Prove that there is a basis for the $n$th degree polynomial vector space consisting $only$ of $n$th degree polynomials.

Question

I tried looking around for an answer to this question, but I couldn't find anything specific enough. I'm thinking I need to use either Cartesian products or subspace sums to produce such a vector space, however I'm still fuzzy on how to actually implement it.

Any help would be great!

Answer 1

Hint:

We know that $\{1, x, x^2, \ldots, x^n\}$ is a basis for your space. But then $$\{1+x^n, x+x^n, x^2+x^n, \ldots, 2x^n\}$$

is also a basis for your space.

Answer 2

Pick any basis $f_1, \ldots, f_n, f_{n+1}$. At least one of these polynomials must have degree exactly $n$, otherwise $x^n$ would not be in the generated subspace. WLOG $\deg f_1 =n$. If some of the other basis vectors have smaller degree, add $f_1$ to them (and only to those with degree strictly less than $n$). That yields a solution you were looking for.