Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 6 down vote accepted

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 $a_0, a_2, \dots, a_{2k}, b_1, b_3, \dots, b_{2l+1} \in\mathbb{F}$?

share|cite|improve this answer
I'm assuming that the above equality only holds if all the coefficients are equal to 0. Is this proper reasoning? – St Vincent Aug 24 '13 at 2:51
That's the correct conclusion. Do you know why it is true? – Michael Albanese Aug 24 '13 at 4:11
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
The general polynomial will only equal to 0 if the coefficients are all 0. Each term in the polynomial will be Independent so the only coefficient that is an element of F that will make the equality true is 0. Is this correct? Thanks a lot for your help! I am new to this website and I love it already! – St Vincent Aug 24 '13 at 6:09
You are correct. You can rearrange the equation in my original hint so that you get a polynomial equal to zero and then apply your conclusion. – Michael Albanese Aug 24 '13 at 6:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.