10

Let $A, B \in \mathcal{M}_n(\mathbb{C})$ such that for all $x,y \in \mathbb{C}, xA+yB$ is diagonalizable.

Show that $AB=BA$.

My idea (not really an attempt) :

It suffices to show that $A$ and $B$ are simultaneous diagonalizable. Then one can use the theorem that says $A$ and $B$ are diagonalizable at the same time if and only if they commute.

Does anyone has a way to prove that $A$ and $B$ are simultaneous diagonalizable ?

Community
  • 1
  • 1
    This was asked on mathoverflow here. – George Shakan Jun 04 '14 at 22:21
  • 1
    It is a paper that studied the more general case. – Jlamprong Jun 04 '14 at 22:24
  • I've deleted my answer. This can't work because I didn't really make use of the assumption that $A+uB$ is diagonalizable (in a sense, I did, but I could run almost the same argument by just looking at one eigenvector/eigenvalue). –  Jun 05 '14 at 02:50
  • In view of George's comment, it's very unlikely that this is an exercise. Where does this question come from? – user1551 Jun 05 '14 at 17:40
  • 1
    @user1551 At usual, its an ENS oral, but this one in particularly infamous. –  Jun 05 '14 at 17:59

1 Answers1

6

Definition: A pair $ (A,B) $ of complex $ (n \times n) $-matrices is said to have Property L if there exist orderings $ (\lambda_{i})_{i = 1}^{n} $ and $ (\mu_{i})_{i = 1}^{n} $ respectively of the eigenvalues of $ A $ and $ B $ such that $$ \forall (x,y) \in \mathbb{C}^{2}: \quad \operatorname{Spectrum}(x A + y B) = \{ x \lambda_{i} + y \mu_{i} \}_{i = 1}^{n}. $$

By Theorems 3 and 4 of T.S. Motzkin, O. Taussky, Pairs of Matrices with Property L. II, Trans. Amer. Math. Soc. 80 (1955) 387-401, if $ \lambda A + \mu B $ is a pencil in which all matrices are diagonalizable, then $ (A,B) $ has Property L, and moreover, $ A $ and $ B $ commute.

EDIT:

Using some concepts in algebraic geometry, Motzkin and Taussky showed that the result above holds for any field $ \mathbb{K} $ (if $ \mathbb{K} $ has finite characteristic, then they further assumed that $ \operatorname{char}(\mathbb{K}) \geq n $). Using complex analysis, Kato (in his book Perturbation Theory for Linear Operators, pp. 82-85) gave another proof that is valid only for $ \mathbb{C} $. Using Kato’s method, Friedland (in A Generalization of the Motzkin-Taussky Theorem, Linear Algebra Appl. 36 (1981) 103-109) and De Seguins Pazzis (in On Commuting Matrices and Exponentials, Proc. Amer. Math. Soc. 141 (3) (2013) 763-774 - An easier access is on arXiv -) showed generalizations of the previous result that are valid only for $ \mathbb{C} $.

  • In your answer on MathOverflow this was restricted to Hermitian matrices. What has happened to that hypothesis (which certainly wasn't there without reason)? – Marc van Leeuwen Jun 06 '14 at 09:05
  • Marc, the OP wondered if, in the case of hermitian matrices, there is a proof not using M.T.'s theorem. –  Jun 06 '14 at 11:23