# vector spaces help

• Oct 22nd 2007, 06:33 AM
mathmonster
vector spaces help
HEY to everyone, never posted here before!!

if given 2v = v, would it be correct, to prove v = 0, by doing the following, (using the axioms of scalar multiplication and addition for a k-vector space)

v = v + (v - v) = (v + v) - v = 2v - v, so

as 2v=v then

2v-v = v-v = 0.

thanl u in advance for ur help!
• Oct 22nd 2007, 06:49 AM
topsquark
Quote:

Originally Posted by mathmonster
HEY to everyone, never posted here before!!

if given 2v = v, would it be correct, to prove v = 0, by doing the following, (using the axioms of scalar multiplication and addition for a k-vector space)

v = v + (v - v) = (v + v) - v = 2v - v, so

as 2v=v then

2v-v = v-v = 0.

thanl u in advance for ur help!

I would do things slightly differently:
$\displaystyle 2v = v$

We know that the additive inverse of v exists and is -v. Thus
$\displaystyle 2v + (-v) = v + (-v)$

$\displaystyle v = 0$

-Dan
• Oct 22nd 2007, 06:51 AM
mathmonster
thanks, did my way still make sense though?
• Oct 22nd 2007, 06:55 AM
topsquark
Quote:

Originally Posted by mathmonster
thanks, did my way still make sense though?

Yes. It's simply that my method (which uses the same principle that yours did) is a bit shorter is all.

-Dan
• Oct 22nd 2007, 06:58 AM
Soroban
Hello, mathmonster!

Welcome aboard . . .

Quote:

If given $\displaystyle 2v = v$, would it be correct, to prove $\displaystyle v = 0$, by doing the following,
using the axioms of scalar multiplication and addition for a k-vector space?

. . $\displaystyle v \;= \;v + (v - v)$

. . . .$\displaystyle = \;(v + v) - v$

. . . .$\displaystyle = \;2v - v$ . . Since $\displaystyle 2v = v$,
. . . . . .
. . . .$\displaystyle = \;\;v - v$

. . . .$\displaystyle =\quad0$

Sure looks good to me!