Linear Algebra: Show Polynomial Is A Subspace Let $\mathbb{P}_n$ be the set of real polynomials of degree at most $n$, and write $p'$ and $p''$ for the first and second derivatives of $p$. Show that 
$S = \{p \in \mathbb{P}_6 : p''(2) + 1\cdot p'(2) = 0\}$
is a subspace of $\mathbb{P}_6$.
I know I need to check 3 things to prove it's a subspace: zero vector, closure under addition and closer under scalar multiplication. How should I do it for this question? Do I integrate the derivatives first? We know that the equation is equal to 0 when the argument of the derivatives is equal to 2, but what about arguments of other value?
Thanks.
 A: Hint: you have to check: 


*

*If $p,q\in S$ then $r:=p+q\in S$ which is easy to do:
$$ r''(2)+1r'(2) = p''(2)+q''(2)+1p'(2)+1q'(2) = 0+0=0$$
so $r\in S$. 

*If $p\in S $ then  $m\cdot p$ is also in $S$ for every scalar $m$. Can you do that?

A: Define $T\colon\mathbb{P}_6\to\mathbb{R}$ defined by $T(p):=(p''+p')(2)$.  Obviously $T$ is linear and $S$ is the kernel of $T$ and as such a subspace.
A: As far as I know the zero polynomial $f(x)=0$ has degree $-\infty$ by definition. So it would not be an element of $\mathbb{P}_6$. But that is no mayor issue, because you could always add the zero polynomial to this set.
[Edit: I just see, that $\mathbb{P}_6$ has degree 'at most' 6. Then what I wrote about is obsolete.]
I show you the closure under addition. The other axioms are similar.
For closure you have to show, that for $f,g\in S$ you have $f+g\in S$.
So let $f,g$ be elements of $S$.
Then it is $(f+g)(x)=f(x)+g(x)$ (This is how addition of functions work)
and $(f+g)'(x)=f'(x)+g'(x)$
and $(f+g)''(x)=f''(x)+g''(x)$
For $f+g$ beeing an element of $S$ it has to hold, that $(f+g)'(2)+(f+g)''(2)=0$
So lets evaluate:
$(f+g)'(2)+(f+g)''(2)=f'(2)+g'(2)+f''(2)+g''(2)=\underbrace{f'(2)+f''(2)}_{=0}+\underbrace{g'(2)+g''(2)}_{=0}=0$
They are equal to zero, since $f$ and $g$ are elements of $S$.
For the other axiom:

$f\in S$, then $\alpha f\in S$ for $\alpha\in\mathbb{K}$

Use, that $(\alpha f)(x)=\alpha f(x)$. 
The proof is similar to the one give above.
