If and are square matrices of the same order over , which of the following statements is/are true?
Ⅰ. If and are simultaneously triangularizable, that is, there exists a similarity matrix such that and are both upper triangular, then and commute.
Ⅱ. If and commute, then they can be simultaneously triangularizable.
Bonus: Generalize this for any number of matrices.
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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 H E − E H = 2 E , where
H = ( 1 0 0 − 1 ) , E = ( 0 0 1 0 ) .
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 A v = λ v . Then, A B v = B A v = B ( λ v ) = λ B v , i.e. B v is eigenvector of A for the same eigenvalue λ . This implies that A -eigenspace V λ is B -invariant, i.e. B V λ ⊆ V λ . Thus, we can restrict B ∣ V λ : V λ → V λ , and since the ground field is C , B ∣ V λ has an eigenvector e in V λ - a common eigenvector with A . Change basis to { e , b 2 , … , b n } and write A and B in that basis:
A = ( λ 0 C A 1 ) , B = ( μ 0 D B 2 ) .
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.