extract maximal set of independent columns from a matrix [closed]

Question

I have a matrix that looks like this:

 1 1 1 1 1 1 1 1 1  1  1  1
 1 1 1 1 1 1 0 0 0  0  0  0
 0 0 1 1 0 0 0 0 1  1  0  0
 1 1 0 0 0 0 1 1 0  0  0  0
 0 0 1 1 0 0 0 0 0  0  0  0
 1 1 0 0 0 0 0 0 0  0  0  0

You can see every two columns are identical, indicating the "group membership" of the design matrix. Now my question is, how can I convert this rank-deficient matrix (rank = 6) into a full-rank matrix automatically in R? This case may be a little bit special, i.e. I can delete duplicate columns manually. I am just curious if there is an approach that solve the problem "more generally". Thanks!

Answer 1

I think the way R does QR decomposition this works (and by works I mean leaves a set of independent columns):

m[, qr(m)$pivot[seq_len(qr(m)$rank)]]

On the example from OP:

m = structure(c(1L, 1L, 0L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 0L, 1L, 1L, 
1L, 1L, 0L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 0L, 1L, 1L, 0L, 0L, 0L, 
0L, 1L, 1L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 1L, 0L, 0L, 1L, 0L, 0L, 
1L, 0L, 0L, 1L, 0L, 1L, 0L, 0L, 0L, 1L, 0L, 1L, 0L, 0L, 0L, 1L, 
0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L), .Dim = c(6L, 12L
))

m[, qr(m)$pivot[seq_len(qr(m)$rank)]]
#     [,1] [,2] [,3] [,4] [,5] [,6]
#[1,]    1    1    1    1    1    1
#[2,]    1    1    1    0    0    0
#[3,]    0    1    0    0    1    0
#[4,]    1    0    0    1    0    0
#[5,]    0    1    0    0    0    0
#[6,]    1    0    0    0    0    0

Answer 2

Try:

X[,duplicated(cor(X))]

cor(x) computes the correlation matrix of x. If two columns are linearly dependent to each other they'll have the same column in the correlation matrix

This will get rid of the columns that are a linear transformations of a single other column.

If you're looking for row reduced echelon form instead, which will show you if a column is a linear combination of multiple other columns, check out this answer:

Reduced row echelon form