Simultaneously Triangularizable

If I. were true, it would imply that any two upper triangular matrices commute (if $A$ and $B$ are upper triangular, then they are simultaneously triangularizable with similarity matrix $P = I$ ). That is obviously not true, for example $HE-EH = 2E$ , where

$H = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\quad E = \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}.$

II., on the other hand is true. The key is that if $A$ and $B$ commute, they share an eigenvector. To prove this consider $v$ such that $Av = \lambda v$ . Then, $ABv = BAv = B(\lambda v) = \lambda Bv$ , i.e. $Bv$ is eigenvector of $A$ for the same eigenvalue $\lambda$ . This implies that $A$ -eigenspace $V_\lambda$ is $B$ -invariant, i.e. $BV_\lambda \subseteq V_\lambda$ . Thus, we can restrict $B|_{V_\lambda}\colon V_\lambda \to V_\lambda$ , and since the ground field is $\mathbb C$ , $B|_{V_\lambda}$ has an eigenvector $e$ in $V_\lambda$ - a common eigenvector with $A$ . Change basis to $\{ e, b_2,\ldots,b_n\}$ and write $A$ and $B$ in that basis:

$A = \begin{pmatrix} \lambda & C \\ 0 & A_1 \end{pmatrix},\quad B = \begin{pmatrix} \mu & D\\ 0 & B_2 \end{pmatrix}.$

Then, $A_1$ and $B_1$ commute, so we can proceed recursively to construct the basis of the whole space in which $A$ and $B$ are both upper triangular.

Generalization for any number of matrices could be:

Let $\mathcal F\subset M_n(\mathbb C).$ If $\mathcal F$ is a commuting family , that is, every pair of matrices in $\mathcal F$ commute, then it is a simultaneously triangularizable family , that is, there is a single nonsingular $S\in M_n(\mathbb C)$ such that $S^{-1}AS$ is triangularizable for every $A\in\mathcal F.$

This is generalized by Lie's theorem.