# How is the Set of all Polynomials Equal to the Following Union?

Given that $P(F)$ is the set containing all polynomials with coefficients from field $F$, I am given the following:

$W_1$ is the set of all polynomials $f(x)$ in $P(F)$ such that for:

$$f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$$

$a_i=0$ if $i$ is even. Similarly, $W_2$ is the set of all polynomials $g(x)$ in $P(F)$ such that for:

$$g(x)=b_nx^n+b_{n-1}x^{n-1}+...+b_1x+b_0$$

$b_i=0$ if $i$ is odd. I am then asked to prove that $P(F)=W_1\oplus W_2$.

I am trying to understand, conceptually, how these two sets can possibly equal one another. The way I am understanding the set $W_1 \oplus W_2$ is that it is a set containing all polynomials with only even powers and all polynomials with only odd powers. However the vector space $P(F)$ contains these polynomials, which exist in $W_1 \oplus W_2$ as well as polynomials with both even and odd powers. This would mean that $P(F)\neq W_1\oplus W_2$. This is, obviously, incorrect because I am being asked to prove that the two sets are equal.

Where is the error in my reasoning?

## 2 Answers

Make sure you understand the definition of $\oplus$; $W_1\oplus W_2$ is very different from $W_1 \cup W_2$! For example, $\mathbb R^2$ is the direct sum of the $x$-axis $=$Span$\left\{\begin{pmatrix}1 \\ 0\end{pmatrix}\right\}$ and the $y$-axis, but it certainly isn't the union of the axes.

• I was interpreting $/oplus$ as an exclusive OR, because that's the only context I've ever seen it in. A quick Google search of "direct sum" clears everything up. Thank you. Commented Jan 26, 2015 at 22:56
• You know, that is confusing! I never thought of that. Commented Jan 26, 2015 at 23:49

You just have to separate the even powers from the odd powers, as in this example:

$$a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5=(a_1x+a_3x^3+a_5x^5)+(a_2x^2+a_4x^4).$$