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!
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The characteristic polynomial for $A$ seems wrong. – Emilio Novati May 24 '17 at 07:32
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A typo..I fixed that ($-4$ in the 13 position). Thanks @Emilio Novati. – boaz May 24 '17 at 07:34
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Do you know how to find the matrix $P$ for a single triangularizable matrix $A$? – Greg Martin May 24 '17 at 08:22
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yes, i know how to do the trianglazation process with a single matrix. – boaz May 24 '17 at 09:03
2 Answers
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.
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Thanks @loup blanc. Two questions: (1) Does the matrices $A_i$ must have a common eigenvectors? (2) What do you mean by "Proceed by recurrence"? – boaz May 24 '17 at 09:08
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Yes, they do have a common eigenvector $u$. Consider a basis in the form $u,\cdots$ and use the same reasoning about the "right-lower" blocks of the new matrices. – May 24 '17 at 09:16
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More explicitly: find a common eigenvector $v_1$ among your given matrices. Now complete a basis arbitrarily $u_2,\dotsc,u_n$. Let $Q = [v,u_1,\dotsc,u_n]$ a matrix. Now $Q^{-1}AQ$, $Q^{-1}BQ$, and $Q^{-1}CQ$ should all have an $(n-1) \times (n-1)$ submatrix in the bottom right with only zeros in the entries of first column directly to the left of it. You then want to find a common eigenvector $w$ among all of these submatrices. Let $U =[u_1,\dotsc,u_n]$. Then $v_2 =Uw$. With this update $Q = [v_1,v_2,u_3,\dotsc,u_n]$ where $u_3,\dotsc,u_n$ may have been rechosen in order to complete the space. You rinse and repeat with $n-2$, $n-3$ and so on.
If you need more information check out the algorithm in this paper: Dubi, C., An algorithmic approach to simultaneous triangularization, Linear Algebra Appl. 430, No. 11–12, 2975–2981 (2009). Zbl 1181.65063.
Abstract
We present and prove the validity of an algorithm constructing a simultaneous triangularization of a set on $N$ matrices in $\mathbb{C}^{n \times n}$. To do so, we first prove that if a set of matrices has a common block decomposition, then the set of matrices has a simultaneous triangularization if and only if the blocks on the diagonal have a simultaneous triangularization.
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