- #1
freddyfish
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The isomorphism of ℝ5 and P4 is obvious for the "standard" inner product space.
The following question arise from an example in my course literature for a course in linear algebra. The example itself is not very difficult, but there is a statement without any proof, that if the inner product is determined by:
<p(t), q(t)>=Ʃp(ti)q(ti)
where tk=-2, -1, 0, 1, 2, respectively, (1≤k≤t)
then the polynomials in P4 are uniquely determined by their values at the given values of tk and each polynomial can be represented as a vector in ℝ5 where the entries of that the vector (from the top) are the polynomials value for each tk (same order as above).
Intuition confirmes this, but as far as I'm concerned intuition won't prove neither the uniqueness nor why the corresponding vectors in ℝ5 is formed from each polynomial's value for each given tk. Can anyone help me to prove this or at least present some idea that might be useful?
I would appreciate it //Freddy
The following question arise from an example in my course literature for a course in linear algebra. The example itself is not very difficult, but there is a statement without any proof, that if the inner product is determined by:
<p(t), q(t)>=Ʃp(ti)q(ti)
where tk=-2, -1, 0, 1, 2, respectively, (1≤k≤t)
then the polynomials in P4 are uniquely determined by their values at the given values of tk and each polynomial can be represented as a vector in ℝ5 where the entries of that the vector (from the top) are the polynomials value for each tk (same order as above).
Intuition confirmes this, but as far as I'm concerned intuition won't prove neither the uniqueness nor why the corresponding vectors in ℝ5 is formed from each polynomial's value for each given tk. Can anyone help me to prove this or at least present some idea that might be useful?
I would appreciate it //Freddy