<s>[INST] Prove $U = \{u(x) \in P_4 | u(0) + u(1) = 0\}$ is a subspace of $P_4$ I am trying to prove $U = \{u(x) \in P_4 | u(0) + u(1) = 0\}$ is a subspace of $P_4$

For U is nonempty I have:

Let $u(x) = 0x^4 + 0x^3 + 0x^2 + 0x + 0$

For U is closed under $+$ I have:

Let $x, y \in U$, then
$x(0) + x(1) = 0$ and $y(0) + y(1) = 0$
$x + t = (x(0) + x(1)) + (y(0) + y(1)) = 0 + 0 = 0$
$\implies x + y \in U$

For $U$ is closed under $\cdot$ I have:

Let $x \in U \implies x(0) + x(1) = 0$
$\alpha x = \alpha ((x(0) + x(1)) = \alpha (0) = 0$
$\implies \alpha x \in U$

Is this sufficient?
 [/INST] $$p(x)=ax^4+bx^3+cx^2+dx+e$$
$$p(0)+p(1)=0$$
$$e+a+b+c+d+e=0$$
$$d=-2e-a-b-c$$
$$ax^4+bx^3+cx^2+dx+e$$
$$ax^4+bx^3+cx^2+x(-2e-a-b-c)+e$$
$$ax^4+bx^3+cx^2-2xe-ax-bx-cx+e$$
$$a(x^4-x)+b(x^3-x)+c(x^2-x)+e(-2x+1)$$
Base for your subspace is:$$(x^4-x,x^3-x,x^2-x,-2x+1)$$ Dimension of your subspace is 4.
So dimension of complementary subspace of your subspace is 1.
Becase $dim P_4=5$. Sum of dimension of subspace and their complementary subspace must be equal to total dimension, and their intersection must be $(0)$. The base for $P_4$ is $(x^4,x^3,x^2,x,1)$ so the base of complementary subspace can be $({1})$. 
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