Fie \( A\in M_{3}(\mathbb{R}) \) o matrice astfel incat \( \det(A^2+I_{3})=0 \). Sa se arate ca:
i) \( \det (A+I_{3})-\det (A-I_{3})=4 \);
ii) \( \tr(A^3)=\tr^3 (A) \).
Cezar Lupu, lista scurta ONM 2006
Matrice ordin 3 cu det(A^2+I_3)=0
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Cezar Lupu
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Matrice ordin 3 cu det(A^2+I_3)=0
Last edited by Cezar Lupu on Wed Feb 20, 2008 5:51 pm, edited 1 time in total.
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\( \det(A^2+I_3)=0\Rightarrow\det(A-iI_3)\det(A+iI3)=0 \)
\( \Rightarrow|\det(A-iI_3)|^2=0 \) si \( |\det(A+iI_3)|^2=0 \)
\( \Rightarrow\det(iI_3-A)=0 \) si \( \det(-iI_3-A)=0 \), ceea ce inseamna ca \( i \) si \( -i \) sunt valori proprii pentru A. Fie \( r \) cea de-a 3-a valoare.
Avem ca \( f_A(\lambda)=\det(\lambda I_3-A)=(\lambda-i)(\lambda+i)(\lambda-r)=(\lambda^2+1)(\lambda-r) \)
i) \( \det(A+I_3)-\det(A-I_3)=-f_A(-1)+f_A(1)= \)
\( -((-1)^2+1)(-1-r)+(1^2+1)(1-r)=4. \)
ii) \( \tr(A^3)=i^3+(-i)^3+r^3=r^3=(i-i+r)^3=\tr(A)^3. \)
\( \Rightarrow|\det(A-iI_3)|^2=0 \) si \( |\det(A+iI_3)|^2=0 \)
\( \Rightarrow\det(iI_3-A)=0 \) si \( \det(-iI_3-A)=0 \), ceea ce inseamna ca \( i \) si \( -i \) sunt valori proprii pentru A. Fie \( r \) cea de-a 3-a valoare.
Avem ca \( f_A(\lambda)=\det(\lambda I_3-A)=(\lambda-i)(\lambda+i)(\lambda-r)=(\lambda^2+1)(\lambda-r) \)
i) \( \det(A+I_3)-\det(A-I_3)=-f_A(-1)+f_A(1)= \)
\( -((-1)^2+1)(-1-r)+(1^2+1)(1-r)=4. \)
ii) \( \tr(A^3)=i^3+(-i)^3+r^3=r^3=(i-i+r)^3=\tr(A)^3. \)
Un lucru este ceea ce este, nu ceea ce pare a fi.