Q : Is this vector subspace? and dimension of vector subspace Q :Is a set A = {p(x)$\in$$\mathbb{P_{2n}}$|p(-x)=-p(x)} vector subspace of Vector space $\mathbb{P_n}$
if the set A is a vector subspace of Vector space $\mathbb{P_n}$ then, what is the dimension of vector subspace?
My thought : $p_1$(x), $p_2$(x)$\in$A and $\alpha$,$\beta$$\in$$\mathbb{R}$ $\Rightarrow$ $p_1$(-x)=-$p_1$(x),$p_2$(-x)=-$p_2$(x) $\Rightarrow$ $\alpha$$p_1$(-x)=-$\alpha$$p_1$(x), $\beta$$p_2$(-x)=-$\beta$$p_2$(x) $\Rightarrow$ $\alpha$$p_1$(-x)+$\beta$$p_2$(-x)=-$\alpha$$p_1$(x)+-$\beta$$p_2$(x) $\Rightarrow$ $\alpha$$p_1$(x)+$\beta$$p_2$(x) $\in$ A
$\Rightarrow$ A is a vector subspace of Vector space $\mathbb{P_n}$
But I don't know about method of the dimension of vector subspace of A
All I know is definition of the dimension. Should I approach it assuming a basis?
Please help me. Thank you.

First this is a vector subspace:
$(\alpha p_1+p_2)(-x) = \alpha p_1(-x)+ p_2(-x)=-\alpha p_1(x)-p_2(x) = -(\alpha p_1+p_2)(x)\Rightarrow$ $A$ is a vector subspace of Vector space $\mathbb{P}_n.$
How $A = \{ p(x) \in \mathbb{P_{2n}}:p(-x)=-p(x) \},$ then a element of $\mathbb{P}_{2n}$ is $p(x) = a_0+a_1x+\cdots a_{2n}x^{2n}$. The condition $p(-x)=-p(x)$ implies $-p(x) = -a_0-a_1x-\cdots -a_{2n}x^{2n} = p(-x) = a_0-a_1x+a_2x^2-a_3x^3+\cdots -a_{2n-1}x^{2n-1}+a_{2n}x^{2n}.$
so comparing the two sides of the above equality, you see that the terms of odd indices cancel each other out, and so
$$ 2a_0+2a_2x^2+2a_4x^4+ \cdots 2a_{2n}x^{2n} = 0,$$
this implies $a_0 = a_2 = a_4 = \ldots =a_{2n}=0.$
so the shape of $p(x)$ is $p(x) = a_1x+a_3x^2+\cdots +a_{2n-1}x^{2n-1}$. The set $\{x,x^3,x^5, \cdots, x^{2n-1}\}$ is basis of $A$, therefore its dimension is $n$.
