Simultaneously Triangularizable

Algebra Level 5

If A A and B B are square matrices of the same order over C \mathbb C , which of the following statements is/are true?

Ⅰ. If A A and B B are simultaneously triangularizable, that is, there exists a similarity matrix P P such that P 1 A P P^{-1}AP and P 1 B P P^{-1}BP are both upper triangular, then A A and B B commute.

Ⅱ. If A A and B B commute, then they can be simultaneously triangularizable.


Bonus: Generalize this for any number of matrices.

Neither Ⅰ and Ⅱ Ⅰ only Ⅱ only

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2 solutions

Denis Husadzic
Feb 2, 2019

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

H = ( 1 0 0 1 ) , E = ( 0 1 0 0 ) . 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 A and B B commute, they share an eigenvector. To prove this consider v v such that A v = λ v Av = \lambda v . Then, A B v = B A v = B ( λ v ) = λ B v ABv = BAv = B(\lambda v) = \lambda Bv , i.e. B v Bv is eigenvector of A A for the same eigenvalue λ \lambda . This implies that A A -eigenspace V λ V_\lambda is B B -invariant, i.e. B V λ V λ BV_\lambda \subseteq V_\lambda . Thus, we can restrict B V λ : V λ V λ B|_{V_\lambda}\colon V_\lambda \to V_\lambda , and since the ground field is C \mathbb C , B V λ B|_{V_\lambda} has an eigenvector e e in V λ V_\lambda - a common eigenvector with A A . Change basis to { e , b 2 , , b n } \{ e, b_2,\ldots,b_n\} and write A A and B B in that basis:

A = ( λ C 0 A 1 ) , B = ( μ D 0 B 2 ) . A = \begin{pmatrix} \lambda & C \\ 0 & A_1 \end{pmatrix},\quad B = \begin{pmatrix} \mu & D\\ 0 & B_2 \end{pmatrix}.

Then, A 1 A_1 and B 1 B_1 commute, so we can proceed recursively to construct the basis of the whole space in which A A and B B are both upper triangular.

Try similar problem: Simultaneously Strictly Triangularizable .

Brian Lie - 2 years, 3 months ago
Brian Lie
Feb 23, 2019

Generalization for any number of matrices could be:

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

This is generalized by Lie's theorem.

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