Re: [理工] [工數]-特徵值與特徵向量
看板Grad-ProbAsk作者ntust661 (Auf Wiedersehen!)時間16年前 (2010/02/14 23:23)推噓1(1推 0噓 2→)留言3則, 2人參與討論串2/3 (看更多)
※ 引述《smallprawn (水中瑕)》之銘言：
: 1.
: [7 -2 -4]
: [3 0 -2]
: [6 -2 -3] 已求出λ=1,1,2
: λ=2 已經確定特徵向量正確
: λ=1,1 這個我計算出
: [1] [2]
: X=[3] ,[0]
: [0] [3]
: 但解答為
: [1] [0 ]
: X=[3] ,[-2]
: [0] [1 ]
: 請問哪個是對的!?如果是解答對請高手解析!!
:
λ=1
[6 -2 -4][x1] [0]
[3 -1 -2][x2] = [0]
[6 -2 -4][x3] [0]
1 3可消
[6 -2 -4][x1] [0]
[3 -1 -2][x2] = [0]
[0 0 0][x3] [0]
3x1 = x2 + 2x3
x1 = (x2)/3 + 2(x3)/3
x2 = α ， x3 = β
[x1] [1/3] [2/3]
[x2] = α[ 1 ] + β[ 0 ]
[x3] [ 0 ] [ 1 ]
if ， x2 = 3x1 - 2x3
x1 = α ， x3 = β
[x1] [ 1 ] [ 0 ]
[x2] = α[ 3 ] + β[-2 ]
[x3] [ 0 ] [ 1 ]
由此可知，特徵向量(表達方式)不唯一唷!
判定小方法，外積會得到相同的結果~(只限維度3啦^^)
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