Vector spaces


by sbo
Tags: spaces, vector
sbo
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#1
Feb8-10, 11:31 PM
P: 5
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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rochfor1
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#2
Feb8-10, 11:40 PM
P: 257
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.
Rasalhague
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#3
Feb9-10, 02:54 AM
P: 1,400
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.

sbo
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#4
Feb9-10, 04:53 AM
P: 5

Vector spaces


thanx now i know that i dont need to prove axioms, they are given. thanks to that.
sbo
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#5
Feb9-10, 05:05 AM
P: 5
thanks now I know that I dont 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 :)
Fredrik
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#6
Feb9-10, 05:27 AM
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I recommend that you start by proving that x+y=x+z implies y=z. The uniquess of the additive inverse follows from that.
HallsofIvy
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#7
Feb9-10, 09:38 AM
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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].
Fredrik
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#8
Feb9-10, 10:36 AM
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Quote Quote by HallsofIvy View Post
[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.
sbo
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#9
Feb12-10, 04:09 PM
P: 5
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?
sbo
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#10
Feb12-10, 04:19 PM
P: 5
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?
Fredrik
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#11
Feb12-10, 05:48 PM
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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?


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