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 ??
Best Answer
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 $\dim (V_0) \cap \dim (V_e)=n.$