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Intarirea unei inegalitati de pe mateforum
Posted: Mon Aug 10, 2009 1:10 pm
by opincariumihai
Fie
\( A, B\in M_{2}(\mathbb{R}) \) doua matrice astfel incat
\( \det(AB-BA)\geq0 \). Sa se arate ca
\( \det(A^{2}+B^{2})\geq(\det A- \det B) ^{2} \).
Mihai Opincariu
Vezi cazul particular de
aici.
Posted: Mon Aug 10, 2009 6:42 pm
by Marius Mainea
Folosim relatia :
\( \det(X+Y)=\det X+\tr(XY^{\ast})+\det Y \)
Asadar \( \det (AB-BA)=\det AB-\tr[(AB)(BA)^{\ast}]+\det BA\ge 0 \),
de unde
\( 2\det AB\ge \tr([(AB)(BA)^{\ast}] \) (***)
Pe de alta parte concluzia se scrie :
\( \det A^2+\tr[A^2(B^2)^{\ast}]+\det B^2\ge \det A^2-2\det AB+\det B^2 \) sau
\( 2\det AB+\tr [A^2(B^2)^{\ast}]\ge 0 \)
Insa folosind (***) e suficient sa aratam ca \( \tr[(AB)(BA)^{\ast}]+\tr[A^2(B^2)^{\ast}]\ge 0 \) care este evidenta.
Posted: Mon Aug 10, 2009 9:44 pm
by opincariumihai
Marius Mainea wrote:
Insa folosind (***) e suficient sa aratam ca \( \tr[(AB)(BA)^{\ast}]+\tr[A^2(B^2)^{\ast}]\ge 0 \) care este evidenta.
De ce e evidenta?
Sper ca nu va referiti la calcul direct...
Posted: Tue Aug 11, 2009 9:38 am
by Marius Mainea
\( \tr[(AB)(BA)^{\ast}]+\tr[A^2(B^2)^{\ast}]=\tr[A(BA^{\ast}+AB^{\ast})B^{\ast}]=\tr[A\tr(AB^{\ast})I_2B^{\ast}]=[\tr(AB^{\ast})]^2\ge 0 \)