Question

T/F:
The dimension of Pn(F) is n.

Answer is false; why? Does it just depend on how many vectors you are using?

Answer 1

If \(P_n(F)\) is a vector space of n degree polynomials, you can form a basis using \(\{1, x, x^2, ... , x^n\}\). Thus\(P_n(F)\) has dimension n+1.

Answer 2

thanks, that makes sense