want- you want (k+ m)u= ku+ mu. Typo?

however, the answer works it out to not hold.

It goes as follows:

Set u = (u1, u2, u3), then
(k+m)u
= (k+m)(u1, u2, u3)
= ((k+m)u1, (k+m)u2, (k+m)u3)

and
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

Yes, exactly, "triples" implies "ordered triples". (1, 2, 1) is NOT the same as (2, 1, 1).

Aha! Thank you... that makes things a lot easier.

[quote=HallsofIvy;503590]Well, that's not what you want- you want (k+ m)u= ku+ mu. Typo?

sorry, typo! Thanks

Seeing then that "triples" implies ordered triples, if we look at the following axiom: with addition defined as (x, y, z) + (u, v, w) = (z+w, y+v, x+u)

0 + u = u + 0 = u and we set
u = (u1, u2, u3)

Then, we get

0 + u = (0, 0, 0) + (u1, u2, u3)
= (0 + u3, 0 + u2, 0 + u1)
= (u3, u2, u1) + (0, 0, 0)
= (u1 + 0, u2 + 0, u3 + 0)
= (u1, u2, u3)
= u

can this be calculated like this??

Thanks a stack!

[QUOTE=mathchickie;504140]Aha! Thank you... that makes things a lot easier.

Originally Posted by HallsofIvy

Well, that's not what you want- you want (k+ m)u= ku+ mu. Typo?

sorry, typo! Thanks

Seeing then that "triples" implies ordered triples, if we look at the following axiom: with addition defined as (x, y, z) + (u, v, w) = (z+w, y+v, x+u)

0 + u = u + 0 = u and we set
u = (u1, u2, u3)

Then, we get

0 + u = (0, 0, 0) + (u1, u2, u3)
= (0 + u3, 0 + u2, 0 + u1)
= (u3, u2, u1) + (0, 0, 0)
= (u1 + 0, u2 + 0, u3 + 0)
= (u1, u2, u3)
= u

can this be calculated like this??

No, Your first two lines are correct:
0 + u = (0, 0, 0) + (u1, u2, u3)
= (0 + u3, 0 + u2, 0 + u1)
so that 0 + u= (u3, u2, u1) and so

Similarly u+ 0= (0, 0, 0)+ (u1, u2, u3)
= (u3, u2, u1) (u1, u2, u3) so .

However, it is true that u+ v= v+ u.

Thanks a stack!