Show something is a vector space.

[hopsonuk]

1. Homework Statement

[PLAIN]http://img87.imageshack.us/img87/9001/150674.png

2. Homework Equations



3. The Attempt at a Solution

I think, I need to write: a(x)[b(+)c]

Then prove:

Distributivity law 1: Unsure of
Distributivity law 2: Proving: a(x)[b(+)c] = [a(x)b](+)[a(x)c]
Associativity law: Proving: a(x)[b(+)c] = [a(x)b](+)c
Identity: Unsure of

Is this along the correct lines?[/CODE][/CODE]

 

[VietDao29]

1. Homework Statement

[PLAIN]http://img87.imageshack.us/img87/9001/150674.png

2. Homework Equations



3. The Attempt at a Solution

I think, I need to write: a(x)[b(+)c]

Then prove:

Distributivity law 1: Unsure of
Distributivity law 2: Proving: a(x)[b(+)c] = [a(x)b](+)[a(x)c]
Associativity law: Proving: a(x)[b(+)c] = [a(x)b](+)c
Identity: Unsure of

Is this along the correct lines?[/CODE][/CODE]

Imageshack is changing some of their policies. Can you move to another host like Photobucket or something along the line? To tell the truth, I cannot see your image. :(

 

[Robert1986]

I'll show you how to prove the first distributive law, and then you can take it from there.

Let r,s be in Q and let v= (b_1,...,b_n) be in R^{n}. Then, you need to show that (r+s)v = rv (x) sv.

So, (r+s)v = (r+s)(b_1,...,b_n) = (b_1^{r+s},...,b_n^{r+s})=(b_1^{r},...,b_n^{r}) (x) (b_1^{s},...,b_n^{s}) = rv (x) sv as desired.

Now, just do similar things for each of the axioms.

 

[hopsonuk]

Imageshack is changing some of their policies. Can you move to another host like Photobucket or something along the line? To tell the truth, I cannot see your image. :(

Ok :) - Here is a re-upload:

 

[hopsonuk]

Imageshack is changing some of their policies. Can you move to another host like Photobucket or something along the line? To tell the truth, I cannot see your image. :(

Is what I have written for distributivity law 2 correct?

 

[Robert1986]

Is what I have written for distributivity law 2 correct?

Yes.

 

[hopsonuk]

I think my associativity is incorrect.

 

[hopsonuk]

If i write this for associativity: b(x)[c(x)a] = (b x c)(x)a

I cannot get the RHS = LHS..

 

[Robert1986]

If i write this for associativity: b(x)[c(x)a] = (b x c)(x)a

I cannot get the RHS = LHS..

Hold on a second. Where does the (x) take place? It is defined as a function from RXR^n to R^n. a,b,c are all elements of R^n. You need to check associativity for (+), not (X), that is:
[a(+)b](+)c = a(+)[b(+)c]

 

[hopsonuk]

Ok, so how do I write this out to check LHS = RHS ?