Let F be a field. By using the division algorithm for polynomials, one can prove

Question

Let F be a field. By using the division algorithm for polynomials, one can prove that any polynomial in P_d(F) having more than d roots must be the zero polynomial. (You may assume this fact for this problem.) Let a_0, a_1, ..., a_d be distinct scalars in F and consider the map T: P_d(F) rightarrow F^d+1 given by T(p) = (p(a_0), p(a_1), ..., p(a_d)). (a) Show that T is a linear transformation. (b) Prove that ker (T) = {0} and deduce that T is an isomorphism. (c) Show that, for any list of d + 1 points (a_0, b_0), ..., (a_d, b_d) with distinct first coordinates, there exists a unique polynomial of degree at most d having the property that p(a_i) = b_i for each 0 lessthanorequalto i lessthanorequalto d.

Explanation / Answer

Let M,P belongs to Pd(F),

Then T(M+P)=(M(ao)+P(ao),M(a1)+P(a1),.....)

=(M(ao),M(a1),..........)+(P(ao)+P(a1)+........ )

=T(M)+T(P) ie,T is a linear transformation.

b)let T(P)=T(M) which implies (P(a0),P(a1),......)=(M(a1),M(a2),......) which implies P(a0)=M(ao),P(a1)=M(a1),... Ie,P=M injective.there clearly ker (T)={0} by a result.a bijective linear map is called an isomorphism.clearly T is onto since

im (T)=Fd+1.therefore it is bijective.ie,T is an isomorphism.

c)by a theorem we know that if U and V are vectorspace {u1,u2,...} are basis of U and v1,v2,. Vn are elements of V then there is a unique linear mapping f such that f(ui)=vi.by applying this theorem in this transformation there exist a unique polynomial of degree atmost d having the property P(ai)=bi.

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