Find $\dim(V_0)$ and $\dim(V_0 ∩ Ve)$

Question

I have some confusion in this question A problem on comparison of dimension between two subspace of polynomial vector space.

Let $V$ be the vector space of all polynomials of degree at most equal to $2n$ with real coefficients. Let $V_0$ stand for the vector subspace $V_0 = \{P ∈ V :P(1) +P(−1) = 0\}$ and $V_e$ stand for the subspace of polynomials which have terms of even degree alone. If $\dim(U)$ stands for the dimension of a vector space $U$, then find $\dim(V_0)$ and $\dim(V_0 ∩ Ve)$.

My attempt :

If i take $n= 2$ , then dim $V=5$ that is polynomial of degree $4$

Now im constructing a polynomial $p(x) = a_0 + a_1 x+ a_2x^2 + a_3x^3 + a_4x^4 $

Now i take $x= 1. x=-1$

Now $V_o = \{ P \in V : P(1) + P(-1) = 0 \} $ that $P(1) + P(-1) = a_0+a_1+a_2 +a_3 +a_4 +a_0 - a_1 + a_2 - a_3 + a_4 = 2a_0 + 2a_2 + 2 a_4 $

so dim $V_0 $ = $3$

in general we can said that dimension $V_0 = 2n-1$ and dim $( V_0 \cap V_e)= 2n-1$

Is its true ??

Answer 1

The polynomials of $V$ have form $ a_0 + a_1 x+ a_2x^2 + a_3x^3 +...+ a_{2n}x^{2n}$ and therefore $\dim (V)=2n+1.$

The relation $P(1) + P(-1) = 0$ reduces the number of degrees of freedom by $1$ and so $\dim (V_0)=(2n+1)-1=2n.$

For $V_e$, we lose $n$ degrees of freedom since $a_1= a_3=...= a_{2n-1}=0$ and so $\dim (V_e)=(2n+1)-n=n+1.$

Then the further reduction of $1$ degree of freedom gives $\dim (V_0) \cap \dim (V_e)=n.$

Answer 2

Note first that, for a polynomial $P\in V_e$, $P(1)=P(-1)$, so that the only polynomials in $V_0\cap V_e$ are those in $V_e$ for which $P(1)=0$.

The dimension of $V_0$ is $\dim V-1=2n$, because the subspace is the kernel of the linear map $V\to\mathbb{R}$, $P\mapsto P(1)+P(-1)$.

Now consider the map $V_e\to\mathbb{R}$, $P\mapsto P(1)$.