homomorphism question

Question

Check $p(x) + p(-x) ∈ P^e$ for every $p(x) ∈ \mathbb{R}[x]$. Check that the map $Ψ:\mathbb{R}[x]\to P^e$ 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

I assume that $$ P^e = \left\{ p(x) \in \mathbb{R}[X] \ \middle| \ p(x)=p(-x) \right\} $$ is the set of even polynomials.

For a $p(x) \in \mathbb{R}[X]$, define $f_p(x)=p(x)+p(-x)$. We calculate $$ f_p(-x) = p(-x)+p(x) = p(x)+p(-x) = f_p(x) $$ and thus $f_p(x) \in P^e$ for every $p(x)$. This implies that $\Psi \colon \mathbb{R}[X] \to P^e$ is well defined.

We check linearity as follows: Let $p(x),q(x) \in \mathbb{R}[X]$, then \begin{align*} \Psi (p(x)+q(x)) &= \frac{p(x)+q(x)+p(-x)+q(-x)}{2} \\ &= \frac{p(x)+p(-x)}{2} + \frac{q(x)+q(-x)}{2} = \Psi (p(x)) + \Psi (q(x)) \end{align*} and for $\lambda \in \mathbb{R}$, $$ \Psi(\lambda p(x)) = \frac{\lambda p(x) + \lambda p(-x) }{2} = \lambda \frac{p(x)+p(-x)}{2} = \lambda \Psi (p(x)) $$ which concludes the proof of the linearity.

We calculate $$ \Psi (\Psi (p(x))) = \Psi \left( \frac{p(x)+p(-x)}{2}\right) = \frac{\frac{p(x)+p(-x)}{2} + \frac{p(-x)+p(x)}{2}}{2} = \frac{p(x)+p(-x)}{2} = \Psi(p(x)), $$ i.e. $\Psi ^2 = \Psi$.

Finally $\Psi (p(x)) = \frac{p(x)+p(-x)}{2} = 0$, if and only if $p(x)+p(-x) =0$, so $$ \ker \Psi = \left\{ p(x) \in \mathbb{R}[X] \ \middle|\ p(-x)=-p(x) \right\}, $$ i.e. the kernel of $\Psi$ is the set of odd polynomials.