Matrice de ordin 2 nilpotenta

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Tudorel Lupu
Euclid
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Joined: Mon Oct 01, 2007 8:58 pm
Location: Constanta

Matrice de ordin 2 nilpotenta

Post by Tudorel Lupu »

a) Fie \( X\in M_{2}(\mathbb{C}) \) o matrice. Demonstrati echivalenta:

\( \det(X+I_{2})=\det(X-I_{2})\Leftrightarrow tr(X)=0 \).

b) Fie \( A\in M_{2}(\mathbb{C}) \) o matrice astfel incat exista \( k\in\mathbb{N}^{*} \) astfel incat sa avem

\( \det(A^{l}+I_{2})=\det(A^{l}-I_{2}) \),

pentru \( l\in\{k, k+1\} \).

Demonstrati ca \( A^{n}=O_{2}, \forall n\in\mathbb{N}, n\geq 2 \).


Nelu Chichirim, Olimpiada locala Constanta, 2008
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Radu Titiu
Thales
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Post by Radu Titiu »

a) Daca \( \det(A-xI)=x^2-Tr(A)x+\det(A) \) atunci din egalitatea \( \det(A+I)=\det(A-I) \) rezulta ca 2Tr(A)=0 de unde concluzia, Tr(A)=0.

b) Din punctul precedent rezulta ca \( Tr(A^k)=Tr(A^{k+1})=0 \Leftrightarrow \) \( x_1^k+x_2^k=0 \) si \( x_1^{k+1}+x_2^{k+1}=0 \), unde \( x_1 \) si \( x_2 \) sunt valorile proprii ale matricei A. Din acest sistem rezulta necesar \( x_1=x_2=0 \) si din teorema lui Caley-Hamilton obtinem \( A^2=0 \).
A mathematician is a machine for turning coffee into theorems.
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