Find the indexes of dependent columns

Question

I have a fairly large singular square matrix, of size 37 and rank 35.

Is there a way to get a vector with the (groups of) indexes of the columns in the original matrix that are linearly dependent?

RowReduce et al won't help me, because they perform an arbitrary number of row swaps.

Answer 1

One way is to start with empty matrix. Add the first column. Then loop, each time adding the next column, and checking if the rank of this matrix has increased from before, if so, keep it, else skip over to the next column. Keep doing this until you reach the last column in the original matrix, or have collected m columns, where m is the rank of the original matrix. (no need to keep trying if found m columns).

function

indepCols[mat_?(MatrixQ[#, NumericQ] &)] := Module[{nRows, nCols, m, idx, 
    i, vecs, candidate},

  {nRows, nCols} = Dimensions[mat];
  m = MatrixRank[mat];
  idx = Table[1, {m}]; (*arrary to collect the index of columns*)
  vecs = {mat[[All, 1]]}; (*first column is always in*)
  Do[
   candidate = Join[vecs, {mat[[All, i]]}];
   If[MatrixRank[candidate] > Length[vecs],(*did the rank increase?*)
    idx[[Length[vecs] + 1]] = i;
    vecs = candidate;
    If[Length[vecs] == m, Break[]](*bail out if got the rank*)
    ],
   {i, 2, nCols}
   ];

  {idx, vecs}
  ]

To use the above function:

First example

(mat = {{1, 0, -2, 1, 0}, {0, -1, -3, 1, 3}, {-2, -1, 1, -1, 3}, 
     {-2, -1, 1, -1, 3}, {0, 3, 9, 0, -12}}) // MatrixForm

The above has rank 3. So there will be 3 L.I. columns

 {idx, out} = indepCols[mat];
 Transpose[out] // MatrixForm

idx
(*{1, 2, 4}*)

Verified using "AdjacencyLists" thanks to Michael E2 above.

sa = Transpose@Unitize@SparseArray@RowReduce[mat];
row2col = Rule @@@ Reverse /@ First /@ GatherBy[sa["NonzeroPositions"], First];
Thread[Range[First@Dimensions@sa] -> (sa["AdjacencyLists"] /. row2col)]

second example

Using example given by Michael E2, which is much larger

SeedRandom[1];
mat = RandomSample[#~Join~Accumulate@RandomSample[#, 2] &@
    RandomInteger[{-5, 5}, {35, 37}]];
{idx, out} = indepCols[mat];
idx

Verified:

sa = Transpose@Unitize@SparseArray@RowReduce[mat];
row2col = Rule @@@ Reverse /@ First /@ GatherBy[sa["NonzeroPositions"], First];
Thread[Range[First@Dimensions@sa] -> (sa["AdjacencyLists"] /. row2col)]

The above gives the index of the L.I. columns. To find the index of the L.D. columns, simply take the complement:

Complement[Range[1, Length[mat]], idx]
(*{36, 37}*)

Answer 2

If RowReduce won't help, then perhaps I don't know what you're looking for. Here's my understanding of the question in which I use RowReduce to get the answer.

Example

A random matrix:

SeedRandom[1];
mat = RandomSample[#~Join~Accumulate@RandomSample[#, 2] &@
    RandomInteger[{-5, 5}, {35, 37}]];
MatrixRank[mat]
(*  35  *)

We can use the SparseArray property "AdjacencyLists" to find for each column, which columns it depends on. I'm not sure what form the final result should be in. The output below indicates for each column, which columns it is a linear of. An entry of the form i -> {i} indicates an independent vector. The entry 21 -> {3, 8} indicates that column 21 is a linear combinations of columns 3 and 8. The leading 1 in each nonzero row of the reduced matrix indicates the column with which the columns with nonzero entries in the row have a relationship. The set of rules row2col basically indicates for each row, the column in which the leading 1 is located.

sa = Transpose @ Unitize @ SparseArray @ RowReduce[mat2];
row2col = 
  Rule @@@ Reverse /@ First /@ GatherBy[sa["NonzeroPositions"], First];
Thread[Range[First @ Dimensions @ sa] -> (sa["AdjacencyLists"] /. row2col)]
(*
{1 -> {1}, 2 -> {2}, 3 -> {3}, 4 -> {4}, 5 -> {5}, 6 -> {6}, 7 -> {7},
 8 -> {8}, 9 -> {9}, 10 -> {10}, 11 -> {11}, 12 -> {12}, 13 -> {13}, 
 14 -> {14}, 15 -> {15}, 16 -> {16}, 17 -> {17}, 18 -> {18}, 
 19 -> {19}, 20 -> {20},
 21 -> {3, 8},
 22 -> {22}, 23 -> {23}, 24 -> {24},
 25 -> {3},
 26 -> {26}, 27 -> {27}, 28 -> {28}, 
 29 -> {29}, 30 -> {30}, 31 -> {31}, 32 -> {32}, 33 -> {33}, 
 34 -> {34}, 35 -> {35}, 36 -> {36}, 37 -> {37}}
*)