# Is this a vector operation

1. Sep 12, 2010

### w3390

1. The problem statement, all variables and given/known data

Consider a new vector operation, 'star' (*), defined for two dimensional vectors by A*B=(A1-B1,A2+B2). Is the resultant object a vector?

2. Relevant equations

3. The attempt at a solution

I want to say that the resultant object would be a vector but I do not know how to start to go about proving it. Any help on where I should start?

2. Sep 13, 2010

### Lanthanum

Re: vectors

It's one of those proofs that's decievingly easy.
Start by defining two general vectors;
let a=<x1,y1>
and b=<x2,y2> where x and y are real.

3. Sep 13, 2010

### w3390

Re: vectors

Okay, but I get confused on what sort of operation I should be doing since it is some sort of made up (*) operation. I already had what you just said. I didn't know where to go from there.

4. Sep 13, 2010

### JonF

Re: vectors

My proof would be something like this:

We define a vector to be an ordered pair in the form of <a,b> where a,b are reals.

Consider the binary operation on vectors A*B =(A1-B1,A2+B2), where A = <A1,A2> and B=<B1,B2>.

A1,B1,A2,B2 are all reals by the definitions of vectors. A1-B1 is a real since addition is a closed operation. A2+B2 is in the reals since addition is a closed operation. Thus (A1-B1,A2+B2) satisfies the definition of a vector.

Or do you mean an operation on a vector space? That question is a bit more fun.

5. Sep 13, 2010

### w3390

Re: vectors

What do you mean by closed operation?

6. Sep 13, 2010

### JonF

Re: vectors

Let ‘*’ be an operation defined over some space. Let a,b be ANY elements of the space. ‘*’ is said to be closed if a*b maps to a unique element in the space.

In normal human language: an operation is called closed over a set if it always produces exactly one element in the set.

But by vector do you mean an object in a vector space? If so there is a lot more to this proof.

7. Sep 13, 2010

### w3390

Re: vectors

Yes, I mean an object in vector space.

8. Sep 13, 2010

### JonF

Re: vectors

clarification: they want you to prove that * makes a vector space, or a subspace correct?

If so, this is much more fun then, but revolves around similar ideas. I would look up the definition of a vector space (should be about 8 properties) and show that the element A*B meets (or doesn’t meet) all 8 of these requirements. Remember it only has to fail one property with any elements in the set to not be a vector space.

Last edited: Sep 13, 2010
9. Sep 13, 2010

Re: vectors

Ok. Thanks.