polinoame de matrici
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polinoame de matrici
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.
n-ar fi rau sa fie bine 
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Radu Titiu
- Thales
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- Location: Mures \Bucuresti
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andy crisan
- Pitagora
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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
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