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problema matrice de ordin 3 cu intrarile complexe

 
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Cezar Lupu
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Joined: 26 Sep 2007
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Location: Pittsburgh/Craiova/Bucharest/Constanta

PostPosted: Mon Nov 07, 2011 12:26 am    Post subject: problema matrice de ordin 3 cu intrarile complexe Reply with quote

Fie A, B\in M_{3}(\mathbb{C}) astfel incat A^2-B^2=I_{3}. Sa se arate ca (AB-BA)^3=O_{3}.
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Mateescu Constantin
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Joined: 21 Apr 2009
Posts: 398
Location: Londra/Pitesti

PostPosted: Wed Nov 09, 2011 7:13 pm    Post subject: Reply with quote

Este clar ca \text{tr}\, \left(AB-BA\right)=0 . Pe de alta parte avem: \left\{\begin{array}{cccc}
\det\, \left(AB-BA\right)=\det\, \left(I_3-\left(A+B\right)\left(A-B\right)\right) \\\\\\\\ 
\det\, \left(BA-AB\right)=\det\, \left(I_3-\left(A-B\right)\left(A+B\right)\right)\end{array}

si cum se stie ca pentru orice X\, ,\, Y\in\mathcal{M}_3\left(\mathbb{C}\right) are loc relatia \det\, \left(I_3-XY\right)=\det\, \left(I_3-YX\right) , obtinem:

\det\, \left(AB-BA\right)=\det\, \left(BA-AB\right)=\left(-1\right)^3\det\, \left(AB-BA\right) de unde rezulta ca \det\, \left(AB-BA\right)=0 .

Mai ramane de aratat ca \text{tr}\, \left(\left(AB-BA\right)^{\ast}\right)=0 si problema se incheie. Echivalent, acest lucru se exprima ca:

\text{tr}\, \left(\left(AB-BA\right)^2\right)=0 sau \text{tr}\, \left(AB\left(AB-BA\right)\right)=0 . Totusi, n-am reusit sa finalizez ... Rolling Eyes
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Claudiu Mindrila
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Joined: 01 Oct 2007
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PostPosted: Mon Nov 14, 2011 8:18 pm    Post subject: Reply with quote

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=353&t=444925
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