Two vector spaces being isomorphic

[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 P_k denote the the vector space of all real polynomials of degree at most k. For which value of n is P_k 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?

 

[Big-T]

How many basis vectors do you need to span P_k?

 

[HallsofIvy]

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.

 

[terryfields]

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?

 

[Dick]

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.

 

[terryfields]

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 P_k isomorphic to n has to be k+1? (crosses fingers and preys)

 

[Dick]

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.

 

[terryfields]

thanks, that's much simpler than it first looked.