# Addition of positive and negative of a object in vector space = 0 related question

• March 22nd 2011, 04:15 PM
x3bnm
Addition of positive and negative of a object in vector space = 0 related question
Sorry if I am posting too frequently. I am learning linear algebra all by myself without
any teacher or TA or without going to college. It's very fun but sometimes the book
gets overly complicated in here and there. That's why these postings.

Now back to the problem. Suppose in a vector space:
$
u = u_1 + u_2\,\,\, and\,\,\, v = v_1 + v_2
$

And addition and subtraction are defined as:
$
u + v = (u_1 + v_1, u_2 + v_2)
$

$
ku = (k.u_1, 0)
$

And a axiom is given:
For each $u$ in $V$, there is an object $-u$ in $V$, called a negative of $u$, such that
$
u + (-u) = (-u) + u = 0
$

We have to prove if the above vector space satisfies this axiom.

So I did it like this:
$
-u = -1.u = (-1.u_1, 0) = (-u_1, 0)
$

Now:
$
u + (-u) = (u_1, u_2) + (-u_1, 0) = (0, u_2)
$

Now how do I get $0 = (0, u_2)$ ? The book says that it satisfies this axiom.
But don't know how to reach it.

Can anyone kindly help me how to get to this relationship?
• March 22nd 2011, 04:51 PM
TheEmptySet
You don't need to (and can't in this case) use the definition of multiplication.

If $u=(u_1,u_2)$ and $v=(v_1,v_2)$

Then $u+v=0 \iff (u_1,u_2)+(v_1,v_2)=(0,0) \iff (u_1+v_1,u_2+v_2)=(0,0)$

this implies that

$v_1=-u_1 \quad v_2=-u_2 \quad v_3=-u_3$

so the additive inverse of u is

$v=-u=(-u_1,-u_2,-u_3)$

This vector satisfies the additive inverse properties.
• March 22nd 2011, 05:08 PM
x3bnm
Quote:

Originally Posted by TheEmptySet
You don't need to (and can't in this case) use the definition of multiplication.

If $u=(u_1,u_2)$ and $v=(v_1,v_2)$

Then $u+v=0 \iff (u_1,u_2)+(v_1,v_2)=(0,0) \iff (u_1+v_1,u_2+v_2)=(0,0)$

this implies that

$v_1=-u_1 \quad v_2=-u_2 \quad v_3=-u_3$

so the additive inverse of u is

$v=-u=(-u_1,-u_2,-u_3)$

This vector satisfies the additive inverse properties.

Roger that. So the keyword is "You don't need to (and can't in this case) use the definition of multiplication." Thanks a lot for helping hand.