Linear algebra, polynomial problem Could someone help me with this question? Because I'm stuck and have no idea how to solve it & it's due tomorrow :(
Let $S$ be the following subset of the vector space $P_3$ of all real polynomials $p$ of degree at most 3:
$$S=\{p\in P_3\mid p(1)=0, p^\prime (1)=0\}$$
where $p^\prime$ is the derivative of $p$.
a) Determine whether $S$ is a subspace of $P_3$
b) determine whether the polynomial $q(x)= x-2x^2 +x^3$ is an element of S
Attempt:
I know that for the first part I need to proof that it's none empty, closed under addition and multiplication right?
will this give me full mark for the part a if I answer like this:
$(af+bg)(1)=af(1)+bg(1)=0+0=0$ and 
$(af+bg)′(1)=af′(1)+bg′(1)=0+0=0$ 
so therefore it's a subspace of $P_3$?
b) i got no idea...
Thank you very much!
 A: Alright, for part a), you have to look at the definition of a subspace. That is, it must contain the additive identity (the zero-polynomial), which is trivial. It must be closed under multiplication by scalar, which it is, since its degree will not change, regardless of what real number $C$ you multiply a polynomial $P$ with. Lastly, we need it to be closed under addition, which it is, by similar argument as for multiplication by scalar. 
For part b), you simply need to check. $q(1) = 1 - 2 + 1 = 0$, $q'(x) = 1 - 4x + 3x^2, q'(1) = 1 - 4 + 3 = 0$. So yes, it is in $S$.
A: (a) To show that $S \le P_3$, we just need to establish closure over addition, and scalar multiplication. We begin by letting $p, q \in S$, and fix $\theta \in \mathbb{R}$ (or any other field you're working with). Since $p$ and $q$ belong in $S$, we have $$p(0) = p'(0) = q(0) = q'(0) = 0.$$
Now, let $t = p \oplus q$. Then, $$t'(0) = p'(0) + q'(0) = 0,$$ so $t \in S$. Lastly, let $s= \theta \otimes p$. Then,
$$r'(0) = \theta p'(0) = 0$$
so that $r \in S$, as required. Hence, $S$ is closed over $\oplus$ and $\otimes$. Hence, $S \le P_3$.
(b) Let $p(x) = x^3 -2x^2 + x$. Then,
$$p(1) = 1 - 2 + 1 = 0,$$
and
$$p'(0) = 3 -4 + 1 = 0.$$
So, we must have $p \in S$.
A: Well, your idea is right. But if you want to receive full marks, you also need to organize you idea in more logically manners.
Here is the answer to part a)
proof: For two elements $p_1$ and $p_2$ in $S$, we have $p_1(1)=0$, $p_1'(1)=0$, $p_2(1)=0$, $p_2'(1)=0$.
Then define $p_3=a*p_1+b*p_2$ where $a\; b$ are scalars from field $K$ which should be defined in question.
$$p_3(1)=a*p_1(1)+b*p_2(1)=0$$
$$p_3'(1)=a*p_1'(1)+b*p_2'(1)=0$$
According to the above two equations, we can conclude that $p_3$ is also in $S$. And $S$ is the subspace of $P^3$.
part b) just proof $q(1)=0$ and $q'(1)=0$ and $q \in S$.
A: The space $P_{3}$ is identified with $\mathbb{R}^{4}$ and $S=\left\{(a,b,c,d):a+b+c+d=0\,,\,3a+2b+c=0 \right\}$.
It is clear that $S\subseteqq \mathbb{R}^{4}$ and $q=(1,-2,1,0)$ which satisfies the constraints  is in $S$.!!
