isomorphic


by terryfields
Tags: isomorphic
terryfields
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#1
Jan5-09, 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 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????
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Big-T
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#2
Jan5-09, 05:21 PM
P: 64
How many basis vectors do you need to span P_k?
HallsofIvy
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#3
Jan5-09, 07:53 PM
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PF Gold
P: 38,879
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
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#4
Jan6-09, 08:21 AM
P: 45

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?
Dick
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#5
Jan6-09, 08:31 AM
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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
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#6
Jan7-09, 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 Pk isomorphic to n has to be k+1? (crosses fingers and preys)
Dick
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#7
Jan7-09, 01:33 PM
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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
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#8
Jan7-09, 05:09 PM
P: 45
thanks, thats much simpler than it first looked.


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