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If there are $6$ matrices in the vector space $M_{6,6}(\Bbb C)$ such that they all satisfy $A^2=0$, does this imply that at least two of them are similar?

My approach: Observation: $A^2=0$ implies that all the eigenvalues of these matrices must equal to zero. Which naturally led me to think about the Jordan forms of these, to try and see if there are less than $6$ types of Jordan matrices corresponding to this property as that would be mean that at least two of these matrices should have the same Jordan matrix, and hence must be similar. I found that there are more than $6$ types of corresponding Jordan matrices, so my current answer is "not necessarily true", but I need a definite true or false answer here. Could someone give me a thought process?

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Your approach is fine. But I don't see how is it that you got more than $6$ types of matrices in Jordan normal for whose square is $0$. There are less than $6$ types and that's why two of them must be similar. Actually, there are only $4$ types: the null matrix and$$\begin{pmatrix}0&1&0&0&0&0\\0&0&0&0&0&0\\0&0&0&0&0&0\\0&0&0&0&0&0\\0&0&0&0&0&0\\0&0&0&0&0&0\end{pmatrix},\begin{pmatrix}0&1&0&0&0&0\\0&0&0&0&0&0\\0&0&0&1&0&0\\0&0&0&0&0&0\\0&0&0&0&0&0\\0&0&0&0&0&0\end{pmatrix}\text{ and }\begin{pmatrix}0&1&0&0&0&0\\0&0&0&0&0&0\\0&0&0&1&0&0\\0&0&0&0&0&0\\0&0&0&0&0&1\\0&0&0&0&0&0\end{pmatrix}.$$There is no other Jordan matrix whose square is $0$.

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  • $\begingroup$ Thank you for replying. Surely there's 6 or more? Because is it not possible that the Jordan matrices could be, say (the numbers denote the sizes of the individual Jordan blocks): a) 1 1 1 1 1 1 b) 1 1 1 1 2 c) 1 1 1 3 d) 1 1 4 e) 1 5 f) 6? $\endgroup$ – user564499 Jun 6 '18 at 7:31
  • $\begingroup$ I assume that there's something about the fact that A^2=0 which forces me to remove some of these options, but I did not notice it. Could you give me some guidance? $\endgroup$ – user564499 Jun 6 '18 at 7:32
  • $\begingroup$ Yes, take your example f. Suppose that $A$ was in Jordan form from the start. Now just take pen and paper and compute $A^2$, $A^3$, $A^4$ etc. It is not a lot of work and very insightful. $\endgroup$ – Vincent Jun 6 '18 at 7:34
  • $\begingroup$ @Vincent I see that when I raise f to some powers, the "diagonal" of 1s will move up rightwards, since f^k = (0+N)^k where N is the matrix of consisting of ones on the first diagonal. I also see that as a result, f^2 =/ = 0. The only thing left is, if A has jordan matrix f, does that mean that A could only have been f from the start? $\endgroup$ – user564499 Jun 6 '18 at 7:40
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    $\begingroup$ @JoséCarlosSantos Ah I see why that is true now, I should've given it more thought before asking. It's because if A= PJP^-1, then J = P^-1AP and hence, J^2 = P^-1 A^2 P and so if the RHS =0 , then J^2 =0 right? $\endgroup$ – user564499 Jun 6 '18 at 7:54

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