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!
 A: 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 $\mu = \lambda$. 
A: You may want to see a proof that can be generalized to more than two vectors.
Consider $\alpha x+\beta y=0$. Then
\begin{align}
\varphi(\alpha x+\beta y)&=0 \\
\mu(\alpha x+\beta y)&=0
\end{align}
which become
\begin{align}
\alpha\lambda x+\beta\mu y&=0 \\
\alpha\mu x+\beta\mu y&=0
\end{align}
Subtract the second from the first:
$$
\alpha(\lambda-\mu)x=0
$$
Since $(\lambda-\mu)x\ne0$ by assumption, we obtain $\alpha=0$ and so $\beta y=0$, from which $\beta=0$.

This can be generalized in a similar way: if $x_k$ is an eigenvector relative to $\lambda_k$, for $k=1,2,\dots,n$ and the eigenvalues are pairwise distinct, then $\{x_1,\dots,x_n\}$ is linearly independent.
The case $n=1$ is obvious. Suppose the thesis holds for $n-1$ eigenvectors relative to distinct eigenvalues. Then, given $\alpha_1x_1+\dots+\alpha_nx_n=0$, we have
\begin{align}
\varphi(\alpha_1x_1+\dots+\alpha_nx_n)&=0 \\
\lambda_n(\alpha_1x_1+\dots+\alpha_nx_n)&=0
\end{align}
which becomes
\begin{align}
\alpha_1\lambda_1x_1+\dots+\alpha_{n-1}\lambda_{n-1}x_{n-1}+\alpha_n\lambda_nx_n&=0 \\
\alpha_1\lambda_nx_1+\dots+\alpha_{n-1}\lambda_nx_{n-1}+\alpha_n\lambda_nx_n&=0
\end{align}
Subtract the second from the first: then
$$
\alpha_1(\lambda_1-\lambda_n)x_1+\dots+
\alpha_{n-1}(\lambda_{n-1}-\lambda_n)x_{n-1}=0
$$
and, by the induction hypothesis,
$$
\alpha_k(\lambda_k-\lambda_n)=0 \qquad (k=1,\dots,n-1)
$$
forcing $\alpha_1=\dots=\alpha_{n-1}=0$. Hence also $\alpha_nx_n=0$ and $\alpha_n=0$.
