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polinoame de matrici

 
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mihai++
Bernoulli


Joined: 28 Nov 2007
Posts: 217
Location: Focsani

PostPosted: Fri Feb 26, 2010 11:25 am    Post subject: polinoame de matrici Reply with quote

Fie  A\in \mathcal{M}_{2,n}(\mathbb{C}),B\in\mathcal{M}_{n,2}(\mathbb{C}). Demonstrati ca f_{BA}=z^{n-2}f_{AB}. Nu stiu daca e adevarata, dar am auzit ca e in culegerea Fadeev-Sominski.
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Radu Titiu
Thales


Joined: 28 Sep 2007
Posts: 181
Location: Mures \Bucuresti

PostPosted: Fri Feb 26, 2010 4:20 pm    Post subject: Reply with quote

Are loc un rezultat mai general .Uite aici.

Si rezultatul s-a mai postat si pe mateforum aici,dar fara nici o demonstratie.
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andy crisan
Pitagora


Joined: 28 Dec 2008
Posts: 87
Location: Pitesti/Londra

PostPosted: Fri Feb 26, 2010 10:50 pm    Post subject: Reply with quote

Am gasit o solutie mai usoara zic eu.
Este suficient sa aratam ca \det(I_n-AB)=\det(I_m-BA),A\in\mathcal{M}_{n,m}(\mathbb{C}),B\in\mathcal{M}_{m,n}(\mathbb{C}) caci putem considera, pentru cazul cu x A\to \frac{1}{x}A.
Consideram n>m
Fie C=\(A\mbox{  } 0_{n-m}) si D=\(B\\0_{n-m}\),C,D\in\mathcal{M}_{n}(\mathbb{C}).
Facand inmultirile obtinem
CD=AB\Rightarrow I_n-AB=I_n-CD\Rightarrow \det(I_n-AB)=\det(I_n-CD)
DC=\(BA\mbox{ }0_{n-m}\\0_{n-m}\mbox{ }0_{n-m}\)\Rightarrow I_n-DC=\(I_m-BA\mbox{ }0_{n-m}\\0_{n-m}\mbox{ }I_{n-m}\)\Rightarrow \det(I_n-DC)=\det(I_m-BA).
De unde trebuie aratat ca \det(I_n-CD)=\det(I_n-DC) care este usor
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