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Wielandt wrote a paper titled "Remarks on diagonable matrices".

According to Mathematische Werke - Mathematical Works : Linear Algebra and Analysis by Helmut Wielandt, Hans Schneider, Bertram Huppert (Editor) page 260 this paper from Wielandt remained unpublished (at least from the 1950s to the 1980s).

Does anyone have a copy of it or an idea of the proof on non defective pencils?

The main theorem states that for $A,B \in \mathcal{M}_n(\mathbb C)$, if in the pencil $\lambda A+ \mu B$ all matrices are diagnosable ($\forall \lambda. \mu \in \mathbb{C}$), then $AB=BA$.

Motzkin and Taussky proved that result (MR0086781 (19,242c)), using algebraic geometry, Kato proved it differently (MR1335452 (96a:47025)), using theory of complex functions in one variable. Wielandt seemed to have given another proof, hence my request.

Thanks

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Could you state precisely the theorem you're looking for? –  Deane Yang Apr 6 '12 at 7:19
    
@Deane, I updated the post with the theorem and references. Thanks –  Portland Apr 6 '12 at 15:01
    
Just curious.. is there an analogous result for the case $AB = -BA$? –  J.A Apr 6 '12 at 17:55
    
Wielandt's notebooks were TeXxed and published online [here][1], but now the page seems to be down, at least from my browser). If the page is not back up by Tuesday, I can ask for information in person --- I work in the same department. Today and Monday are bank holiday days in Germany so the admins are not there for sure, sorry. [1]: www4.math.tu-berlin.de/numerik/mt/Wielandt/index_en.html –  Federico Poloni Apr 6 '12 at 17:57
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Are you looking for Wielandt's proof or just a proof that uses only linear algebra? Have you already looked at Gantmacher's books on theory of matrices? –  Deane Yang Apr 11 '12 at 8:30
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2 Answers

up vote 4 down vote accepted

Now that the server is back up, I am posting this as a real answer.

With some work, you might be able to find the proof in Wielandt's notebooks, which were TeXxed and put online here.

The TeX source files are also published, so you can download them and use an automated search tool. Nevertheless, there's lots of material there, so it is not an easy task if you have no idea which period to look at.

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Perfect, thank you Federico! –  Portland Apr 12 '12 at 16:26
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The Mutzin Taussky theorem states in fact an equivalence between the two propositions you gave. If you understand french, here is a paper presenting a problem ( starting from page 16 till 22) whose purpose is to prove that theorem. http://agreg.org/Rapports/rapport2009.pdf

have fun!

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