Fie \( A,B \in M_n(C) \) doua matrice care comuta si fie \( \alpha \in C \) o valoare proprie pentru \( A+B. \) Demonstrati ca exista \( \lambda \in C \) si \( \beta \in C \) valori proprii pentru \( A \), respectiv \( B \), astfel incat \( \alpha= \lambda + \beta \).
C:2579, GM 12/2002
Valori proprii ale sumei a doua matrice
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Bogdan Cebere
- Thales
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Bogdan Cebere
- Thales
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Radu Titiu
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Fie \( x \) vectorul propriu corespunzator valorii proprii \( \alpha \). Astfel avem relatia \( Cx=\alpha x \Leftrightarrow (A+B)x=\alpha x \Rightarrow (A-\lambda_k I_n)x=(\alpha I_n-B-\lambda_k I_n)x \)
Astfel avem: (inductiv)
\( O_n=f_A(A)\cdot x=\left(\prod_{k=1}^n (A-\lambda_k I_n)\right)x=\left(\prod_{k=1}^n(\alpha I_n-B-\lambda_k I_n)\right)x \)
Deoarece \( x \neq O_n \), rezulta ca \( \det \left( \prod_{k=1}^n(\alpha I_n-B-\lambda_k I_n) \right) =0 \), rezulta ca exista k a.i.
\( \det((\alpha-\lambda_k )I_n-B)=0 \) deci \( \beta =\alpha-\lambda_k \), de unde rezulta concluzia.
Astfel avem: (inductiv)
\( O_n=f_A(A)\cdot x=\left(\prod_{k=1}^n (A-\lambda_k I_n)\right)x=\left(\prod_{k=1}^n(\alpha I_n-B-\lambda_k I_n)\right)x \)
Deoarece \( x \neq O_n \), rezulta ca \( \det \left( \prod_{k=1}^n(\alpha I_n-B-\lambda_k I_n) \right) =0 \), rezulta ca exista k a.i.
\( \det((\alpha-\lambda_k )I_n-B)=0 \) deci \( \beta =\alpha-\lambda_k \), de unde rezulta concluzia.
Last edited by Radu Titiu on Sat May 31, 2008 12:36 pm, edited 2 times in total.
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