O matrice \( A \in M_{n}(C) \) este singulara daca si numai daca exista o matrice \( B \in M_{n}(C) \), nenula, astfel incat pentru orice \( p \in N^{*},\ (A+B)^{p}=A^{p}+B^{p} \).
Ion Savu
O alta matrice singulara...
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bogdanl_yex
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O alta matrice singulara...
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Beniamin Bogosel
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Pentru cealalta implicatie: presupunem ca \( A \) este inversabila.
-\( p=2 \Rightarrow AB+BA=0 \)
-\( p=3 \Rightarrow (A+B)(A^2+B^2)=A^3+B^3 \Rightarrow AB^2+BA^2=0 \).
Deci \( AB^2=-BA^2=ABA \) si analog(daca consideram produsul de la p=3 in ordine inversa \( B^2A=-A^2B=ABA \)
Deci \( AB^2=B^2A \).
\( AB^2+BA^2=-BAB-ABA \). Deci \( ABA+BAB=0\Rightarrow ABA-AB^2=0\Rightarrow BA=B^2\Rightarrow \)
\( \Rightarrow BA^2=ABA\Rightarrow AB=BA=0\Rightarrow B=0 \). Contradictie. Deci \( A \) nu este inversabila.
-\( p=2 \Rightarrow AB+BA=0 \)
-\( p=3 \Rightarrow (A+B)(A^2+B^2)=A^3+B^3 \Rightarrow AB^2+BA^2=0 \).
Deci \( AB^2=-BA^2=ABA \) si analog(daca consideram produsul de la p=3 in ordine inversa \( B^2A=-A^2B=ABA \)
Deci \( AB^2=B^2A \).
\( AB^2+BA^2=-BAB-ABA \). Deci \( ABA+BAB=0\Rightarrow ABA-AB^2=0\Rightarrow BA=B^2\Rightarrow \)
\( \Rightarrow BA^2=ABA\Rightarrow AB=BA=0\Rightarrow B=0 \). Contradictie. Deci \( A \) nu este inversabila.