Find "a'' basis for $W = [ p(x) ∈ P_5(R) | p(−x) = p(x)]$

Question

$W = [ p(x) ∈ P_5(R) | p(−x) = p(x)]$

If W is a subspace, then find a basis for W.

Hi. So far, I proved that it is a subspace of W but i don't know how to find the basis. What would be my approach?

Answer 1

NOTE: The condition $p(x)=p(-x)$ is for even functions. Thus only polynomials that will be even functions (linear combination of even functions) will survive.

But here is a rather detailed way of doing this.

A general fifth degree (or lower) polynomial will have the form $$p(x)=ax^5+bx^4+cx^3+dx^2+ex+g.$$ For $p(x) \in W$ we need $$ p(-x) = -ax^5+bx^4-cx^3+dx^2-ex+g=\color{red}{ax^5+bx^4+cx^3+dx^2+ex+g=p(x)}. $$ This gives $$ax^5+cx^3+ex=0 \qquad \forall x \in \mathbb{R}.$$ This can happen only when $a=c=e=0$. Thus the $p(x) \in W$ has to be of the form $$p(x)=bx^4+dx^2+g.$$ So a basis for $W$ is $\{1,x^2,x^4\}$.

Answer 2

Consider the linear map $T\colon P_5(\mathbb{R})\to P_5(\mathbb{R})$, where for $p\in P_5(\mathbb{R})$, $T(p)=p(x)-p(-x)$. (Linearity is an easy verification.)

Then $W$ is the kernel of $T$, hence a subspace. The matrix of $T$ with respect to the basis $\{1,x,x^2,x^3,x^4,x^5\}$ is $$ A=\begin{bmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$ Gaussian elimination leads to the reduced row echelon form $$ \begin{bmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$ so a basis for the null space of $A$ is given by the vectors $$ \begin{bmatrix}1 \\ 0 \\ 0 \\ 0 \\ 0\end{bmatrix} \quad \begin{bmatrix}0 \\ 0 \\ 1 \\ 0 \\ 0\end{bmatrix} \quad \begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \\ 1\end{bmatrix} $$ and so a basis for $W$ is $\{1,x^2,x^4\}$.