[FIXED] How do I make an increasing minimum index?

Issue

let's consider this is the intended matrix; I need to find the location of the minimum in each column, but there is a condition that the ind(min1) < ind(min2) < ind(min3) < ..... <ind(minN)

where IND is the index so in another word is the location of the minimum. This matrix is okay, but I want to make an automatized process for future use.

[[  0.91018066   4.48975932   2.09120921]
 [  0.99666765   6.20469762   4.39193871]
 [255.09412294   1.0510132    3.18827888]
 [ 10.04494426   1.02542159   2.41095802]
 [  1.10101048   7.75462488   1.55075049]
 [  1.16764376   1.56933637   2.33618634]
 [  0.67289877   4.27517897   2.53459737]
 [  4.82191354   0.717983     3.14042209]
 [  0.35376553   3.01928107   2.05515314]
 [ 11.29192073   0.4509206    4.48405362]
 [  2.77928849   2.57319267   0.44103186]
 [  1.53813319   3.11950628   4.82088164]
 [  1.67829804   2.40947038   5.49557038]
 [ 69.7214563    2.96405783   4.52767726]
 [  1.97205265   4.57095239   8.45765019]
 [ 13.12256504   1.45513654  35.58973711]
 [  8.96582583   7.37919581   1.63986824]
 [  2.01843202   3.73197058   1.8612331 ]
 [  6.54770837   2.43552378   6.16662901]
 [  6.34760671   2.42883926   6.23632993]
 [ 12.1397713    2.27797639   4.18868256]
 [  5.58218562   2.82203753   3.41404581]
 [  3.66370152   2.97445241   3.34320906]
 [  2.56132298   2.81115595   1.13722405]
 [  4.302606     3.24668351   3.81259428]
 [  3.3940013    2.32417808  38.78896462]
 [ 72.86244959   3.75110755   6.08713172]
 [ 17.04329424   1.49873191   3.69710865]
 [  2.13217024  13.5293613    4.24590484]
 [  4.11955718   2.60752147  95.89256472]]

Solution

I think the example is a bit odd if i understand your question correctly. It seems that you want to find the minimum of the whole first column (index 9, value 0.35376553) and then the minimum of column 2 index 9:end (index 10, value 0.4509206) etc. So it seems, that your indices are 9, 10 and 11:

I called you matrix M:

minimumRow = 1;
for iColumn = 1:size(M,2)
    [a,iRow]=min(M(minimumRow:end,iColumn));
    minimumRow = minimumRow + iRow -1
end

Answered By - Solstad