# Prove that $P(\mathbb{F}) = U_e \oplus U_o$ (sum of two subspaces equals the vector space of all polynomials with coefficients in $\mathbb{F}$).

I need help with the following:

Consider the vector space $P(\mathbb{F})$ of all polynomials with coefficients in $\mathbb{F}$.

Let $U_e$ denote the subspace of $P(\mathbb{F})$ consisting of all polynomials $p$ of the form $p(z) = a_{0}z^{0} + a_{2}z^{2} + \cdots + a_{2m}z^{2m}$, and let $U_o$ denote the subspace of $P(\mathbb{F})$ consisting of all polynomials $p$ of the form $p(z) = a_{1}z + a_{3}z^{3} + \cdots + a_{2m+1}z^{2m+1};$ here $m$ is a nonnegative integer and $a_0, \ldots, a_{2m+1} \in \mathbb{F}$.

Prove that $P(\mathbb{F}) = U_e \oplus U_o$.

I understand that adding odd and even powers of $z$ will get us general polynomial of the vector space. The subspace $U_e$ being the space of all polynomials with even powers and $U_o$ being the subspace of all polynomials with odd powers.

In my textbook, there are two conditions that must be verified in order for the sum of subspaces to be considered a direct sum.

$V = U_e + U_o$ is satisfied. (We obtain our deserved vector space by adding the the two subspaces).

The second part, I am not so sure how to show. I must show that $U_e \cap U_o = \{0\}$. (Subspaces cannot be disjoint because they all contain the zero vector). In other words, I need to show that the two subspaces have no elements in common besides the zero vector.

Thanks for the help.

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Hint: If $$a_0 + a_2z^2 + \dots + a_{2k}z^{2k} = b_1z + b_3z^3 + \dots + b_{2l+1}z^{2l+1},$$ what can you say about $a_0, a_2, \dots, a_{2k}, b_1, b_3, \dots, b_{2l+1} \in\mathbb{F}$?
Further hint: If $c_0 + c_1z + \dots + c_mz^m = 0$, what can you say about $c_0, c_1, \dots, c_m \in \mathbb{F}$? –  Michael Albanese Aug 24 '13 at 5:06