Question

If a, b ∈ R with a not equal to 0, show that the infinite set {1,(ax + b),(ax + b)² ,(ax + b)³ , · · · } of polynomials is a basis for F[x].

Answer 1

We know the standard basis of F[x] is,

B = {1,x,x²,x³,x⁴,.....}

Here the basis is, B' = {1,(ax+b),(ax+b)²,(ax+b)³,(ax+b)⁴,....}

To show that B' is a basis of F[x], we will define a bijective map from B onto B'

Define, f : B ------> B' by,

f(1) = 1 and, f(xⁿ) = (ax+b)ⁿ for all n > 0

Clearly, f is well defined.

f is one-to-one, because, f(xⁿ) = f(x^m) implies,

(ax+b)ⁿ = (ax+b)^m

So, n = m

So, xⁿ = x^m

So, f is one-to-one.

Again, for any (ax+b)ⁿ in B' we have, xⁿ in B such that,

f(xⁿ) = (ax+b)ⁿ

So, f is onto.

Hence, f is a bijection.

Since, B is a basis of F[x] so B' is a basis of F[x] as well.