Answer to Question #95197 in Linear Algebra for Khadag singh

Question

Check p(x) + p(−x) ∈ P
(e)
for every p(x) ∈ R(x). Check that the map
ψ : R[x] → P
(e) given by ψ(p(x)) = p(x)+p(−x)
2
is a linear map. Further, check that
ψ
2 = ψ. Determine the kernel of ψ.

Answer 1

1)Let $p(x)\in \mathbb R[x]$ and $q(x)=p(x)+p(-x)$ . We have $q(-x)=p(-x)+p(-(-x))=p(-x)+p(x)=q(x)$ , so $q(x)\in P(e)$

2)Check that $\psi$ is a linear map. Let $p(x)\in \mathbb R[x]$, $q(x)\in\mathbb R[x]$ and $\alpha\in\mathbb R$, then

$\psi(p(x)+q(x))=\frac{\bigl(p(x)+q(x)\bigr)+\bigl(p(-x)+q(-x)\bigr)}{2}=$

$=\frac{p(x)+p(-x)}{2}+\frac{q(x)+q(-x)}{2}=\psi(p(x))+\psi(q(x))$ and

$\psi(\alpha p(x))=\frac{\alpha p(x)+\alpha p(-x)}{2}=\alpha \frac{p(x)+p(-x)}{2}=\alpha\psi(p(x))$

So $\psi$ is a linear map

3)Check that $\psi^2=\psi$. Let $p(x)\in \mathbb R[x]$, then

$\psi^2(p(x))=\psi\bigl(\psi(p(x))\bigr)=\frac{\psi(p(x))+\psi(p(-x))}{2}=$

$=\frac{\frac{p(x)+p(-x)}{2}+\frac{p(-x)+p(-(-x))}{2}}{2}=\frac{p(x)+p(-x)}{2}=\psi(p(x))$

So $\psi^2=\psi$

4)Let $p(x)\in\ker\psi$ , then $0=\psi(p(x))=\frac{p(x)+p(-x)}{2}$ . We have $p(x)=-p(-x)$, that is $p(x)$ is an odd polynomial. So $\ker\psi$ is the set of odd polynomials.