Grabbing the independent columns of a matrix A

Question

I found a nice way to generate a random $5\times 7$ matrix with rank 3.

MatrixForm[A = RandomInteger[5, {5, 3}].RandomInteger[5, {3, 7}]]
MatrixRank[A]

Let me store an example output, because each time we use RandomInteger, the matrix changes. So, we'll work with this one, an output from the strategy above.

A = {{18, 29, 20, 17, 37, 22, 38}, {12, 16, 10, 4, 14, 14, 16}, {6, 
   16, 12, 17, 28, 9, 27}, {10, 11, 9, 6, 18, 10, 17}, {14, 20, 17, 
   18, 39, 15, 36}}

If we use:

MatrixForm[RowReduce[A]]

Then we can identify the pivot columns. Then we can select the pivot columns and store them in matC.

MatrixForm[matC = A[[All, 1 ;; 3]]]

I am just wondering if there is a cute Mathematica command I haven't encountered that will automatically select the independent columns or the independent rows of a matrix.

Answer 1

I believe that the following code selects the columns of the matrix that provides an approximate "Interpolative Decomposition". This means that all the other columns can be written as linear combinations of the selected columns with minimal coefficients. There are some interesting results (I believe) on how small these coefficients need be (never greater that 2 if I recall correctly).

{q, r, p} = QRDecomposition[N[A], Pivoting -> True];
id = A.p[[All, 1 ;; MatrixRank[A]]]

Pivoting only seems to take effect for inexact arithmetic.

If you want the coefficient matrix, this can be computed using

coeff = LinearSolve[id,A]

so that

id.coeff == A // FullSimplify