Simultaneous triangularization of matrices

Question

Let $\mathcal{F}=\{A_1,A_2,\ldots,A_r\}$ be a triangulable commuting family of $n\times n$ matrices (that is, each $A_i$ is triangulable and $A_iA_j=A_jA_i$ for every $i,j$). I know that $\mathcal{F}$ can be simultaneous triangularization, but what is the algorithm of finding the invertible matrix $P$ such that $P^{-1}A_iP$ is triangular? As a working example consider the matrices $$ A= \begin{pmatrix} -3 & 2 & -4 \\ -1 & 0 & -1\\ 2 & -2 & 3 \end{pmatrix}\qquad B= \begin{pmatrix} 3 & -2 & 2 \\ -1 & 2 & -1\\ -2 & 2 & -1 \end{pmatrix}\qquad C= \begin{pmatrix} 1 & 0 & 1 \\ 1 & 0 & 1\\ 0 & 0 & 0 \end{pmatrix} $$ Here $AB=BA$, $AC=CA$ and $BC=CB$. In addition, the characteristic polynomials of these matrices are $$ f_A(x)=x(x-1)(x+1)\\ f_B(x)=(x-2)(x-1)^2\\ f_C(x)=x^2(x-1) $$ so each one of them is triangulable. Thanks!

Answer 1

Here, it is easy. Since $A$ has $3$ distinct eigenvalues and $B,C$ commute with $A$, we can deduce that $B,C$ are polynomials in $A$ and it suffices to triangularize $A$.

In the general case.

Step 1. Find a common eigenvector of $A,B,C$.

Step 2. Proceed by recurrence.

Moreover, you can choose $P$ as an orthogonal matrix.