On Wed, 2 Jul 2014, Riccardo (Jack) Lucchetti wrote: > On Wed, 2 Jul 2014, Sven Schreiber wrote: > >>> The problem is unlikely to be the rank of your contraint matrix. If you >>> fill a 9x9 matrix with 36 zeros at random, the probability if it being >>> singular is zero (it shouldn't be difficult to prove this formally) >>> although of course the event is not impossible (nice example of an event >>> with 0 Lebesgue measure). >> >> Jack, I don't think this is true, for example the probability of getting >> a column with all (9) zeros by chance may be small, but certainly not >> zero if you allocate 36 zeros randomly. If I were good in combinatorics >> I could give you the exact numbers, but I'm not. >> >> Apart from that the constraints are not chosen at random. A line has >> measure zero in a plane, but still I can pick exactly points on the line. >> >> So I'm not saying the matrix is singular, but without further >> investigation/thinking I cannot rule out that it is. > > You're absolutely right, my friend. After all, the number of ways you can put > n*(n-1)/2 zeros into a square matrix with n rows is finite, and some of those > do give rise to singular matrices, so yes, it isn't a Lebesgue 0. I stand > corrected.
I believe Prob(one column is all zeros) = 9 * (1/9)^9 = 2.3e-08 when inserting 9 zeros at random; it would be higher for 36 zeros. Allin
