Splitting Polynomials into Even and Odd Parts: A Unique Direct Sum Decomposition

[Gregg]

1. Let $$\mathbb{R}[x]_n^+$$ and $$} \mathbb{R}[x]_n^-$$ denote the vector subspaces of even and odd polynomials in $$\mathbb{R}[x]_n$$

Show $$\mathbb{R}[x]_n=\mathbb{R}[x]_n^+ \oplus\mathbb{R}[x]_n^-$$

3. For every $$p^+(x) \in \mathbb{R}[x]_n^+$$ $$\displaystyle p^+(x)=\sum_{m=0}^n a_m x^m=p^+(-x)$$

So $$a_m = 0$$ for $$m=2k+1, k=0,1,2,...$$ else $$a_m \in \mathbb{R}$$. Similarly, $$a_m=0$$ for $$m=2k, k=0,1,2,...$$ if the function is odd.

$$p^+(x)=a_0+a_2x^2+a_4x^4+\cdots, a_m\in\mathbb{R}$$

$$p^-(x)=a_1x+a_3x^3+a_5x^5+\cdots a_m\in\mathbb{R}$$

$$p(x)=a_0+a_2x^2+a_4x^4+\cdots+a_1x+a_3x^3+a_5x^5+\cdots$$ for every $$p(x)\in \mathbb{R}[x]_n$$. So every $$p(x)$$ is some $$p^+(x)$$ with some $$p^-(x)$$. Is this enough? Is it better to find a basis for the two subspaces and show that the union of the two basis sets spans $$\mathbb{R}[x]_n$$ ?

 

[Dick]

All that stuff about odd and even powers is true, but it's probably easiest to prove it if you use the projection operators p+(x)=(p(x)+p(-x))/2 and p-(x)=(p(x)-p(-x))/2.

 

[Gregg]

Say p+(x)=(p(x)+p(-x))/2 and p-(x)=(p(x)-p(-x))/2. Then say that p(x)= p+(x)+p-(x) for all p(x) in R[x]_n?

 

[Dick]

Say p+(x)=(p(x)+p(-x))/2 and p-(x)=(p(x)-p(-x))/2. Then say that p(x)= p+(x)+p-(x) for all p(x) in R[x]_n?

Sure. That splits a polynomial into an even part and an odd part, which is exactly the power split you are talking about. Then I suppose you would want to show that the split is unique. That if p(x)=e(x)+o(x) where e(x) is even and o(x) is odd, then e(x)=p+(x) and o(x)=p-(x), right? Might depend on how exactly your definition of 'direct sum' is phrased.