# Vector space

1. Feb 18, 2006

### stunner5000pt

decide whether this is a vector space or not
a(x,y,z) = (2ax,2ay,2az)

all the addition axoims hold easily
for the scalar multiplications axioms
for some real scaral a
$$a(x,y,z) = (ax,ay,az) \in 2(ax,ay,az)$$
$$a(x_{1}+x_{2},y_{1}+y_{2},z_{1}+z_{2}) = a(x_{1},y_{1},a(z_{1}) + a(x_{2},y_{2},a(z_{2}) \in (2ax,2ay,2az)$$ this doesnt seem to hold because it would only be possible if the wo vecotrs being added were not distinct.

$$(a+b)(x,y,z) = (ax+bx,ay+by,az+bz) \in (2az,2ay,2az)$$

$$a(bx,by,bz) = (ab)(x,y,z)$$ this would seem t ohold if b was 2... not not anytrhing else? Not sure here?

multiplcation by 1 gives us what we want. so the last axiom holds
i can post hte axioms if u want

my text says it is not a vector space because of the failure of scalara multiplication where i have stated the doubts mysefl. Are those hte valid reasons for that??

Last edited: Feb 18, 2006
2. Feb 18, 2006

### matt grime

what the hell are x,y,z, or a?

3. Feb 18, 2006

### HallsofIvy

Staff Emeritus
a(x,y,z)= (2ax,2ay,2az) doesn't make sense. Are you asked to show that the set of all (x,y,z) with ordinary addition but scalar multiplication defined by a(x,y,z)= (2a,2ay,2az) is a vector space?

4. Feb 19, 2006

### stunner5000pt

that is what i am asked to prove

5. Feb 19, 2006

### matt grime

Multiplication by must send v to v, ie 1.v=v for all v in the space.

6. Feb 19, 2006

### HallsofIvy

Staff Emeritus
I don't believe for a moment that your problem says exactly
"decide whether this is a vector space or not a(x,y,z) = (2ax,2ay,2az)"
There isn't even a set that could be a vector space there!