[線代] linearly independent

看板Math作者 (風切羽狂)時間14年前 (2011/05/08 18:53), 編輯推噓1(105)
留言6則, 4人參與, 最新討論串4/4 (看更多)
※ 引述《mqazz1 (無法顯示)》之銘言: : let A be a 3*3 matrix and let x1, x2 and x3 be vector in R^3 : show that if the vectors : y1 = Ax1 : y2 = Ax2 : y3 = Ax3 : are linearly independent, then the matrix A must be nonsingular and the vector : x1, x2 and x3 must be linearly independent : 請問這要怎麼證明呢? : 我想了兩天左右 還不太清楚怎麼證..謝謝 Let y1,y2,y3 be column vetor. Note that [y1 y2 y3] = [Ax1 Ax2 Ax3]= A[x1 x2 x3] all of them are 3x3 matrix Since y1,y2,y3 are linearly independent ,so det[y1 y2 y3] = det(A[x1 x2 x3]) = det(A)det[x1 x2 x3] ≠ 0 thus, det(A) ≠ 0 and det[x1 x2 x3] ≠ 0 which implies that matrix A must be nonsingular and the vector x1, x2 and x3 must be linearly independent Done. -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 218.160.250.142 ※ 編輯: sm008150204 來自: 218.160.250.142 (05/08 18:55)

05/08 19:16, , 1F
你怎麼不用回文= =
05/08 19:16, 1F

05/08 21:06, , 2F
我打到一半突然當機,第一次用暫存檔不太會用 抱歉
05/08 21:06, 2F

05/08 21:08, , 3F
真的是超尷尬的 還是我現在應該複製 然後重回一篇?
05/08 21:08, 3F

05/08 21:11, , 4F
把標題改成跟原文一樣,程式就會判成相關文章了
05/08 21:11, 4F

05/08 21:11, , 5F
頂多不會變成黃色罷了
05/08 21:11, 5F

05/09 18:47, , 6F
謝謝
05/09 18:47, 6F
文章代碼(AID): #1DndOMNX (Math)
文章代碼(AID): #1DndOMNX (Math)