Vector Space Q: Is Additive Identity Unique?

[bjgawp]

Just wondering. Suppose we some plane, any plane like S = \{ (x_1, x_2, x_3) \in F^{3} \ : \ x_1 + 5x_2 + 3x_3 = 0 \} where F is either \mathbb{R} or \mathbb{C} . We know that S is a vector space (passes the origin).

We know that (0,0,0) is the additive identity and it should be unique by virtue of the field we're working with. But say we had any arbitrary vector (a,b,c). Wouldn't (-a, -b, -c) or (0, 3, -5) also be counted as an additive identity as well?

 

[waht]

(-a, -b, -c) would be the inverse under addition.

the other (0,3,-5), forgot what's it called, I think the kernel if I'm not mistaken

 

[bjgawp]

But wouldn't they also be the additive identity /zero vector as well, violating the fact zero vectors of vector spaces are unique?

 

[waht]

Given an identity 0, and a vector v

v + 0 = v

but

v + (0,3,-5) does not equal vso it's not an identity, it's a vector that maps to zero, but it's not zero.

 

[bjgawp]

Doesn't it though for v in S?

Let v = (a,b,c) = a + 5b + 3c \in S. Then (a + 5b + 3c) + (0 + 5(3) + 3(-5)) = (a + 5b + 3c) + (0) = a + 5b + 3c = v

So we found an element w = (0, 3, -5) such that v + w = v which by definition, we call w the additive inverse?

 

[waht]

I was wrong about the kernel thing, because you need to have a map for that.

You just have a vector space that is constrained to S:

<br /> \ x_1 + 5x_2 + 3x_3 = 0 \<br />meaning that the components of every element of S have to satisfy this equation.

 

[bjgawp]

Not quite sure if I entirely understand. Wouldn't any element in S then be called an additive identity since adding it to any vector v would just simply give us v.?

 

[jasonRF]

bjgawp,

I think you may be a little confused. Since you are in a 3 dimensional space, each vector contains three components, including the zero vector. So let
\mathbf{v} = \left(a,b,c\right) \in S, and let
\mathbf{w} = \left(0,3,-5\right), where of course
\mathbf{w} \in S as well. When we add these, we get
\mathbf{v} + \mathbf{w} = \left(a+0, b+3, c-5 \right).

For example, if \mathbf{v} = \left(1,-2,3\right)
then \mathbf{v} + \mathbf{w} = \left(1, 1, -2 \right).

Also note that if \mathbf{u} = \left(0,-3,5\right),
then
\mathbf{v} + \mathbf{u} = \left(0+0, 3-3, -5+5 \right) <br /> = \left(0,0,0 \right) = \mathbf{0}. Note that the
vector \mathbf{0} \in S
is not the same as the scalar 0 \in F.

I hope that helps,

Jason

Doesn't it though for v in S?

Let v = (a,b,c) = a + 5b + 3c \in S. Then (a + 5b + 3c) + (0 + 5(3) + 3(-5)) = (a + 5b + 3c) + (0) = a + 5b + 3c = v

So we found an element w = (0, 3, -5) such that v + w = v which by definition, we call w the additive inverse?

 

[HallsofIvy]

Doesn't it though for v in S?

Let v = (a,b,c) = a + 5b + 3c \in S.

NO! (a, b, c) is NOT equal to a+ 5b+ 3c and a+ 5b+ 3c is NOT in S. "(a,b,c)= a+ 5b+ 3c doesn't even make sense. One is a member of R³, the other of R¹- they can't be equal! And a+ 5b+ 3c is not in S because S only contains things in R³ and a+ 5b+ 3c is not.

[/quote]Then (a + 5b + 3c) + (0 + 5(3) + 3(-5)) = (a + 5b + 3c) + (0) = a + 5b + 3c = v

So we found an element w = (0, 3, -5) such that v + w = v which by definition, we call w the additive inverse?[/QUOTE]
IF v= (a, b, c) is in S, then a+ 5b+ 3c= 0. If w= (0, 3, -5) then v+ w= (a, b+3, c- 5) which is NOT equal to (a, b, c). Frankly, it appears that you have no clue what
S = \{ (x_1, x_2, x_3) \in F^{3} \ : \ x_1 + 5x_2 + 3x_3 = 0 \}
means.