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I have found the following example in Horn's and Johnson's Matrix Analysis book. Let $$A=\begin{bmatrix} 0&1&0&0\\ 0&0&0&0\\ 0&0&0&1\\ 0&0&0&0 \end{bmatrix}$$ $$B=\begin{bmatrix} 0&1&0&0\\ 0&0&1&0\\ 0&0&0&0\\ 0&0&0&0 \end{bmatrix}$$ The two matrices have the same eigenvalues (characteristic polynomial), trace, determinant and rank. But since $A^2=0$ and $B^2\neq0$, we can conclude that they are not similar. Can you explain me the last claim i.e. why it implies that they cannot be similar? Please cite relevant theorems or lemmas.

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  • $\begingroup$ similar matrices have the same minimal polynomial $\endgroup$ – Will Jagy Feb 20 '18 at 3:53
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Suppose $A^2=0$ and $B^2 \neq 0$ but $B=P^{-1}AP$

then $B^2=P^{-1}A^2P=0$ which is a contradiction.

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