Eigenvectors of linear transformation

Question

Assume that a linear transformation $T$ has two eigenvectors $x$ and $y$ belonging to distinct eigenvalues $\lambda$ and $\mu$. If $ax + by$ is an eigenvector of $T$, prove that $a=0$ or $b=0$.

Answer 1

$(x,y)$ are linearly independent vectors. Indeed, if $x=ky$ for example, $Tx=\lambda_xx=k\lambda_yy=\lambda_yx$, impossible because $\lambda_x\neq\lambda_y$ and $x\neq 0$.

Suppose $a\neq 0$. $T(ax+by)=\lambda(ax+by)=\lambda_xax+\lambda_yby$. As $(x,y)$ are independant, $\lambda a = \lambda_xa$; $\lambda= \lambda_x$, and $\lambda_x b = \lambda_yb$, hence $b=0$.