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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.

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    $\begingroup$ Why would that definition not make sense? $\endgroup$
    – Mariano Suárez-Álvarez
    Mar 12, 2015 at 21:05
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    $\begingroup$ 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. $\endgroup$
    – Mariano Suárez-Álvarez
    Mar 12, 2015 at 21:16
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    $\begingroup$ 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. $\endgroup$
    – Mariano Suárez-Álvarez
    Mar 12, 2015 at 21:19
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    $\begingroup$ @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. $\endgroup$
    – Calculon
    Mar 13, 2015 at 1:04
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    $\begingroup$ I will just add this here as it may be relevant: en.wikipedia.org/wiki/Joint_spectral_radius $\endgroup$
    – Calculon
    Jan 12, 2020 at 21:10

1 Answer 1

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There is a simple characterization indeed, very similar to what exists for just one nilpotent matrix.

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).

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  • $\begingroup$ 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. $\endgroup$
    – Calculon
    Mar 16, 2015 at 13:50
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    $\begingroup$ @Calculon as I find the question you just asked interesting, I put it in another MSE question : math.stackexchange.com/questions/1192450/… $\endgroup$
    – Ewan Delanoy
    Mar 16, 2015 at 15:10
  • $\begingroup$ Thank you! I appreciate it. $\endgroup$
    – Calculon
    Mar 16, 2015 at 15:11
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    $\begingroup$ @Calculon Your question was recently answered at math.stackexchange.com/questions/1193815/… It might still be however that you have a different finite-time algorithm to decide if matrices are jointly nilpotent $\endgroup$
    – Ewan Delanoy
    Mar 18, 2015 at 8:37

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