A problem on comparison of dimension between two subspace of polynomial vector space.

Question

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 $\dim(V_0 ∩ Ve)$.

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From the condition $P(1) +P(−1) = 0$, we have $a_{2n}+a_{2n-2}+..+a_0=0$. So dimension of $V_0$ is $2n$.
Obviously dimension of $V_e$ is $n+1$.
Am I right here?
but after that how should I find $\dim(V_0 ∩ Ve)$

Answer 1

You are just missing one dimension everywhere because of the way the degree matches up with dimension. So $\dim(V) = 2n + 1, \dim(V_0) = 2n$ and $\dim(V_e) = n + 1$.

EDIT:

We have $$ V_0 \cap V_e = \{P \in V: P \text{ has terms of even degree}, P(1) + P(-1) = 0\} = \{P \in V: a_{2n - 1} = a_{2n - 3} = \dotsb = a_3 = a_1 = 0, a_{2n} + a_{2n - 2} + \dotsb + a_2 + a_0 = 0\} $$ So the coefficients $a_{2n - 1} = a_{2n - 3} = \dotsb = a_3 = a_1 = 0$ and $a_0$ is determined by the coefficients $a_{2n}, a_{2n - 2}, \dotsc, a_4, a_2$. So you are only free to choose the coefficients $a_{2n}, a_{2n - 2}, \dotsc, a_4, a_2$. Since there are $n$ of them, we have $\dim(V_0 \cap V_e) = n$. Can you make the argument rigorous by actually finding a basis?

Try choosing polynomials with coefficients so that $a_{2n}, a_{2n - 2}, \dotsc, a_4, a_2$ are all $0$ except one of them (choose that one coefficient to be $1$ for simplicity). Then show that the set of such polynomials form a basis.