[線代] 關於rank(A^T A)=rank(A)的證明?

[ck6fuz516 (不是一就是二)]

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


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→ deflife     : 應該就是列簡梯陣
→ deflife     : 可以看高斯消去法求pivot
→ deflife     : 剛剛打的不太正確
→ deflife     : 1.你的想法是對的 有些矩陣的確不行
→ deflife     :  抱歉 我在想一下好了 有點混亂
→ wohtp       : let v be a vector, if Av = 0, then (A^T)Av = 0
→ wohtp       : if Av != 0, then |Av| != 0
→ wohtp       : then v^T A^T A v != 0
→ wohtp       : then A^T A v != 0
→ wohtp       : 所以不只是rank一樣，整個null space根本都一樣