From Hoffman and Kunze.

Let $N_1$ and $N_2$ be 6 X 6 nilpotent matrices over the field F. Suppose that $N_1$ and $N_2$ have the same minimal polynomial and the same nullity. Prove that $N_1$ and $N_2$ are similar. Show that this is not true for 7 X 7 nilpotent matrices.

Right so nilpotent matrices have $N^k= 0$ for some k. Since these have the same nullity and minimal polynomial $N_1^k = N_2^k = 0$ from some k right? The same minimal polynomial implies the same characteristic values but thats not enough to say they have the same jordan form and are thus similar right?

and how would I get started in proving this breaks for 7x7 nilpotent matrices?

Thanks in advance for any assistance.

  • 1
    $\begingroup$ Dear Avatar, What you have written down (that there is a $k$ such that $N_1^k = N_2^k = 0$) is weaker than $N_1$ and $N_2$ having the same minimal polynomial, and you haven't incorporated at all in your discussion the fact that they have the same nullity. How about you try the $2 \times 2$ case first, and then the $3\times 3$ case, and so on, working your way up to the $6\times 6$ and $7\times 7$ cases, so that you build up some intuition for the question. Regards, $\endgroup$ – Matt E May 8 '13 at 2:04

Since the matrices are nilpotent, all eigenvalues are zero, so we only need to consider the size of the Jordan blocks. The possible sizes correspond to partitions of $n=6$. Since the matrices have the same minimal polynomials, they must have the same largest Jordan block. Can you see what the nullity assumption requires of their Jordan decompositions? Together, these requirements should be enough to show that their Jordan blocks are the same, so they are similar.

With the extra bit of freedom offered by matrices of size $7$, there are counterexamples. If you understand how to prove the case for matrices of size $6$, a counterexample should come fairly easily.


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