# 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 $$\dim(V_0 ∩ Ve)$$.

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 $$\dim(V_0 ∩ Ve)$$

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 $\dim(V_e) = n + 1$.
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.