How to find dependent column vectors of a matrix for a given column vector?

For example if I have a matrix m where:

m=[1 2 3 4; ...
   1 5 3 6; ...
   1 7 3 8];

if I put as an input column 1 ,m(:,1), I receive the result:

r=[3 3 3]'

thanks,

[Image Analyst]

I don't understand. If you pass some function column 1 of m, which is [1;1;1], then how are you getting the result of [3,3,3]'??? Does it have something to do with m having 3 rows? So the function I would guess would recognize that [1;1;1] is column 1, and would get, say, foundColumnNumber = 1 (because it found it in column #1), but then do you do something like

r = m(1, foundColumnNumber) * ones(size(m, 1), 1);

where it takes the first element in the found column and replicates it for as many rows as you have???

Answer 1

[Pavithra Ashok Kumar]

One of the ways to find this would be to do an element-wise ratio and check if the multiple is constant. The same can also be achieved by using the cross product of the vectors.

 for i = 1:ncols
    b = A(:,i)./X;   %Do element-wise division
    if((b(1)*ones(nrows,1)) == b) % check if it is a constant
          disp(i);  % This would give the indices of the dependent columns 
    end
 end

However, If you could give more information on the use-case, the community might be able to suggest better alternatives. Hope this helps.

Answer 2

[Stephen23]

Guessing a bit because the question is not totally clear... here is a function that returns the column indices of any columns that are integer multiples of the input column vector:

>> m = [1,2,3,4;1,5,3,6;1,7,3,8];
>> fun = @(c)all(~diff(bsxfun(@rdivide,m,c)));ú
>> fun([1;1;1])
ans =
     1     0     1     0
>> fun([2;3;4])
ans =
     0     0     0     1
>> fun([2;5;7])
ans =
     0     1     0     0

And you can extract those columns by using these indices:

>> m(:,fun([2;3;4]))
ans =
     4
     6
     8
>> m(:,fun([1;1;1]))
ans =
     1     3
     1     3
     1     3