Re: [理工] [線代]-對稱矩陣
※ 引述《crazykk (JK)》之銘言：
: ※ 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|
我這一步 不太懂
請問要怎麼降階...
: =(-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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