Question: Consider a $n \times n$ matrix $A=(a\_{ij})$ with $a\_{12}=1$, $a\_{ij} =0 \ \forall \ (i,j) \neq (1,2)$. Prove that there is no invertible matrix $P$ such that $PAP^{-1}$ is a diagonal matrix. 

Any help?

Answer: 
If there's some invertible $P$ such that
$$PAP^{-1}=D$$ 
is diagonal, then 
$$PA^2P^{-1}=D^2$$ 
$$\Rightarrow P0P^{-1}=D^2$$
$$\Rightarrow D=0$$
So
$$A=0$$
a contradiction.
For the sake of exploring different methods of proof, here's my take: if $PAP^{-1}$ is equal to some diagonal matrix $D$, then $D\neq0$. Yet, $\operatorname{trace}(PAP^{-1})=0$. Hence $D$ has at least two nonzero diagonal entries. Consequently, $\operatorname{rank}(D)\ge2>1=\operatorname{rank}(A)$, which is a contradiction because rank is preserved upon multiplication of invertible matrices.