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One of the consequences of the well-known Motzkin-Taussky theorem (https://www.jstor.org/stable/1990825) is the following : if two complex matrices $A, B$ generate a vector space of diagonalisable matrices, then $A$ and $B$ commutes and in particular are simultaneously diagonalisable.

Does the result hold for nilpotent matrices : Let $A$ and $B$ two (complex) matrices such that $sA+tB$ are nilpotent for all $s,t\in \mathbb{C}$ are they simultaneously triangularisable ?

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The matrices $$A = \begin{pmatrix} 0&1&0 \\ 0&0&1 \\ 0&0&0 \end{pmatrix}, B = \begin{pmatrix} 0&0&0 \\ 1&0&0 \\ 0&-1&0 \end{pmatrix}$$ satisfy $(sA+tB)^3=0$ for all $s,t$, but $$AB = \begin{pmatrix} 1&0&0 \\ 0&-1&0 \\ 0&0&0 \end{pmatrix},$$ which is not nilpotent, so $A$ and $B$ are not simultaneously triangularisable (as strictly upper triangular matrices).

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  • $\begingroup$ Thank you ! I have two sub-questions from your answer : 1) How did you find this counter-example ? 2) Can we find a counte-example with 3 matrices ( in $M_3(\mathbb{C})$ ) ? More precisely, If there are 3 matrices in $M_3(\mathbb{C})$ that generate a vector space $V$ of nilpotent matrices (or $\frac{n(n-1)}{2}$ matrices in $M_n(\mathbb{C})$) is $V$ conjugated to the strictly upper triangular matrices ? $\endgroup$
    – user9099103
    Commented Feb 9, 2022 at 16:15

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