1. ## 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!

2. 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

3. thanks, did my way still make sense though?

4. 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

5. Hello, mathmonster!

Welcome aboard . . .

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!