Is $f'(x)-3f(x) = 0$ subspace of differentiable functions $f\colon (0,1)\to \mathbb{R}$ $V$ is space of differentiable functions $f(0,1) \to \mathbb{R}$ and $W$ is a subset of $f$ that meets $f'(x) - 3f(x) = 0$ for all $x\in (0,1).$
Is subset $W$ a subspace of $V$?
I know that I have to prove that it's closed under scalar multiplication and under addition. But I forgot all the stuff about differentiable functions, so I don't know how to do this. But my intuition tells me that if there's a zero it's valid subspace :) If anyone could elaborate on this I would be grateful.
 A: Certainly. Let $f, g$ be two differentiable functions satisfying the given conditions (i.e., $f, g \in W$, and let $c$ be a scalar in $\mathbb{R}$. Then we can see:
$$(f+g)'(x) - 3(f+g)(x) = f'(x)+g'(x) - 3(f(x) + g(x)) = (f'(x) - 3f(x)) + (g'(x) - 3g(x)) = 0 + 0 = 0,$$ i.e., $(f+g)(x) \in W$, and
$$(cf)'(x) - 3(cf(x)) = cf'(x) - 3cf(x) = c(f'(x) - 3f(x)) = c(0) = 0,$$
i.e., $cf(x) \in W$. So $W$ is clearly closed under addition and scalar multiplication. Observing that $f(x) = 0$ is obviously in $W$, we are done. 
A: Let $f,g\in W$ so $f'(x)-3f(x)=0,~g'(x)-3g(x)=0,~~x\in (0,1)$. Now try to verify if $$[\alpha f(x)+g(x)]'-3[\alpha f(x)+g(x)]=0,~~~\alpha\in\mathbb R$$
A: Taking the derivative is a linear map from the space of differentiable functions to the space of functions. This is just fancy speak for $(f+g)=f'+g'$ and $(cf)'=cf'$.
Multiplying with a constant is also linear, the sum of linear maps is linear.
So finally $f\mapsto f'-3f$ is a linear map from the space of differentiable functions $(0,1)\to\mathbb R$ to the space of functions $(0,1)\to\mathbb R$ and your $W$ is simply the kernel (maybe this is where your "if there's a zero, it's a valid subspace" intuition comes from) of this linear map, hence a subspace.
