
#1
Jan509, 11:03 AM

P: 45

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 R^{n}. 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 dimR^{n})=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
Jan509, 05:21 PM

P: 64

How many basis vectors do you need to span P_k?




#3
Jan509, 07:53 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

For example, P_{1} 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 R^{2}.




#4
Jan609, 08:21 AM

P: 45

isomorphic
so for a second degree polynomial you would have ax^{2} +bx +c so is the answer just n1 because the dimension of a polynomial is always one higher than the degree?




#5
Jan609, 08:31 AM

Sci Advisor
HW Helper
Thanks
P: 25,165

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
Jan709, 11:58 AM

P: 45

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)




#7
Jan709, 01:33 PM

Sci Advisor
HW Helper
Thanks
P: 25,165

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
Jan709, 05:09 PM

P: 45

thanks, thats much simpler than it first looked.



Register to reply 
Related Discussions  
prove not isomorphic?  Linear & Abstract Algebra  2  
Isomorphic groups?  Linear & Abstract Algebra  3  
Isomorphic Help!  Calculus & Beyond Homework  4  
isomorphic  Calculus & Beyond Homework  4  
Isomorphic: PLease Help  Calculus & Beyond Homework  7 