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O identitate simpla cu det si tr
Posted: Fri Sep 03, 2010 2:04 pm
by Mateescu Constantin
Aratati ca daca \( A\in\mathcal{M}_2(\mathbb{C}) \) atunci exista relatia : \( \fbox{\ \det\ \left\(A^3+A^2+A+I_2\right\)=\left\[1+\tr A+\det A\right\]\ \cdot\ \left\[\left\(1-\det A\right\)^2+\tr^2 A\right\]\ } \) .
Posted: Fri Sep 03, 2010 7:49 pm
by Marius Mainea
\( (\lambda_1^3+\lambda_1^2+\lambda_1+1)(\lambda_2^3+\lambda_2^2+\lambda_2+1)=(1+\lambda_1+\lambda_2+\lambda_1\lambda_2)[(1-\lambda_1\lambda_2)^2+(\lambda_1+\lambda_2)^2] \)
Posted: Fri Sep 03, 2010 11:14 pm
by Mateescu Constantin
Sa incercam mai elegant
Fie
\( P(X)=\det (A-XI_2)=X^2-a\cdot X+b \) polinomul caracteristic al matricei
\( A \) , unde
\( a=\tr A \) ,
\( b=\det A \) .
Atunci :
\( \left\|\ \begin{array}{cccc}
P(\mbox{i}) & = & -1-\mbox{i}\cdot a+b \\\\\\\\\\
P(-1) & = & 1+a+b \\\\\\\\\\
P(-\mbox{i}) & = & -1+\mbox{i}\cdot a+b\end{array}\ \right|\ \bigodot\ \Longrightarrow\ P(\mbox{i})\cdot P(-1)\cdot P(-\mbox{i})=(1+a+b)\cdot\left\[(1-b)^2+a^2\right\]\ \Longleftrightarrow \)
\( \det\left[(A+I_2)\cdot (A-\mbox{i}\cdot I_2)\cdot (A+\mbox{i}\cdot I_2)\right\]=(1+a+b)\cdot\left\[(1-b)^2+a^2\right\]\ \Longleftrightarrow\ \det\ \left\(A^3+A^2+A+I_2\right)=(1+a+b)\cdot\left\[(1-b)^2+a^2\right\] \)
Observatie. In aceeasi maniera se poate arata ca pentru o matrice
\( A\in\mathcal{M}_3(\mathbb{C}) \) are loc relatia :
\( \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \fbox{\ \det\ (A^3+A^2+A+I_3)=-\left\(1+\tr(A)+\tr(A^{\ast})+\det(A)\right\)\ \cdot\ \left\[\left\(\tr(A)-\det(A)\right\)^2+\tr^2(A^{\ast})\right\]\ } \)