## rand matrix for FM

on 31 Mar 2015
Latest activity Edited by James Tursa

### James Tursa (view profile)

on 1 Apr 2015
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

Stephen Cobeldick

### Stephen Cobeldick (view profile)

on 31 Mar 2015
Integer values only, or with decimal fraction?

on 31 Mar 2015
Both for integer and decimals.............

### Products ### Roger Stafford (view profile)

on 31 Mar 2015
Edited by James Tursa

### James Tursa (view profile)

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

### John D'Errico (view profile)

on 31 Mar 2015
And of course, this is the best answer.

### Zoltán Csáti (view profile)

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

### John D'Errico (view profile)

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

### Zoltán Csáti (view profile)

on 1 Apr 2015
Yes, you are right I didn't think of this aspect.

### Brendan Hamm (view profile)

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(:))

on 31 Mar 2015
Brendan,,,,,,,,, can you please tell how I can put the condition............
Roger Stafford

### Roger Stafford (view profile)

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

### Brendan Hamm (view profile)

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(:)));