Re: [理工] [線代]-對稱矩陣
※ 引述《mdpming (阿阿 要加油)》之銘言：
: [ 1 0 3 0]
: [ 0 0 0 0] = Q
: [ 3 0 1 0]
: [ 0 0 0 0]
: 1 .find P orthogonally diagonalizes of Q..
: T
: 2.P QP
: 恭喜上榜的各位 都好利害
1. [ 1 0 3 0]
令Q=[ 0 0 0 0]
[ 3 0 1 0]
[ 0 0 0 0]
|1-x 0 3 |
PQ(x)=det(Q-xI)=-x| 0 -x 0 |=-x[(1-x)^2 (-x) - 9(-x)]
| 3 0 1-x|
=(-x)^2 (-8-2x+x^2)=(-x)^2 (x-4)(x+2)
λ(Q)={-2,0,4}
[ 3 0 3 0] [-1] [-1/√2]
V(-2)=ker(Q+2I)=ker[ 0 2 0 0]=span{[ 0]} 單位化 [ 0 ]
[ 3 0 3 0] [ 1] => [ 1/√2]
[ 0 0 0 2] [ 0] [ 0 ]
[ 1 0 3 0] [0] [0]
V(0)=ker(Q)=ker[ 0 0 0 0]=span{[1]，[0]}
[ 3 0 1 0] [0] [0]
[ 0 0 0 0] [0] [1]
[0] [0] [0] [0]
先做Gram-Schmidt => [1]，[0] 再單位化=> [1]，[0]
[0] [0] [0] [0]
[0] [1] [0] [1]
[-3 0 3 0] [1] [1/√2]
V(4)=ker(Q-4I)=ker[ 0 -4 0 0]=span{[0]} 單位化 [ 0 ]
[ 3 0 -3 0] [1] => [1/√2]
[ 0 0 0 -4] [0] [ 0 ]
[-1/√2 0 0 1/√2] [-2 0 0 0]
取P=[ 0 1 0 0 ] 使得P^T Q P =[ 0 0 0 0]
[ 1/√2 0 0 1/√2] [ 0 0 0 0]
[ 0 0 1 0 ] [ 0 0 0 4]
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◆ From: 60.244.37.56
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※ 編輯: crazykk 來自: 60.244.37.56 (03/20 16:39)
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※ 編輯: crazykk 來自: 60.244.37.56 (03/20 16:57)
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