al matrix with inequality

little monster

2010-05-08 11:43 pm

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)

回答 (1)

用戶40344

2010-05-09 3:35 am

✔ 最佳答案

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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