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Fie \( A,B\in\mathcal{M}_2(\mathbb{R}) \) si \( a\in\mathbb{R^{\ast}} \) astfel incat \( A^2+a^2B^2=aAB \).

Demonstrati ca \( (AB-BA)^2=O_2 \)

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Daca notam \( C=aB \), avem identitatea \( A^2+C^2=AC \), si avem de aratat ca \( \frac{1}{a^2}(AC-CA)^2=O_2 \), deci \( (AC-CA)^2=O_2 \).

Din Hamilton-Cayley, \( (AC-CA)^2=\tr(AC-CA)\cdot (AC-CA)-\det(AC-CA)\cdot I_2 \). Cum \( \tr(AC-CA)=\tr(AC)-\tr(CA)=0 \), mai ramane sa demonstram ca \( \det(AC-CA)=0 \).

Fie \( f(x)=\det(AC(1-x)+CAx)=\det(AC+x(CA-AC))=\det(AC)+\alpha x +x^2 \det(CA-AC) \)
\( f(i)=\det(AC(1-i)+CAi)=\det(A^2+C^2-i(AC-CA))= \)
\( \det[(A+Ci)(A-Ci)]=\det(A+Ci)\det(A-Ci)=|\det(A+Ci)|^2 \in \mathbb{R} \)
Din \( f(i) \in \mathbb{R} \) \( \Rightarrow \det(AC)+\alpha i -\det(CA-AC) \in \mathbb{R} \), deci \( \alpha=0 \).
\( \Rightarrow \det(AC(1-x)+CAx)=\det(AC)+x^2 \det(CA-AC) \).
Luand \( x=1 \), obtinem \( \det(CA)=\det(AC)+\det(CA-AC) \).
\( \Rightarrow \det(AC-CA)=\det(CA-AC)=0 \).