[線代] 關於rank(A^T A)=rank(A)的證明?

看板Math作者 (不是一就是二)時間8年前 (2015/10/15 23:30), 編輯推噓0(0010)
留言10則, 2人參與, 最新討論串1/3 (看更多)
關於rank(A^T A)=rank(A)for any A m×n的證明如下: Since elementary operations do not change the rank of a matrix. We have rank(ATA)=rank(ETATAE) , where E is a multiplication of several elementary operations which make AE=[A1,A2], where A1 is a column full rank matrix with rank(A1)=rank(A). Thus we can find a matrix B such that A1B=A2 and AE=[A1,A1P]=A1[I,P] Thus rank(ETATAE)=rank(A1[I,P])T(A1[I,P]) In this equation, the four matrices are all full rank and the rank equals rank(A) , so rank(ATA)=rank(A), completing the proof. 原文網址:http://goo.gl/A4Mg3l 我有疑問的是: 1.A是 m×n rank(A)可能小於min(m,n) AE=[A1,A2]這一定辦的到嗎? 2.為何rank(A1[I,P])T(A1[I,P])會等於rank(A)? 以上兩點請高手指點迷津 謝謝 -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 36.224.85.148 ※ 文章網址: https://www.ptt.cc/bbs/Math/M.1444923023.A.BDF.html

10/16 00:38, , 1F
應該就是列簡梯陣
10/16 00:38, 1F

10/16 00:40, , 2F
可以看高斯消去法求pivot
10/16 00:40, 2F

10/16 02:27, , 3F
剛剛打的不太正確
10/16 02:27, 3F

10/16 02:27, , 4F
1.你的想法是對的 有些矩陣的確不行
10/16 02:27, 4F

10/16 03:06, , 5F
抱歉 我在想一下好了 有點混亂
10/16 03:06, 5F

10/16 04:31, , 6F
let v be a vector, if Av = 0, then (A^T)Av = 0
10/16 04:31, 6F

10/16 04:33, , 7F
if Av != 0, then |Av| != 0
10/16 04:33, 7F

10/16 04:33, , 8F
then v^T A^T A v != 0
10/16 04:33, 8F

10/16 04:33, , 9F
then A^T A v != 0
10/16 04:33, 9F

10/16 04:34, , 10F
所以不只是rank一樣,整個null space根本都一樣
10/16 04:34, 10F
文章代碼(AID): #1M7yQFlV (Math)
文章代碼(AID): #1M7yQFlV (Math)