Proof Vector Spaces: Unique Vector Satisfying "u + 0 = u

  • Thread starter soul
  • Start date
  • Tags
    Proof
In summary: The following is a summary of the content. In summary, the author is trying to prove that for the vector spaces, there is a unique vector that satisfies "u + 0 = u". He used contradiction technique. The last point that he reached is u + 0_1 = u + 0_2. However, he does not know whether he can say 0_1 = 0_2 after this statement or there are some other operations that he must do (like this statement needs a proof as well?).
  • #1
soul
62
0
Hi eveyone,

I was trying to prove that for the vector spaces, there is a unique vector that satisfy "u + 0 = u" and I used contradiction technique. The last point that I reached is [tex]u + 0_1 = u + 0_2[/tex]. However, I don't know whether I can say [tex]0_1 = 0_2[/tex] after this statement or there are some other operations that I must do (like this statement needs a proof as well?).

Thank you.
 
Last edited:
Physics news on Phys.org
  • #2
? Why do you need to prove that? Isn't that an axiom that every vector space has to satisfy? Namely that every vector space has a unique zero vector? What axioms do you start off with?

You can "prove" this by noting that along with your last step, -u also exists in the same vector space.
 
  • #3
Sorry, I guess I wrote my question wrong. I was trying to prove there is a unique identitiy element in summation and what I did is to select two different vectors and at the end of it, to show they are the same. I used the axiom in the question.
 
  • #4
Uniqueness of the identity is an axiom in groups but can be proved in a vector space.

In any vector space, there exist additive inverses and addition is commutative. Add the additive inverse of u to both sides of your equation.
 
  • #5
HallsofIvy said:
Uniqueness of the identity is an axiom in groups but can be proved in a vector space.

In any vector space, there exist additive inverses and addition is commutative. Add the additive inverse of u to both sides of your equation.

Is this true? Almost every group theory book I have looked at proves the uniqueness of the identity as a theorem.
 
  • #6
HallsofIvy said:
Uniqueness of the identity is an axiom in groups but can be proved in a vector space.

In any vector space, there exist additive inverses and addition is commutative. Add the additive inverse of u to both sides of your equation.

The axioms say there must exist a zero vector. It does not say it is unique or must be unique. You prove that it is unique if there exists such a vector.
 
  • #7
It is impossible to reply. You do not said definition of the vector space, and do not said about preceding procedure.
 

FAQ: Proof Vector Spaces: Unique Vector Satisfying "u + 0 = u

What is a proof vector space?

A proof vector space is a mathematical concept that refers to a set of vectors that satisfy certain axioms, or rules. These axioms include properties such as vector addition and scalar multiplication, which allow for the manipulation and combination of vectors in a consistent and logical manner.

Why is it important for a vector space to have a unique vector satisfying "u + 0 = u"?

The axiom "u + 0 = u" is known as the identity element property and is an essential property for vector spaces. It ensures that there is always a zero vector in the space, which is useful for many calculations and proofs. Additionally, having a unique vector satisfying this property ensures that the vector space is well-defined and consistent.

How do we prove that a vector space has a unique vector satisfying "u + 0 = u"?

To prove that a vector space has a unique vector satisfying "u + 0 = u", we must show that there is only one vector that satisfies this property. This can be done through logical reasoning and mathematical manipulation, using the axioms of vector spaces. The proof typically involves assuming the existence of two different vectors that satisfy the property and showing that they must be equivalent, thus proving uniqueness.

Can a vector space have more than one vector satisfying "u + 0 = u"?

No, a vector space can only have one vector satisfying "u + 0 = u" by definition. This is because the identity element property states that there is only one vector in the space that behaves like the zero vector, and all other vectors must be equivalent to this vector.

How does the uniqueness of the vector satisfying "u + 0 = u" impact the properties of a vector space?

The uniqueness of the vector satisfying "u + 0 = u" is crucial for the consistency and coherence of vector spaces. It ensures that the properties of vector addition and scalar multiplication hold true and allows for the use of these operations in mathematical calculations and proofs. Without this uniqueness, the properties of vector spaces would not be well-defined, and the space would not be a valid mathematical concept.

Similar threads

I Question from a proof in Axler 2nd Ed, 'Linear Algebra Done Right'
Replies
3
Views
2K
MHB Direct Sum Property: Proving Uniqueness
Replies
2
Views
2K
I Proving inequality about dimension of quotient vector spaces
Replies
0
Views
846
I Proving Convexity of the Set X = {(x, y) E R^2; ax + by <= c} in R^2
Replies
5
Views
2K
I Showing that a set of differentiable functions is a subspace of R
Replies
19
Views
5K
I When is a subset a subspace in a vector space?
Replies
7
Views
2K
I Question about an eigenvalue problem: range space
Replies
1
Views
1K
I How is uniqueness about the determinant proved by this theorem?
Replies
10
Views
2K
B Vector Proof: Solving for IuI and IvI using Dot Product Properties
Replies
10
Views
2K
MSSM Higgs Potential Homework: Get Equ. (1.70)
Replies
4
Views
1K

Hot Threads

Recent Insights

Back
Top