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# rand matrix for FM

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Offroad Jeep on 31 Mar 2015
Closed: MATLAB Answer Bot on 20 Aug 2021
Hi all,I want to generate a random matrix so that the sum of all the elements is zero. I mean the random numbers are so selected that the total sum of the matrix goes to zero.Regards. Early reply will be highly appreciated.
for example
a = rand(3)
a =
0.3000 0.5000 0.8000
-0.1000 -0.4000 -0.6000
-0.4000 0.8000 -0.9000
>> sum(sum(a))
ans =
0
Stephen23 on 31 Mar 2015
Integer values only, or with decimal fraction?
Offroad Jeep on 31 Mar 2015
Both for integer and decimals.............

Roger Stafford on 31 Mar 2015
Edited: James Tursa on 1 Apr 2015
If your x values are subject to common upper and lower bounds, you can use my 'randfixedsum' function in the File Exchange, located at:
It is designed to give a uniform distribution on the hyperplane of values satisfying the condition of a predetermined sum - in your case a sum of zero.
John D'Errico on 31 Mar 2015
And of course, this is the best answer.

Zoltán Csáti on 31 Mar 2015
I recommend you to generate the matrix of the required size and then modify one element of it so that the sum holds. E.g.
A = rand(3);
totalSum = sum(sum(A));
A(end,end) = A(end,end) - totalSum;
Then the sum will give you zero, aside from the round-off error.
John D'Errico on 31 Mar 2015
Edited: John D'Errico on 31 Mar 2015
It depends on how you want the elements themselves to be distributed. See that if they should originally be bounded in the interval [-1,1], then by the final shift, they often will no long be so bounded.
For example...
A = rand(3) *2 - 1;
A = A - sum(A(:))/numel(A)
A =
0.53571 -0.78404 0.51328
-0.81271 -0.68212 -0.24218
-0.32253 0.77053 1.0241
sum(A(:))
ans =
-2.2204e-16
See that while the sum is now zero, that now one of the elements actually exceeded 1, even though the original elements fell inside [-1,1].
Zoltán Csáti on 1 Apr 2015
Yes, you are right I didn't think of this aspect.

Brendan Hamm on 31 Mar 2015
How about you create a random matrix and then subtract from each element the sum(matrix(:))/numel(matrix).
n = 4;
A = rand(4);
s = sum(A(:))/numel(A);
A = A - s;
sum(A(:))
Roger Stafford on 1 Apr 2015
The vagueness has to do with how you expect the set of vectors for which the sum has a given value, to be statistically distributed. Consider each vector as a point in n-dimensional vector space. What kind of statistical density function do you desire for the hyperplane therein where the vector sums are a given value? Is it to be uniform, gaussian, etc., and is it bounded?
Brendan Hamm on 1 Apr 2015
They are currently U([-c c]) where c i s the sum of the originally sampled elements divided by the number of elements. You want them to be U([-1 1]), then just divide by the magnitude of the largest resulting element now:
A = A / max(abs(A(:)));