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I have a question concerning simultaneous triangularisability (i. e. if a family of matrices $M_{\lambda}$ with $\lambda \in K$ commute and if there is one matrix $S \in GL(n, K)$ for all $M_{\lambda}$ such that $SM_{\lambda}S^{-1}$ is an upper triangular matrix then this family of matrices is simultaneously triangularizable).

  1. Why do the matrices need to commute? What goes wrong if they don't?

  2. I had show that a family of matrices is simultaneously triangularizable. In order to do so, I calculated the characteristic polynomial and got $\chi_{M_{\lambda}}(X) = X(X-1)^2$. So since there are only linear factors in $\chi$, the matrices are triangularizable. Is it necessary that all $\lambda$ cancel each other out? Would those matrices still be triangularizable if e. g. $\chi (X) = X(X-1)(X+ \lambda)$?

Edit: I suppose I can find a counterexample for question 1. But is there any intuitive explanation?

  • supposing $\mathbb K$ is algebraically closed, then actual the claim is: a sufficient condition for simultaneous triangularization: a collection of matrices forms an abelian (semi)group. Your question (1) is why it is necessary for the (semi)group to be abelian, which is a non-sequitor. – user8675309 May 14 '24 at 18:17

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For 1), (I think given the assumption that $K$ is algebraically closed, or at least the characteristic polynomials of your matrices split), the key idea is that commuting matrices preserve each other's eigenspaces, see here for instance. The converse isn't necessarily true (though it is for simultaneous diagonalizability, see here), there is a counterexample on Wikipedia here.

Alex Pawelko
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