# Two vector spaces being isomorphic

1. Jan 5, 2009

### terryfields

sadly not been able to put much effort into this one! was a lecture i missed towards the end of term and didnt get the notes on it, however here is the question.
for K>or equal to 1 let Pk denote the the vector space of all real polynomials of degree at most k. For which value of n is Pk isomorphic to Rn. Give a brief reason for your answer.

Now from what i have found on two vector spaces being isomorphic they need to have equal dimensions (dimu=dimv) so knowing that we have dimRn)=n is as far as i have got. Not really understanding this one, surely they would be isomorphic at any value of n as long as it's between 1 and k????

2. Jan 5, 2009

### Big-T

Re: isomorphic

How many basis vectors do you need to span P_k?

3. Jan 5, 2009

### HallsofIvy

Staff Emeritus
Re: isomorphic

For example, P1 is the space of all first degree polynomials which can be written in the form ax+ b which, in turn, can be mapped to (a,b) in R2.

Last edited: Jan 5, 2009
4. Jan 6, 2009

### terryfields

Re: isomorphic

so for a second degree polynomial you would have ax2 +bx +c so is the answer just n-1 because the dimension of a polynomial is always one higher than the degree?

5. Jan 6, 2009

### Dick

Re: isomorphic

A single polynomial doesn't have a 'dimension'. The point is that the set of ALL degree two or less polynomials can be represented as linear combinations of the three linearly independent functions 1, x and x^2. They form a basis. The 'dimension' of a space is the number of elements in a basis.

6. Jan 7, 2009

### terryfields

Re: isomorphic

Ok, so the basis formed by the polynomial is always going to be one higher than the degree of the polynomial, therefore value of n that will make Pk isomorphic to n has to be k+1? (crosses fingers and preys)

7. Jan 7, 2009

### Dick

Re: isomorphic

Preys? I think you want to pray. Sure, P_k has dimension k+1. R^n has dimension n. Two finite dimensional vector spaces are isomorphic if they have the same dimension.

8. Jan 7, 2009

### terryfields

Re: isomorphic

thanks, thats much simpler than it first looked.

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