※ 引述《ilmvm0679 (映雪)》之銘言：
: 如題 想求 rank(A)=rank(A^t)的證明
: 不過不能用到"rank(PAQ)=rank(A),其中P,Q是可逆矩陣"來證明~
: 我看課本的證法必須要用到這點~我想了很久也都想不到>"<
: 想麻煩各位大大看看能不能想到別的做法
: 謝謝~~~
a11 a12 a13 a14 A1 (row 1 vector)
a21 a22 a23 a24 = A2 (row 2 vector)
a31 a32 a33 a34 A3 (row 3 vector)
(i) elementay row computation does not change row rank. obvious
(ii) elementary column computation does not change row rank.
Proof of (ii)
(1) two coulumns exchange. obvious
(2) one column multiplied by a nonzero constant
c*a11 a12 a13 a14 A1'
c*a21 a22 a23 a24 = A2'
c*a31 a32 a33 a34 A3'
A1',A2',A3' are linearly dependent iff A1,A2,A3 are linearly dependent
(3) one column multiplied by a nonzero constant and added to another coulumn
a11 c*a11+a12 a13 a14 A1"
a21 c*a21+a22 a23 a24 = A2"
a31 c*a31+a32 a33 a34 A3"
A1",A2",A3" are linearly dependent iff A1,A2,A3 are linearly dependent
hence, elementray column computation DOES NOT change row rank
However, every m*n matrix can be reduced to the standard form
1 0 0 0 0...
0 1 0 0 0...
0 0 1 0 0...
0 0 0 0 0...
We can see that rank(A)=rank(A^T)
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◆ From: 27.147.57.77
※ 編輯: JohnMash 來自: 27.147.57.77 (07/02 12:22)