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

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?

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 $0$.
• 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. – Vincent Jun 6 '18 at 7:34