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al matrix with inequality
Let A be a 3x3 matrix with property that A^2=O or A^2=I,where O and I are the 3x3 zero matrix and 3X3 identity matrix, respectively. Prove that det(A+I)>=det(A-I)
最佳解答:
If AA = I 0 <= [det (A+I)]^2 = det (A+I) (A+I) = det (AA + 2A + I) = det (2A+2I) = 2 det (A+I) If AA = O 0 <= [det (A/2+I)]^2 = det (A/2+I) (A/2+I) = det (AA/4 + A + I) = det (A+I) Therefore for any matrix A, if AA =O or AA = I, then det(A + I) => 0 Now as (– A) (– A) = AA = I or O, det [I + (– A)] => 0 Then det(A – I) = - det (I – A) = - det [I + (– A)] <=0 as (– A) also satisfies (– A) (– A) = I or (– A) (– A) = O hence det(A + I) => 0 => det(A – I) 2010-05-08 22:36:41 補充： 小錯誤 det(2A+2I) = det(2 I) det (A+I) = 8 det(A + I) for 3x3 matrix
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al matrix with inequality
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發問:Let A be a 3x3 matrix with property that A^2=O or A^2=I,where O and I are the 3x3 zero matrix and 3X3 identity matrix, respectively. Prove that det(A+I)>=det(A-I)
最佳解答:
If AA = I 0 <= [det (A+I)]^2 = det (A+I) (A+I) = det (AA + 2A + I) = det (2A+2I) = 2 det (A+I) If AA = O 0 <= [det (A/2+I)]^2 = det (A/2+I) (A/2+I) = det (AA/4 + A + I) = det (A+I) Therefore for any matrix A, if AA =O or AA = I, then det(A + I) => 0 Now as (– A) (– A) = AA = I or O, det [I + (– A)] => 0 Then det(A – I) = - det (I – A) = - det [I + (– A)] <=0 as (– A) also satisfies (– A) (– A) = I or (– A) (– A) = O hence det(A + I) => 0 => det(A – I) 2010-05-08 22:36:41 補充： 小錯誤 det(2A+2I) = det(2 I) det (A+I) = 8 det(A + I) for 3x3 matrix
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