Proving Vector Subspaces Question 1:
The set $\mathbb R^3$ of all column vectors of length three, with real entries, is a vector space. Is the subset $$B=\{xyz \in \mathbb R^3 \mid xy+yz= 0\}$$ a subspace of $\mathbb R^3$? 
Justify your answer.
Attempted answer:
$(0)$: let $0$ vector be in set $B$, then $0\cdot0 + 0\cdot0 = 0$, hence $B$ is non-empty.
$(1)$ A1: Let $u = \begin{bmatrix}0&0&1\end{bmatrix}$ and $V = \begin{bmatrix}0&1&0\end{bmatrix}$, which are both in $B$, but $u +v = \begin{bmatrix}0&1&1\end{bmatrix}$ is not in $B$, since $0\cdot1 +1\cdot1 \neq 0$. Hence not closed under multiplication.
$(2)$ S1: I know its irrelevant since its not closed under A1...but is this correct--> let $u = \begin{bmatrix}a&b&c\end{bmatrix}$, then $k\cdot u = \begin{bmatrix}ka&kb&kc\end{bmatrix}\implies (ka)(kb) + (kb)(kc) = 0 \implies k^2(ab) + k^2(bc) = 0$, hence must clearly be in set $B$?
Question 2
Show that the set of all twice differentiable functions $f:\mathbb R\to \mathbb R$ satisfying the differential equation $\sin(x)f''(x) + x^2f(x) = 0$ is a vector space with respect to the usual operations of addition of functions and multiplication by scalars. Note that $f''$ is the second derivative of $f$.
Attempted answer: The zero vector in $F$ space has the zero function $0(x) = 0 \in \mathbb R$ and we have $0''(x) = 0'(x) = 0$. Assume that the set of solutions for the equation is a subset of $F$ space, hence it contains the zero function and so $0(x) = \sin(x)\cdot 0''(x) + x^2\cdot0(x) = 0+0 = 0$ so set of solutions is in subspace.
 A: Your first answer is correct, just replace the "not closed under multiplication" with "not closed under addition".
For the second answer, you only proved that $F$ is not empty. What you still have to prove is two things:


*

*Take any two functions $f,g\in F$. You must show that $f+g\in F$.

*Take any function $f\in F$ and scalar $\alpha\in\mathbb R$. You must show that $\alpha f\in F$.


Edit:
In your own answer, you started with the line $$\sin(x) (f+g)'' + x^2(f+g)(x) = 0,$$
which is not the way you go about these things. You can only ever write something you have proven, but you have not yet proven the line above. What you know is that both $f$ and $g$ are elements of $F$. This means that $$\sin(x) f''(x) + x^2 f(x) =0\\\sin(x) g''(x) + x^2g(x)=0.$$ Now, what you want to prove is that if the equations above hold, then the equation
$$\sin(x) (f+g)''(x) + x^2(f+g)(x) = 0$$
also holds. Because $(f+g)'' = f'' + g''$, the expressions simplifies to
$$\sin(x) (f''(x)+g''(x))+ x^2(f(x)+g(x)) = \sin(x) f''(x) + \sin(x) g''(x) + x^2 f(x) + x^2 g(x).$$
Can you see now why this expression will equal $0$? If you can, then you have proven that $f+g$ is an element of $F$.
Once you understand why $f+g\in F$, proving the second part (multiplication) should not be hard.
A: So for question 2:
A1: $\sin(x)(f+g)'' + x^2 (f+g)(x)= 0 \implies 
\sin(x) [f''(x) + g''(x)] + x^2\cdot f(x)+x^2\cdot g(x) = 0$ $ \implies \sin(x)\cdot f''(x) + \sin(x)\cdot g''(x) + x^2\cdot f(x) +x^2\cdot g(x)$ so clearly $f +g \in F$.  
S1: $\alpha \cdot (\sin(x)\cdot f''(x) + x^2\cdot f(x)) \implies \alpha \sin(x)\cdot f''(x) + \alpha \cdot x^2\cdot f(x)$.....stuck here, how do i tell if its in the subset or not ??? really confused in this regard.
A: Question 1) 
To prove U (some arbitrary subspace) is a subspace of V (some arbitrary vector space) you need to prove  a) the zero vector is in U b) U is closed by addition c) U is closed by scalar multiplication by the field V is defined by (in your case any real number) d) for every $ u \in U$, $u \in V$.  
a) Obviously true since when x, y, z = 0, xy + yz = 0. And {0,0,0} is the zero vector in $R^3$.
b) Show that if $v_1 = {x_1, x_2, x_3}$ and $v_2= {y_1, y_2, y_3}$ then $x_1+y_1, x_2+y_2, x_3+y_3$ satisfies xy + yz = 0.
c) Show if $v_1= {x_1, x_2, x_3}$ then $\lambda$$x_1$, $\lambda*x_2$, $\lambda*x_3 \in U$ (or in other words satisfies xy+yz= 0).
d) This is obviously true.
Question 2) Do the same for number 2.  The idea that the derivative function is a linear function is abundantly shown with quick googling.
