Vector space axioms

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diogomgf

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[MATH](E, +, \cdot)[/MATH] is a vector space [MATH]\in \mathbb{R}[/MATH] with [MATH]u+v = u-v[/MATH] and [MATH]a\cdot u= -a \cdot u[/MATH], for any [MATH]u,v \in E[/MATH] and any [MATH]a,b \in \mathbb{R}.[/MATH]
I tried to see if the multiplication axioms are satisfied by this definitions:

-Ax1: [MATH]a(u+v) = au + av \Leftrightarrow a(u+v) = -a(u+v) \Leftrightarrow au+av = -au-av \Leftrightarrow -2au = 2av \Leftrightarrow -u= v.[/MATH] So axiom 1 isn't satisfied by the definition.
-Ax2: [MATH](a+b)u = au + bu \Leftrightarrow (a+b)u = -au-bu \Leftrightarrow au+bu = -au-bu \Leftrightarrow 2au = -2bu \Leftrightarrow au = -bu \Leftrightarrow a = -b [/MATH] So axiom 2 isn't satisfied by the definition.
-Ax3: [MATH](ab)u = a(bu) \Leftrightarrow -(ab)u = -a(bu) \Leftrightarrow -abu = -abu [/MATH] So axiom 3 is satisfied by the definition.
-Ax4: [MATH] 1u = u \Leftrightarrow 1u= -1u \Leftrightarrow u = -u [/MATH] So axiom 4 isn't satisfied by the definition.

Is this correct?
 
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Axiom 1: ax(u+v) = ax(u-v) = -a(u-v)=-au+av
axu + axv = -au + -av = -au - -av = -au + av
 
I did not check all of them. It is very difficult for me to read what you wrote because you use the same symbol to represent different things. For example some + signs mean one thing and other + signs mean something else. Maybe use + to mean the unusual + sign and + to mean the usual + sign. It really is painful to follow your work especially with a headache which I have. If I have time later and am feeling better I will go through your work more carefully.
 
diogomgf said:
[MATH](E, +, \cdot)[/MATH] is a vector space [MATH]\in \mathbb{R}[/MATH] with [MATH]u+v = u-v[/MATH] and [MATH]a\cdot u= -a \cdot u[/MATH], for any [MATH]u,v \in E[/MATH] and any [MATH]a,b \in \mathbb{R}.[/MATH]
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A vector space(Linear Space) is a pair \((\mathcal{X},\mathcal{F})\) where \(\mathcal{F}\) is a field and \(\mathcal{X}\) is an Abelian group with respect to the operation. But the given vector operation is not commutative.
 
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pka said:
A vector space(Linear Space) is a pair \((\mathcal{X},\mathcal{F})\) where \(\mathcal{F}\) is a field and \(\mathcal{X}\) is an Abelian group with respect to the operation. But the given vector operation is not commutative.
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Not sure what you mean by that...
 
@Jomo I think i'm too confused on this subject...
Could you help me prove the axiom 2 in this case?

The O.P symbols are
[MATH]u\oplus v = u+(-v)[/MATH] and [MATH]a\odot u = -au[/MATH]
 
Where are you stuck? What are you trying to show. This is a help forum where we help students solve their problems on their own with our help.
So please state what you want to show and then try to show it so we can see where you are making a mistake.
 
Jomo said:
Where are you stuck? What are you trying to show. This is a help forum where we help students solve their problems on their own with our help.
So please state what you want to show and then try to show it so we can see where you are making a mistake.
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If E is a vector space under the real numbers, with addition and multiplication defined as:
[MATH] u\boxplus v = u+(-v); \space a \boxdot u = (-a)u[/MATH]For any [MATH]a,b \in \mathbb{R}; \space u,v \in E [/MATH]
How can I prove the multiplication distributive axioms:
[MATH]a \boxdot (u\boxplus v) = a\boxdot u \boxplus a \boxdot v [/MATH][MATH](a+b)\boxdot u = a\boxdot u \boxplus b \boxdot u[/MATH]
You already corrected my attempt at proving the first axiom :)
The second one I think I can do it now, thank you
 
How can you prove/disprove a⊡(u⊞v)=a⊡u⊞a⊡v? By definition!
a⊡(u⊞v) = a⊡(u-v)= -a(u-v)
a⊡u⊞a⊡v=(a⊡u)⊞(a⊡v) =-au ⊞ -av = -au+ av= -a(u-v)

Now you try the next one. All I used was the definition which you supplied me with.
 
diogomgf said:
Not sure what you mean by that...
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Actually I meant exactly that. A vector space is a pair, \((\mathcal{X},\mathcal{F})\).
Where \(\mathcal{X}\) is Abelian group with respect to its operation; and \(\mathcal{F}\) is simply a field.
Now if you have a textbook that uses a different definition then you need to not post here or else find someone willing to abandon almost two hundred years of usage to work with you.
BTW: The common usage is to say \(\mathcal{X}\) is a vector space over the field \(\mathcal{F}\).
 
Jomo said:
How can you prove/disprove a⊡(u⊞v)=a⊡u⊞a⊡v? By definition!
a⊡(u⊞v) = a⊡(u-v)= -a(u-v)
a⊡u⊞a⊡v=(a⊡u)⊞(a⊡v) =-au ⊞ -av = -au+ av= -a(u-v)

Now you try the next one. All I used was the definition which you supplied me with.
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Yep, thank you. ?
I'll present the proofs tomorrow, its very late here
 
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pka said:
Actually I meant exactly that. A vector space is a pair, \((\mathcal{X},\mathcal{F})\).
Where \(\mathcal{X}\) is Abelian group with respect to its operation; and \(\mathcal{F}\) is simply a field.
Now if you have a textbook that uses a different definition then you need to not post here or else find someone willing to abandon almost two hundred years of usage to work with you.
BTW: The common usage is to say \(\mathcal{X}\) is a vector space over the field \(\mathcal{F}\).
Click to expand...
The notation is correct, its just that the definition is not presented that way in the book. There is no definition for an Abelian group, nor is it mentioned. It doesn't even mention commutative groups when presenting the definition of a vector space. After searching online for Abelian group, and comparing with the axioms in the definition of a vector space, I can see that you and the book are saying the same thing, just differently. Maybe because that notation is used mostly in abstract algebra?

Anyway, sorry I spelled under instead of over. In portuguese we say A 'sobre' B, wich means A over B as you mentioned. But when saying A under B we say A 'sob' B. They are very similar and maybe thats where I got confused...

Cheers.
 
@Jomo

[MATH](a+b)\boxdot u = -(a+b)\cdot u = (-a-b)\cdot u = -au-bu[/MATH][MATH]a\boxdot u \boxplus b\boxdot u = (-au) \boxplus (-bu) = -au - (-bu) = -au + bu[/MATH]
This axiom isn't satisfied by the definition.
 
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