Linear Algebra, proving that two eigenvectors are linearly independent Let $E$ be a vector space and $\varphi: E \to E$ be a linear map. Let $x, y \in E \setminus \{0\}$ and $\lambda, \mu \in F$ such that
$\varphi(x) = \lambda x$ and $\varphi(y) = \mu y$. Prove that if $\lambda \neq \mu$ then $\{x, y\}$ is linearly independent.
This proof seems like it should be on the simpler side. But perhaps I am over thinking it. This is what I have:
Proof :
Suppose $\{x,y\}$ is not linearly independent.
Then, there exists scalars $a,b$ s.t. $ax+by=0$ where $a=b=0$.
So, $0=ax+by=\varphi(x)+\varphi(y)= \varphi(x+y)$
And this is where I am stuck. I can use the fact that $\varphi$ is linear, and show that $\varphi(0)=0$, but the map is not necessarily injective, so there might be more elements in the null space. I can't use $\varphi(x+y)=c(x+y)$ because this assertion would only hold in dimension $1$. What am I missing? Am I going about this in the right way?
Thank you in advance!

Suppose that $\{x,y\}$ is linearly dependent. Then either $x$ is a multiple of $y$ or vice versa. Without loss of generality, assume that $x = cy$. Then
$$\lambda x = \phi(x) = \phi(cy) = c\phi(y) = c\mu y = \mu (cy) = \mu x$$
Thus, we have that $\mu x = \lambda x$. Since $x\neq 0$, it follows that $\mu = \lambda$. 
