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:

$$\ x_1 + 5x_2 + 3x_3 = 0 \$$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.