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Rank of matrix without echelon form

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1

$begingroup$

I need to compute the rank of the following matrix:
$$beginbmatrix
1 1 1 0 0 0 0 0 0\
0 0 0 1 1 1 0 0 0\
0 0 0 0 0 0 1 1 1\
1 0 0 1 0 0 1 0 0\
0 1 0 0 1 0 0 1 0\
0 0 1 0 0 1 0 0 1\
1 0 0 0 1 0 0 0 1\
0 1 0 0 0 1 0 0 0\
0 0 0 1 0 0 0 1 0\
0 0 1 0 1 0 1 0 0\
0 1 0 1 0 0 0 0 0\
0 0 0 0 0 1 0 1 0\
endbmatrix$$

Is there a fast way to do it without reducing to row echelon form?

linear-algebra

share|cite|improve this question

asked Mar 22 at 3:08

dxdydzdxdydz

48110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is this some arbitrary matrix, or did it come from somewhere?
  $endgroup$
  – Minus One-Twelfth
  Mar 22 at 3:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Rank of given matrix is ≥4. This is clearly visible (from 4th entry of first column, 5th entry of second column and 6th entry of third column it is clear that first three columns are linearly independent. further, from second entry of 4th column it is clear that, 4th column does not belongs to span of first three and hence first four columns are linearly independent)
  $endgroup$
  – Akash Patalwanshi
  Mar 22 at 4:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Row reduction is fast especially for $0-1$ matrices. To make it faster, take the transpose and then perform row operations; or what amounts to the same thing, perform column operations instead. The rank is the same in both cases.
  $endgroup$
  – Chrystomath
  Mar 22 at 5:10
  

  
  
  
  

add a comment | 

1

$begingroup$

I need to compute the rank of the following matrix:
$$beginbmatrix
1 1 1 0 0 0 0 0 0\
0 0 0 1 1 1 0 0 0\
0 0 0 0 0 0 1 1 1\
1 0 0 1 0 0 1 0 0\
0 1 0 0 1 0 0 1 0\
0 0 1 0 0 1 0 0 1\
1 0 0 0 1 0 0 0 1\
0 1 0 0 0 1 0 0 0\
0 0 0 1 0 0 0 1 0\
0 0 1 0 1 0 1 0 0\
0 1 0 1 0 0 0 0 0\
0 0 0 0 0 1 0 1 0\
endbmatrix$$

Is there a fast way to do it without reducing to row echelon form?

linear-algebra

share|cite|improve this question

asked Mar 22 at 3:08

dxdydzdxdydz

48110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is this some arbitrary matrix, or did it come from somewhere?
  $endgroup$
  – Minus One-Twelfth
  Mar 22 at 3:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Rank of given matrix is ≥4. This is clearly visible (from 4th entry of first column, 5th entry of second column and 6th entry of third column it is clear that first three columns are linearly independent. further, from second entry of 4th column it is clear that, 4th column does not belongs to span of first three and hence first four columns are linearly independent)
  $endgroup$
  – Akash Patalwanshi
  Mar 22 at 4:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Row reduction is fast especially for $0-1$ matrices. To make it faster, take the transpose and then perform row operations; or what amounts to the same thing, perform column operations instead. The rank is the same in both cases.
  $endgroup$
  – Chrystomath
  Mar 22 at 5:10
  

  
  
  
  

add a comment | 

$begingroup$

I need to compute the rank of the following matrix:
$$beginbmatrix
1 1 1 0 0 0 0 0 0\
0 0 0 1 1 1 0 0 0\
0 0 0 0 0 0 1 1 1\
1 0 0 1 0 0 1 0 0\
0 1 0 0 1 0 0 1 0\
0 0 1 0 0 1 0 0 1\
1 0 0 0 1 0 0 0 1\
0 1 0 0 0 1 0 0 0\
0 0 0 1 0 0 0 1 0\
0 0 1 0 1 0 1 0 0\
0 1 0 1 0 0 0 0 0\
0 0 0 0 0 1 0 1 0\
endbmatrix$$

Is there a fast way to do it without reducing to row echelon form?

linear-algebra

share|cite|improve this question

asked Mar 22 at 3:08

dxdydzdxdydz

48110

$endgroup$

share|cite|improve this question

asked Mar 22 at 3:08

dxdydzdxdydz

48110

share|cite|improve this question

asked Mar 22 at 3:08

dxdydzdxdydz

48110

asked Mar 22 at 3:08

dxdydzdxdydz

48110

asked Mar 22 at 3:08

dxdydzdxdydz

48110