# Equivalent sets in different vector spaces?

1. Aug 25, 2006

### pivoxa15

{1, x, 2x^2} is a basis for V (the polynomial vector space with maximum power 2)

then could I say that the coordinate vectors with respect to V, which form the set {(1,0,0), (0,1,0), (0,0,2)} for R^3 is equivalent to the above set in V?

Although the word equivalent is not defined. But it is true that any property one set has automatically implies to the other set? If yes then it seems it is much easier to work with the set in R^3.

2. Aug 25, 2006

### mathwonk

what are you trying to understand?

3. Aug 26, 2006

### matt grime

You can formally identify V and any copy of R^3 by, or any 3d vector space over the same field, by identifying basis vectors, though I don't see why you'd want to do it in this case. They are after all just vector spaces and the only invariants ofa vector space are the underlying basefield and the dimension: i.e. given any two vector spaces of the same dimension over F they are (non-canonically) isomorphic.

4. Aug 26, 2006

### HallsofIvy

Of course, R3 and V are isomorphic.

I think what you are saying is that the function $f:V \rightarrow R^3$ defined by f(a+ bx+ cx2)= a(1, 0, 0)+ b(0, 1, 0)+ (c/2)(0, 0, 2) is an isomorphism. But then if two vector spaces are isomorphic then must have the same dimension. And then any function which maps a basis of one bijectively to a basis of the other is an isomorphism.

5. Aug 26, 2006

### shmoe

What you mean by "equivalent" is what they've called "isomorphic".

They will share all the properties that follow from their usual vector space qualities.

It's not really much easier to work with R^3 than the polynomials, a little less to write I suppose.

6. Aug 27, 2006

### pivoxa15

isomorphic is what I was looking for in equivalence. So properties in one set implies the other. That is what I needed to know.

7. Aug 28, 2006

### HallsofIvy

But note that the term "isomorphic" refers to the entire space, not just the two bases.

8. Aug 28, 2006

### ObsessiveMathsFreak

Although it is true that the two spaces are isomorphic under linear operations( addition, multiplication by scalars), they will not be isomorphic under more complicated operations. They certianly won't be isomorphic if you throw multiplication into the mix. x * x^2 = x^3 is not an element of the original space. If you like, that space is not closed under multiplication. As you can guess, there is no corressponding operation in R^3 that is going to do this; at least, not a nice neat one.

If all you're doing is adding, and multiplying by scalars, then yes, use the R^3 representation. This is what you will eventually have to do anyway if you were to say, represent this on a computer.

9. Aug 28, 2006

### HallsofIvy

The question was about vector spaces. The only operations defined for a vector space are addition and scalar multiplication.

10. Aug 28, 2006

### pivoxa15

Just to add to that I am only considering linear transformations. The one you were pointing to isn't a linear transformation. But does that make a nonlinear transformation? In what space could it occur?

11. Aug 29, 2006

### HallsofIvy

My point was that allowing multiplication of vectors in the way ObsessiveMathFreak was talking about gives an "Algebra", not a "Vector Space" which is what we were discussing. Multiplication is a "binary" operation (a*b requires a and b) while a "transformation" is "unary": a single vector is mapped into another. If you were working in an algebra in which multiplication of vectors was defined the such transformations as
A(v)= u*v (for fixed v) would be linear but B(v)= v*v would not.