Any two matrices M and N which have the same characteristic and minimal polynomial are similar ?

Is there any easier way to explain ?

I tried as M and N have same Jordan forms. So let Jordan form J of M implies J = P^-1MP and Jordan form K of N give K = Q^-1NQ & J=K gives after solving N = R^-1MR where R = PQ^-1. Gives M and N are similar.


marked as duplicate by user228113, TravisJ, choco_addicted, Alex M., Daniel W. Farlow May 9 '16 at 12:08

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    $\begingroup$ If their size is $\le 3$, yes. If it's $\ge 4$, not necessarily. $\endgroup$ – user228113 May 9 '16 at 11:15
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    $\begingroup$ The same procedure can be applied to matrices over any algebraically closed field. Moreover, since two matrices in $M_n(\Bbb K)$ are similar if and only if they are similar in $M_n\left(\overline{\Bbb K}\right)$, the result can be extended to arbitrary fields. $\endgroup$ – user228113 May 9 '16 at 11:25
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    $\begingroup$ Well, technically, just the counterexample in $M_4(\Bbb C)$ (which actually works for $M_4(\Bbb K)$) solves your question. As for the remaining part, in my humble opinion, I think the use of Jordan canonical form, for the complex case, is somehow needed, because the proof I am aware of ultimately boils down to "if you classify the possible Jordan forms of a $n\times n$ complex matrix, it turns out that there are not that many of them for $n\le 3$, while for $n>3$ there are". $\endgroup$ – user228113 May 9 '16 at 12:34
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    $\begingroup$ I don't know if the generalization to arbitrary fields was within your interests, but I think that, as algebraic as the details may be, the lemma on similarity that I quoted (at least in the version I used) is somehow needed to face the full problem. If you were interested only in complex matrices, you can ignore it. $\endgroup$ – user228113 May 9 '16 at 12:39