# Jointly nilpotent matrices

We say that a matrix $J \in \mathbb{R}^{n \times n}$ is nilpotent if $J^n = 0$. This is equivalent to the statement that $\forall x \in \mathbb{R}^{n} \quad \exists k \in \mathbb{Z}^+$ such that $J^kx = 0$. What I would like to do is to extend this notion to pairs of matrices in the following way.

$J_1, J_2 \in \mathbb{R}^{n \times n}$ is a nilpotent pair if $$\forall x \in \mathbb{R}^{n} \quad \exists N \in \mathbb{Z}^+, \{i_1,k_1,\ldots,i_N,k_N\} \in {\mathbb{Z}^+}^{2N}$$ such that $$J_1^{i_N}J_2^{k_N}\ldots J_1^{i_1}J_2^{k_1}x = 0$$

I am not sure if this definition makes sense. What I would like to ask is how one would attempt to characterize such a notion, i.e. what kind of properties the matrices $J_1,J_2$ should satisfy to be called a nilpotent pair. My apologies if this trivially follows from some existing result.

• Why would that definition not make sense? Commented Mar 12, 2015 at 21:05
• Definitions are judged according to whether they capture an idea. You should tell us what you want a «nilpotent pair» to be, and then we can tell if the definition matches that. As for characterizing the pairs satisfying the conditin you wrote, that looks rather impossible to me. Commented Mar 12, 2015 at 21:16
• A more standard notion would be that the ideal generated by $J_1$ and $J_2$ in the matrix algebra $M_n(\mathbb R)$ be nilpotent, or nil. I doubt your condition is equivalent to this. Commented Mar 12, 2015 at 21:19
• @Ivan It is a counterexample to "if $J_1$ is invertible then $J_2$ is nilpotent". You can see that the example that I gave is a nilpotent pair. Commented Mar 13, 2015 at 1:04
• I will just add this here as it may be relevant: en.wikipedia.org/wiki/Joint_spectral_radius Commented Jan 12, 2020 at 21:10

A pair $(J_1,J_2)$ is jointly nilpotent iff one of the products $J_1^{i_N}J_2^{k_N}\ldots J_1^{i_1}J_2^{k_1}$ is the zero matrix.
Because if all those products are non-zero, then all their kernels are strict subpaces of ${\mathbb R}^n$. So the union of all those kernels (which is a countable union) cannot cover the whole of ${\mathbb R}^n$ (for example it is nowhere dense by the Baire category theorem).
• Thanks for your answer. I have to look into it a bit to convince myself. But I have a question. In the case of one nilpotent matrix it is sufficient to check the analogous condition up to the $n^{th}$ power where $n$ is the dimension of the matrix. But in the case of two matrices you have to do that for all $N$ and for all possible sequences $\{i_1,k_1,\ldots,i_N,k_N\}$. Is that true? If that is the case, then I cannot check the joint nilpotency of two matrices in finite time. Commented Mar 16, 2015 at 13:50