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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).

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.

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).

Source Link
wcb
  • 356
  • 1
  • 4

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.