1

I have a very sparse matrix A. I need to find out a smallest number of rows of A which satisfies the following conditions:

1). Let us suppose the number of rows form a sub-matrix B. In another word, for a given sparse tall matrix A, we need to find out sub-matrix B formed by a smallest number of rows of A;

2). Matrix B must contain the first row (should be non-zero row) of A;

3). The rank of B must be equal to the number of non-zero columns (a non-zero column is defined as a column containing at least one non-zero element in the column) of B.

4). The rank of B must be smaller than of at worst equal to the rank of matrix A.

For example, A = [

 1     0    -1     0     0
 0    -1     0     2     0
 0     0     2    -1    -1
 2     0     1     0     0

];

The answer is obvious. The matrix B is formed by the first row and the last row of A:

B = [

 1     0    -1     0     0
 2     0     1     0     0

];

The condition is satisfied: rank(B) = 2, number of non-zero columns of B = 2.

How about the following example?

A = [

 1    -1     0     0     0     0     0     0     0
 1     0     0     0     0     0     0     0    -1
 0     0     0     0     0     0     0     0     0
 0     0    -1    -1     0     0     0    -1     0
 0     0     1     0     0     0     0     0     0
 0     0     0     1     0     0     0     0     0
-1     0     0     0     4    -1    -1    -1     0
 0     0     0     0    -1     1     0     0     0
 0     0     0     0    -1     0     1     0     0
 0     0     0     0    -1     0     0     2     0
 0     0     0     0     0     0     0     0     0
 3    -1     0     0    -1     0     0     0    -1
-1     1     0     0     0     0     0     0     0
-1     0     0     0     0     0     0     0     1

];

Please help to find out a sub-matrix B for a given sparse tall matrix A.

Thanks a lot.

Benson

Benson
  • 11
  • I tried to search all the combinations of rows in A to obtain submatrix B. But it was too time-wasting, because the size of my matrix could be 10000x10000. – Benson Jan 27 '21 at 02:58

0 Answers0