Linear Algebra: Show that $V_1$ is a subspace

Question

Show that $V_1 = \{f(x) \in \mathbb{C}^{\infty}(R) \mid f'(x) − f(x) = 0\}$ is a subspace of $\mathbb{C}^{\infty}(R)$.

I know you have to prove it's non-empty, closed under addition, and closed under multiplication. But what exactly from this am I trying to prove is all of those three, and how do I go about it?

Answer 1

$0\in V_{1}$ because $0'-0=0$. Given $f,g\in V_{1}$, then $f,g\in C^{\infty}$ then so is $f+g\in C^{\infty}$. Now the calculus rules give $(f+g)'-(f+g)=f'+g'-(f+g)=(f'-f)+(g'-g)=0$, so $f+g\in V_{1}$.

For scalar $a$, can you complete in showing that $af\in V_{1}$?

Answer 2

Hint: the map $T\colon C^{\infty}(\mathbb{R})\to C^{\infty}(\mathbb{R})$ defined by $T(f)=f'-f$ is linear. What's its kernel?