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I have large sparse square matrix n by n, its rank is slightly below n, let's say m. I want to make it non-singular by removing rows and columns by a certain rule. The rule is that if you remove ith row, you must remove ith column as well, so that the matrix is always square. This is effectively removing a node in an adjacency graph.

My first question is: does there always exist such a combination of n-m rows and columns I can remove such that the remaining m by m submatrix is structurally non singular.

My second questions is: is there an effective algorithm to obtain a p by p non-singular submatrix without removing excessive amount of rows and columns

To provide more context, the matrix I'm dealing with is about 1000 by 1000 with sparsity close to 0.05

Dennis
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  • where does this rule come from? You can instead do a sparse LU or qr depending on the required cost – percusse Jan 29 '16 at 16:56
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    because I'm effectively operating on a graph. If I remove a row, I'm effectively removing a node and all its edges. Therefore the corresponding column will be gone as well. – Dennis Jan 29 '16 at 17:57
  • Is it a binary matrix (just asking because of the graph context)? – Daniel Jan 29 '16 at 18:27
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    No, the matrix is not binary. Actually, what I have is the matrix binary pattern. The actual numbers in each entry could change. My goal is to extract a non-singualr submatrix (a non-singualr adjacency matrix) by removing rows and columns (nodes). – Dennis Jan 29 '16 at 18:31

2 Answers2

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1 is not true. here's a example.

[1 0 0 0;
 0 1 0 0;
 0 0 0 1;
 0 0 0 0]

The rank is clearly 3 which happens to be the number of nonzero rows/columns). You can't remove rows 1,2,3 nor columns 1,2,4. So 1 to 4 are covered.

hiandbaii
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The first one is not true. As has been answered by hiandbali. I managed to solve the second problem by doing a DFS. The interior adjacency matrices are not singular.

Dennis
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