# Find $\dim(V_0)$ and $\dim(V_0 ∩ Ve)$

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 ??

• If $V_0$ and $V_0\cap V_e$ have the same dimension then we must have $V_0=V_0\cap V_e$ or $V_0 \subset V_e$ which is not true. Sep 9, 2019 at 8:40
• Actually the second one is $n-1$. Sep 9, 2019 at 8:42

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 $$\dim (V_0) \cap \dim (V_e)=n.$$

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 $$V_e\to\mathbb{R}$$, $$P\mapsto P(1)$$.