# Need help understanding Real Vector Spaces

• July 29th 2010, 04:57 PM
Sabo
Need help understanding Real Vector Spaces
I cannot figure out whether a set is a vector space or not given certain operations.

There are 10 vector space axioms.
1: If u and v are objects in V, then u + v is in V.
2: u + v = v + u
3: u + (v + w) = (u + v) + w
4: There is an object 0 in V, called a zero vector for V, such that 0 + u = u + 0 = u for all u in V.
5: For each u in V, there is an object -u in V, called a negative of u, such that u + (-u) = (-u) + u = 0.
6: If k is any scalar and u is any object in V, then ku is in V.
7: k(u + v) = ku + kv
8: (k + l)u = ku + lu
9: k(lu) = (kl)u
10: 1u = u

In the exercies I'm to determine which sets are vector spaces under the given operations, and for those that are not I'm to list all the axioms that fail.

An exercise.
The set of all tripples of real numbers (x, y, z) with the operations (x, y, z) + (x', y', z') = (x + x', y + y', z + z') and k(x, y, z) = (0, 0, 0).

According to the answers this is not a vector space since axiom 10 fails.
Am I understanding it correct when I do 1(x, y, z) = (x, y, z) != (0, 0, 0), if either x != 0, y != 0 or z != 0?
How do I test axiom 7? Or 8 and 9 as well. I can't get my head around it.
• July 29th 2010, 05:47 PM
Ackbeet
Quote:

Am I understanding it correct when I do 1(x, y, z) = (x, y, z) != (0, 0, 0).
Better would be the following:

According to the definition of your candidate space,

$1(x, y, z) = (0, 0, 0)\not=(x,y,z).$

It does not matter what x, y, or z are. This causes axiom 10 to fail.

Now, axiom 7 holds, actually. Everything on both sides there will be the zero vector. The same is true of axioms 8 and 9.
• July 29th 2010, 07:57 PM
Sabo
Why does it not matter what x, y or z are? If all of them are 0, then axiom 10 would pass, no?

You say that axiom 7, 8 and 9 all holds. What I need help with is why. I don't know where to begin when trying to work these out. It's all "greek" to me, which is a bit strange since I usually have some sort of idea from which I can proceed.
• July 29th 2010, 11:37 PM
CaptainBlack
Quote:

Originally Posted by Sabo
Why does it not matter what x, y or z are? If all of them are 0, then axiom 10 would pass, no?

You say that axiom 7, 8 and 9 all holds. What I need help with is why. I don't know where to begin when trying to work these out. It's all "greek" to me, which is a bit strange since I usually have some sort of idea from which I can proceed.

For axiom 10 to hold the requires that $1.\bf{u}=\bf{u}$ for all ${\bf{u}} \in V$

CB
• July 30th 2010, 02:25 AM
HallsofIvy
Quote:

Originally Posted by Ackbeet
Better would be the following:

According to the definition of your candidate space,

$1(x, y, z) = (0, 0, 0)\not=(x,y,z).$

It does not matter what x, y, or z are. This causes axiom 10 to fail.

Yes, it does matter what x, y, and z are. If x= y= z= 0, then the $\ne$ is incorrect. Of course, as Captain Black pointed out, it is only necessary that the axiom "1(x, y, z)= (x, y, z)" fail for a single triple in order that this NOT be a vector space.

Quote:

Now, axiom 7 holds, actually. Everything on both sides there will be the zero vector. The same is true of axioms 8 and 9.