Let $M$ be a square matrix. Assume at $x$ is an eigenvector for $M$ with eigenvalue $\lambda$, and that $y$ is an eigenvector for $M$ with eigenvalue $\mu$. Let $a$ and $b$ be real numbers.
How does can be proven that if the vector $v=ax+by$ is an eigenvector of $M$, then we have $a=0$ or $b=0$ ?
Initial assumptions : We must assume $\lambda \neq \mu$ ; thus, as a consequence, $x$ and $y$ are independent (non collinear).
Otherwise, we could fall in the case of an eigenspace (associated with $\lambda$) with dimension $\geq 2$ and the result to be established wouldn't be true.
Proof: Let us reason by contradiction.
Let us assume the contrary of ''$a=0$ or $b=0$ i.e., ''$a \neq 0$ and $b \neq 0$".
Let us write that $v:=ax+by$ is an eigenvector of $M$:
$$\tag{1} \exists \nu \ \ \ \text{such that} \ \ \ Mv=\nu v \iff Mv=\nu(ax+by)$$
Besides:
$\tag{2}Mv=M(ax+by)=aMx+bMy=a \lambda x+b\mu y$
Equating (1) and (2) would give
$$\tag{3}a(\lambda-\nu)x=b(\nu-\mu)y$$
As $x$ and $y$ are assumed independent (see initial assumptions), and as $a$ and $b$ are $\neq 0$, the only possibility for (3) to hold is that :
$$\tag{4}\lambda-\nu=\nu-\mu=0 \ \ \ \ \ \iff \ \ \ \ \ \lambda=\mu=\nu$$
But $\lambda \neq \mu$ (our initial assumption). Contradiction.