Show something is a vector space.

In summary: Ok, so how do I write this out to check LHS = RHS ?In the most general way, there are a few steps to proving an equality:1. Start by writing out the left-hand side (LHS) and right-hand side (RHS) of the equality, using the definitions and assumptions given in the problem.2. Work on one side at a time, using known rules and properties to simplify the expression as much as possible. This may involve using properties of addition, multiplication, and other operations, as well as assumptions about the objects involved.3. Once you have simplified both sides as much as possible, compare them to see if they are equal. If they are, then you have proven the equality. If
  • #1
hopsonuk
16
0
1. Homework Statement

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

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]
 
Last edited by a moderator:
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  • #2
hopsonuk said:
1. Homework Statement

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

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. :(
 
Last edited by a moderator:
  • #3
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 [tex]v= (b_1,...,b_n)[/tex] be in [tex]R^{n}[/tex]. Then, you need to show that (r+s)v = rv (x) sv.

So, [tex](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[/tex] as desired.

Now, just do similar things for each of the axioms.
 
  • #4
VietDao29 said:
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:

Wtc.png
 
  • #5
VietDao29 said:
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?
 
  • #6
hopsonuk said:
Is what I have written for distributivity law 2 correct?

Yes.
 
  • #7
I think my associativity is incorrect.
 
  • #8
If i write this for associativity: b(x)[c(x)a] = (b x c)(x)a

I cannot get the RHS = LHS..
 
  • #9
hopsonuk said:
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]
 
  • #10
Ok, so how do I write this out to check LHS = RHS ?
 

1. What are the defining properties of a vector space?

A vector space is a set of elements that can be added together and multiplied by scalars, usually real or complex numbers. The defining properties of a vector space include closure under addition and scalar multiplication, associativity, commutativity, existence of a zero vector, existence of additive inverses, and compatibility with field multiplication.

2. How do you prove that a set is a vector space?

To prove that a set is a vector space, you must show that it satisfies all of the defining properties. This can be done by checking each property individually, using the given operations and elements of the set. If all properties are satisfied, the set is considered a vector space.

3. Can a vector space have more than one zero vector?

No, a vector space can only have one zero vector. This is because the zero vector must satisfy the property of being the additive identity, meaning that when added to any other vector, it results in that same vector. If there were multiple zero vectors, this property would not hold.

4. What is the difference between a vector space and a subspace?

A subspace is a subset of a vector space that also satisfies all of the defining properties of a vector space. In other words, a subspace is a smaller vector space contained within a larger vector space. All vector spaces have at least two subspaces: the zero subspace, which consists of only the zero vector, and the original vector space itself.

5. Can a set of functions be a vector space?

Yes, a set of functions can be a vector space as long as it satisfies all of the defining properties. For example, the set of all continuous functions on a certain interval is a vector space, with the operations of addition and scalar multiplication defined in the usual way for functions. However, not all sets of functions will be vector spaces, as they may not satisfy all of the necessary properties.

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