In legatura cu o problema de la ONM 1996

opincariumihai · Post by **opincariumihai** » Tue May 19, 2009 1:31 am

Fie A, B doua matrice de ordinul 2 cu elementele reale astfel incat \( \det(AB-BA)=\det(AB+BA) \). Aratati ca \( \det(A^2+B^2)\geq0 \)

Bogdan Cebere · Post by **Bogdan Cebere** » Tue May 19, 2009 11:13 am

Banuiesc ca este in legatura cu problema propusa de Cristinel Mortici la ONM 1996:
Fie \( A,B \in M_2 (R) \) doua matrice a.i. \( \det(AB+BA) \leq 0 \). Demonstrati ca avem inegalitatea \( \det(A^2+B^2) \geq 0 \).

Aici, ca si acolo, pornim de la relatia \( \det(X+Y)+ \det(X-Y)=2( \det X+ \det Y) \) si folosim polinomul \( f(X+tY)=(\det Y) t^2 +at+ \det X \).

opincariumihai · Post by **opincariumihai** » Tue May 19, 2009 5:54 pm

Acum mai ramane de studiat cazul det(AB+BA)>0 care este echivalent, folosind ipoteza, cu det(AB-BA)>0. Incercati http://www.mateforum.ro/viewtopic.php?t=297.

mihai++ · Post by **mihai++** » Thu Jun 25, 2009 12:02 pm

Ar putea sa posteze cineva (va rog) o solutie la Problema de la ONM 1996?

Post by **Laurian Filip** » Thu Jun 25, 2009 12:22 pm

Bogdan Cebere wrote:Aici, ca si acolo, pornim de la relatia \( \det(X+Y)+ \det(X-Y)=2( \det X+ \det Y) \) si folosim polinomul \( f(X+tY)=(\det Y) t^2 +at+ \det X \).

Pentru \( X=A^2+B^2 \) si \( Y=AB+BA \)
\( 0 \leq \det^2(A+B)+\det^2(A-B)=2(\det(A^2+B^2)+\det(AB+BA)) \)

Marius Mainea · Post by **Marius Mainea** » Thu Jun 25, 2009 7:08 pm

Inca \( \det (A+iB)(A-iB)=\det[A^2+B^2+i(BA-AB)]\ge 0 \) si \( \det (A-iB)(A+iB)=\det[A^2+B^2-i(BA-AB)]\ge 0 \), deci \( 2(\det(A^2+B^2)+\det[i(BA-AB)]\ge 0 \) (1)
\( \det(A^2+B^2)-\det(AB+BA)\ge 0 \) (2)

Prin adunarea lui (1) impartita la 2 cu (2) rezulta concluzia.

mihai++ · Post by **mihai++** » Thu Jun 25, 2009 9:15 pm

La ultimul pas ati schimbat un semn si nu e corect.