# satisfying axioms to 'prove' vector space

• Apr 29th 2010, 05:20 AM
mathchickie
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 (Happy)
• Apr 29th 2010, 06:44 AM
HallsofIvy
Quote:

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?

Quote:

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 (Happy)
Yes, exactly, "triples" implies "ordered triples". (1, 2, 1) is NOT the same as (2, 1, 1).
• Apr 30th 2010, 12:32 AM
mathchickie
Aha! Thank you... that makes things a lot easier. (Happy)

[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!
• Apr 30th 2010, 02:40 AM
HallsofIvy
[QUOTE=mathchickie;504140]Aha! Thank you... that makes things a lot easier. (Happy)

Quote:

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 \$\displaystyle 0+ u\ne u\$

Similarly u+ 0= (0, 0, 0)+ (u1, u2, u3)
= (u3, u2, u1)\$\displaystyle \ne\$ (u1, u2, u3) so \$\displaystyle u+ 0\ne u\$.

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

Quote:

Thanks a stack!