Vector Space Q: Is Additive Identity Unique?

In summary: It does NOT mean that every vector in S is equal to 0. It's definition says that a vector is in S precisely when the numbers that make up the vector obeys the equation x_1+ 5x_2+ 3x_3= 0. If you are given a vector, (a, b, c), it is NOT equal to a+ 5b+ 3c. They are different kinds of things. And it is NOT true that (a, b, c)+ (0, 3, -5)= (a, b+ 3, c- 5)= (a, b, c) so it is NOT true that v+w= v
  • #1
bjgawp
84
0
Just wondering. Suppose we some plane, any plane like [tex]S = \{ (x_1, x_2, x_3) \in F^{3} \ : \ x_1 + 5x_2 + 3x_3 = 0 \} [/tex] where F is either [tex]\mathbb{R}[/tex] or [tex]\mathbb{C}[/tex] . We know that S is a vector space (passes the origin).

We know that [tex](0,0,0)[/tex] 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 [tex](a,b,c)[/tex]. Wouldn't [tex](-a, -b, -c)[/tex] or [tex](0, 3, -5)[/tex] also be counted as an additive identity as well?
 
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  • #2
(-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
 
  • #3
But wouldn't they also be the additive identity /zero vector as well, violating the fact zero vectors of vector spaces are unique?
 
  • #4
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.
 
  • #5
Doesn't it though for v in S?

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

So we found an element [tex]w = (0, 3, -5)[/tex] such that [tex]v + w = v[/tex] which by definition, we call w the additive inverse?
 
  • #6
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:

[tex]
\ x_1 + 5x_2 + 3x_3 = 0 \
[/tex]meaning that the components of every element of S have to satisfy this equation.
 
  • #7
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.?
 
  • #8
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
[tex] \mathbf{v} = \left(a,b,c\right) \in S[/tex], and let
[tex] \mathbf{w} = \left(0,3,-5\right) [/tex], where of course
[tex] \mathbf{w} \in S [/tex] as well. When we add these, we get
[tex] \mathbf{v} + \mathbf{w} = \left(a+0, b+3, c-5 \right) [/tex].

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

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

I hope that helps,

Jason

bjgawp said:
Doesn't it though for v in S?

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

So we found an element [tex]w = (0, 3, -5)[/tex] such that [tex]v + w = v[/tex] which by definition, we call w the additive inverse?
 
  • #9
bjgawp said:
Doesn't it though for v in S?

Let [tex]v = (a,b,c) = a + 5b + 3c \in S[/tex].
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 [tex](a + 5b + 3c) + (0 + 5(3) + 3(-5)) = (a + 5b + 3c) + (0) = a + 5b + 3c = v[/tex]

So we found an element [tex]w = (0, 3, -5)[/tex] such that [tex]v + w = v[/tex] 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
[tex]S = \{ (x_1, x_2, x_3) \in F^{3} \ : \ x_1 + 5x_2 + 3x_3 = 0 \} [/tex]
means.
 

1. What is a vector space?

A vector space is a mathematical structure that consists of a set of objects, called vectors, and a set of operations, such as addition and scalar multiplication, that can be performed on the vectors. These operations must follow certain rules, such as the commutative and associative properties, to maintain the structure of a vector space.

2. What is the additive identity in a vector space?

The additive identity in a vector space is the vector that, when added to any other vector, results in that same vector. In other words, it does not change the value of the vector when added. This is similar to the number 0 in regular arithmetic, where any number added to 0 remains unchanged.

3. Why is the additive identity unique in a vector space?

The additive identity is unique in a vector space because it is the only vector that satisfies the definition of the additive identity. This means that there is only one vector that, when added to any other vector, results in that same vector. If there were more than one additive identity, it would violate the rules of a vector space.

4. How is the uniqueness of the additive identity proven in a vector space?

The uniqueness of the additive identity can be proven using mathematical proofs and the properties of a vector space. For example, one can show that if there were two different additive identities, then they would have to be equal to each other, which contradicts the definition of uniqueness.

5. Why is the uniqueness of the additive identity important in a vector space?

The uniqueness of the additive identity is important in a vector space because it is a fundamental property that ensures the consistency and structure of the space. Without a unique additive identity, the operations in a vector space would not follow the necessary rules and the space would not be considered a vector space. This property allows for the use of algebraic techniques and theorems in vector spaces.

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