Doua matrice de ordin 3 diferite
Moderators: Bogdan Posa, Laurian Filip, Beniamin Bogosel, Radu Titiu, Marius Dragoi
-
Cezar Lupu
- Site Admin
- Posts: 612
- Joined: Wed Sep 26, 2007 2:04 pm
- Location: Bucuresti sau Constanta
- Contact:
Doua matrice de ordin 3 diferite
Fie \( A, B\in M_{3}(\mathbb{R}) \) doua matrice astfel incat \( A\neq B \). Daca \( A^3= B^3 \) si \( A^{2}B=B^{2}A \), aratati ca matricea \( A^{2}+B^{2} \) nu este inversabila.
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
-
Ciprian Oprisa
- Pitagora
- Posts: 55
- Joined: Tue Feb 19, 2008 8:01 pm
- Location: Lyon sau Cluj sau Baia de Cris
\( (A^2+B^2)(A-B)=A^3-A^2B+B^2A-B^3=O \)
Presupunem k \( \det(A^2+B^2)\neq0 \)
\( \Rightarrow rg(A^2+B^2)=n \)
Din inegalitatea lui Sylvester, \( rg(XY)\geq rg{X}+rg{Y}-n \)
\( \Rightarrow 0=rg(A^2+B^2)(A-B)\geq rg(A^2+B^2)+rg(A-B)-n=rg(A-B) \)
\( \Rightarrow rg(A-B)=0 \ \Rightarrow A=B \) -absurd
Presupunem k \( \det(A^2+B^2)\neq0 \)
\( \Rightarrow rg(A^2+B^2)=n \)
Din inegalitatea lui Sylvester, \( rg(XY)\geq rg{X}+rg{Y}-n \)
\( \Rightarrow 0=rg(A^2+B^2)(A-B)\geq rg(A^2+B^2)+rg(A-B)-n=rg(A-B) \)
\( \Rightarrow rg(A-B)=0 \ \Rightarrow A=B \) -absurd
Un lucru este ceea ce este, nu ceea ce pare a fi.