Proving Vector Space Axioms: (-1)u=-u

In summary, rochfor1 is trying to prove that the additive inverse of a vector is unique. He is using the properties of vector spaces and the distributive law to do so. He is also worrying about the negative of a vector and trying to prove that it is unique.
  • #1
sbo
5
0
Hi. please anyone help me with vector spaces and the way to prove the axioms.

like proving that (-1)u=-u in a vector space.
 
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  • #2
Ok, here goes: -u is the unique element such that u + (-u) = 0 = (-u) + u, so all we have to do is show that (-1) u has this property. That's not too bad: u + (-1)u = (1 + -1)u = 0u = 0. (The other case is identical.) The first equality follows from the distributivity of scalar multiplication. The third equality follows from this computation: 0u = (0 + 0)u = 0u + 0u, so adding -0u to both sides, we get 0u = 0.
 
  • #3
Just a word about the jargon: axioms are rules that are given and don't need proving. Theorems are what you prove from the axioms. For example, rochfor1's proof uses axioms such as the distributivity of scalar multiplication to prove the theorem that (-1)u = -u.
 
  • #4
thanx now i know that i don't need to prove axioms, they are given. thanks to that.
 
  • #5
thanks now I know that I don't need to prove axioms, they are given. thanks to that.

Im still worried about this vector thing and ill try to prove that the negative of a vector in V is unique, thaks all :)
 
  • #6
I recommend that you start by proving that x+y=x+z implies y=z. The uniquess of the additive inverse follows from that.
 
  • #7
I would also like to make sure you understand what the difference between [itex](-1)\vec{u}[/itex] and [itex]-\vec{u}[/itex] and why we need to prove they are equal.

[itex](-1)\vec{u}[/itex] is the is the additive inverse of the multiplicative identity in the field of scalars (the real numbers if you like) multiplied by the vector [itex]\vec{u}[/itex]. [itex]-\vec{u}[/itex] is the additive inverse of vector [itex]\vec{u}[/itex]. It is not at all obvious that those two things have to be the same!

To show that they are the same you use the basic properties (axioms) of vector spaces: specifically that -1 and 1 are addivitive inverses in the field of scalars, that [itex]1\vec{u}= \vec{u}[/itex], that [itex]0\vec{u}= \vec{0}[/itex], and the distributive law [itex](a+ b)\vec{u}= a\vec{u}+ b\vec{u}[/itex].

[itex](1+ -1)\vec{u}= 0\vec{u}= \vec{0}[/itex]
and [itex](1+ -1)\vec{u}= 1\vec{u}+ (-1)\vec{u}= \vec{u}+ (-1)\vec{u}[/itex].

Since those are both equal to [itex](1+ -1)\vec{u}[/itex] they are equal to each other and [itex]\vec{u}+ (-1)\vec{u}= \vec{0}[/itex], which is precisely the definition of "additive inverse: [itex](-1)\vec{u}[/itex] is equal to the additive inverse of [itex]\vec{u}[/itex].
 
  • #8
HallsofIvy said:
[itex] 0\vec{u}= \vec{0}[/itex]
This step requres proof as well. We have

[tex]0\vec u=(0+0)\vec u=0\vec u+0\vec u[/tex]

which implies that

[tex]0\vec u+\vec 0=0\vec u+0\vec u[/tex]

and now the result [itex]0\vec u=\vec 0[/itex] follows from the theorem I mentioned in #6.
 
  • #9
if you are required to prove that -(-u)=u
can you say that proving that condition is equivalent to proving that -(-u)+(-u)=0 since you would have added a negative of a vector -u and worked it through until you arrive there?
 
  • #10
if you are required to prove that -(-u)=-u
can you say that proving that condition is equivalent to proving that -(-u)+(-u)=0 since you would have added a negative of a vector -u and worked it through until you arrive there?
 
  • #11
There are too many minus signs in what you wrote. You probably want to prove that [itex]-(-\vec u)=\vec u[/itex], i.e. that the additive inverse of [itex]-\vec u[/itex] is [itex]\vec u[/itex]. The axiom [itex]-\vec u+\vec u=\vec 0[/itex] says that [itex]\vec u[/itex] is an additive inverse of [itex]-\vec u[/itex], and if you have already proved that the additive inverse is unique, you can safely conclude that [itex]\vec u[/itex] is the additive inverse of [itex]-\vec u[/itex]. As I said before, the uniqueness of follows from #6. Have you proved that one yet?
 

1. What is a vector space axiom?

A vector space axiom is a mathematical statement that defines the properties that a set of vectors must have in order to be considered a vector space. These axioms include closure under addition and scalar multiplication, as well as the existence of an additive identity and inverse for each vector.

2. How is (-1)u=-u used to prove vector space axioms?

The equation (-1)u=-u is one of the axioms that is used to prove that a set of vectors is closed under scalar multiplication. It states that multiplying a vector by -1 is the same as taking the additive inverse of that vector. This is an important property in order for a set of vectors to be considered a vector space.

3. Why is proving vector space axioms important?

Proving vector space axioms is important because it allows us to determine whether or not a set of vectors satisfies all of the necessary properties to be considered a vector space. This is essential for understanding the fundamental properties of vectors and their operations, and is a crucial concept in many areas of mathematics and science.

4. Can (-1)u=-u be used to prove all vector space axioms?

No, (-1)u=-u is only one of the many axioms that are used to prove that a set of vectors is a vector space. In order to fully prove that a set of vectors is a vector space, all of the axioms must be satisfied. Some other axioms include associativity and distributivity of scalar multiplication over vector addition.

5. Are all vector spaces proven using the same set of axioms?

Yes, all vector spaces are proven using the same set of axioms. These axioms are defined and accepted by the mathematical community and are used as the standard for determining whether a set of vectors is a vector space. However, different sets of vectors may require different methods of proof to show that they satisfy the axioms.

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