1

Just trying to confirm if I did this right. I first lift them to $\mathbb{C}$ and then can put them into Jordan form. Since they commute, they can be simultaneously put into Jordan form with say matrix $P$, and since they are nilpotent, all eigenvalues are $0$, so their Jordan forms are actually identical with all $0$'s on the diagonal and same sized blocks. Then we have $PT_1...T_nP^{-1} = PT_1P^{-1}PT_2P^{-1}...PT_nP^{-1} = PJ^nP^{-1}$. Clearly multiplying a strictly upper-triangular matrix $n$ times gives $0$. Then $T_1...T_n = P^{-1}0P=0$. Since all $T_i$'s are real matrices, if they multiply to $0$ in $\mathbb{C}$, they multiply to $0$ in $\mathbb{R}$ as well.

user21417
  • 195
  • 1
    It is not true that commuting matrices can simultaneously be put into Jordan normal form: https://math.stackexchange.com/questions/49378/when-can-two-linear-operators-on-a-finite-dimensional-space-be-simultaneously-jo – Qiaochu Yuan Aug 03 '22 at 22:39
  • 3
    However, commuting matrices over an algebraically closed field can be simultaneously upper triangularized and that's enough here. – Qiaochu Yuan Aug 03 '22 at 22:43
  • 2
    Furthermore, the assertion that the Jordan forms must be identical isn't justified. – Greg Martin Aug 03 '22 at 22:43
  • They are supposed to preserve generalized eigenspaces. Doesn't that mean they have the same block sizes? And they have the same eigenvalues. – user21417 Aug 03 '22 at 22:48
  • Simultaneous upper triangularization does solve this. Since it would still have the same characteristic polynomial, the main diagonal is completely 0. Then every time I multiply I get another diagonal of $0$'s, so doing it n times gives the $0$ matrix. – user21417 Aug 03 '22 at 22:51

0 Answers0