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.
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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.
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Why should $D^{2}=0 \implies D=0$. – mmt Apr 13 '13 at 09:28
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2@mmt: Because $D$ is diagonal. multiply a diagonal matrix by itself to see the result. – Apr 13 '13 at 09:30
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Observe that $A$ is upper triangular, so that you can read off the eigenvalues of $A$. What are the eigenvalues of $A$, then? What would this fact imply about $PAP^{-1}$ if such a $P$ existed?
Branimir Ćaćić
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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.
user1551
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