jordan canonical form of 6x6 matrix

Question

Let Nl and N2 be 6 X 6 nilpotent matrices over the field F. Suppose that N1 and N2 have the same minimal polynomial and the same nullity. Prove that N1 and N2 are similar. Show that this is not true for 7 X 7 nilpotent matrices.

Well, if Nl and N2 are 6 X 6 nilpotent matrices, the characteristic polynomial of them, are p1(x)=p2(x)=x^6 and the minimal m1(x)=m2(x)=x^i with i=1,...,6. As these matrices has same nullity, (call it "s") then rank(N1)=rank(N2)=6-s. Now, should i do? Should i fix the minimal and consider the possibilities for the rank and confirming that all have the same Jordan form?

Answer 1

You have to show that they have the same Jordan form, that is the same number of Jordan blocks (equals nullity), Jordan blocks of the same size.

Now you make case distinction with respect to the degree $d$ of the minimal polynomial. List every possible configuration of Jordan blocks and check whether there is any ambiguity

• $d=6$ one Jordan block of size 6
• $d=5$: one Jordan block of size 5, one Jordan block of size 1
• $d=4$: one Jordan block of size 4, either two Jordan blocks of size 1 or one of size two. Hence 2 or 3 Jordan blocks, however number of Jordan blocks is determined by nullity, hence $N_1$ and $N_2$ have the same Jordan form.

Hope you can complete this...

For the $7\times 7$ matrices, the case of $d=3$ and nullity$=3$ leaves room for different configuration of Jordan blocks ($3+3+1$ vs. $3+2+2$).