看板 Grad-ProbAsk 關於我們 聯絡資訊
※ 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^^) -- -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 122.122.218.114
smallprawn:感謝@@..判定方式不懂= =||| 02/14 23:30
ntust661:就是兩個特徵向量外積相同,他們就視同一掛的(同一個面) 02/14 23:32
smallprawn:嗯嗯~暸改了!感謝!! 02/14 23:32