If every non-zero vectors be the eigenvector of a real matrix $A$, prove that $A$ is the scalar matrix $\lambda I_n$. 
If every non-zero vectors in $\mathbb{R}^n$ be the eigenvector of a real $n \times n$ matrix $A$ corresponding to a real eigenvalue $\lambda$, prove that $A$ is the scalar matrix $\lambda I_n$.

I have no idea to start with. Please help me to solve it. 

Let $x$ be a non zero vector, $A(x)=\lambda x$, If $y=hx$, $A(y)=A(hx)=hA(x)=h\lambda x=\lambda y$, if $x,y$ are linearly independent,  $A(y)=dy$, $A(x+y)=e(x+y)=A(x)+A(y)=\lambda x+dy$. This implies that $(e-\lambda)x+(e-d)y=0$ since $x,y$ are linearly independent, $e-\lambda=0$ and $e-d=0$, thus $e=\lambda=d$. done.
