alexyflemming
Oct14-11, 05:00 PM
EQUALITY OF ROW AND COLUMN RANK (o'Neil's proof) Is there smt wrong?
http://www.mediafire.com/imageview.php?quickkey=znorkrmk3k1otjd&thumb=6
Theorem 7.9: EQUALITY OF ROW AND COLUMN RANK
Proof: Page 210.
It writes:....
so the dimension of this column space is AT MOST r (equal to r if these columns are linearly independent, less than r if they are not)
I THINK THIS IS WRONG. Look at the r vectors:
1 0
0 1
: 0
0 :
BETAr+1,1 BETAr+1,2
:
BETAm1 BETAm2
The first r columns of these r vectors are e1,e2,...er. Hence, they are DEFINITELY LINEARLY INDEPENDENT.
There is no way to obtain 1 in the first coordinate of the first of the r vectors from the remaining r-1 vectors since the 1st coordinate of all of the remaining r-1 vectors are all 0.
Hence, the correct one should be:
so the dimension of this column space is EXACTLY r.
Where am I wrong? or O'neil's is really wrong as I indicated.
http://www.mediafire.com/imageview.php?quickkey=znorkrmk3k1otjd&thumb=6
Theorem 7.9: EQUALITY OF ROW AND COLUMN RANK
Proof: Page 210.
It writes:....
so the dimension of this column space is AT MOST r (equal to r if these columns are linearly independent, less than r if they are not)
I THINK THIS IS WRONG. Look at the r vectors:
1 0
0 1
: 0
0 :
BETAr+1,1 BETAr+1,2
:
BETAm1 BETAm2
The first r columns of these r vectors are e1,e2,...er. Hence, they are DEFINITELY LINEARLY INDEPENDENT.
There is no way to obtain 1 in the first coordinate of the first of the r vectors from the remaining r-1 vectors since the 1st coordinate of all of the remaining r-1 vectors are all 0.
Hence, the correct one should be:
so the dimension of this column space is EXACTLY r.
Where am I wrong? or O'neil's is really wrong as I indicated.