How to create a symmetric random matrix?

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Muhammad Shafique on 29 Mar 2014
Commented: Irfan Ahmed on 11 Apr 2020
Hello,
I need to create a random matrix meeting the following conditions:
- The values on the main diagonal are between a given range (e.g., 0 to 1000000)
- Each value on the diagonal is randomly distributed/spread in its corresponding row and column vectors.
- The matrix is symmetric (that is to say, corresponding values in upper and lower triangles are the same)
Any help will be highly appreciated.
Best,
M

Roger Stafford on 30 Mar 2014
Muhammad, could you please expound at greater length on that second condition: "randomly distributed/spread in its corresponding row and column vectors"? What exactly does that mean? Is that a condition on the sum of the elements in the corresponding row and column, and if so, just what is the condition? What does "randomly distributed/spread" mean in this context? Perhaps a short example of what you have in mind would help.
Muhammad Shafique on 30 Mar 2014
Roger, thanks for the question. I meant to say that the values on the row and column must be between 0 and the value on the diagonal. For instance, a random value is chosen within the given range for any element on the diagonal and this value becomes the upper bound of the range for random number generation for the corresponding row/column. This also means that any value in the triangle must not be larger than any of the values on the diagonal that are incident to it.

Roger Stafford on 30 Mar 2014
Let the random matrix to be generated be called M and its size be NxN.
d = 1000000*rand(N,1); % The diagonal values
t = triu(bsxfun(@min,d,d.').*rand(N),1); % The upper trianglar random values
M = diag(d)+t+t.'; % Put them together in a symmetric matrix
If you want whole numbers, apply the 'floor' function to 'd' and then after computing 't', apply it to 't'.

Muhammad Shafique on 30 Mar 2014
Thank you very much, Roger. That is what I needed. It was great help!
Youssef Khmou on 30 Mar 2014
efficient proposition.

Brian Crafton on 3 Dec 2018
Just came up with this gem and wanted to share it :
A = rand(4)
A .* A'
This will generate a random 4x4 matrix and its clear why.

1 Comment

Irfan Ahmed on 11 Apr 2020
I think this should not be element-wise multiplication, instead, it should be A*A'

Walter Roberson on 29 Mar 2014
When the matrix A is square, (A + A')/2 is symmetric (and positive definite)

1 Comment

John D'Errico on 22 Mar 2019
Actually, the statement shown here is incorrect. Given a square matrix A, (A+A')/2, MAY be positive definiite. But there is no such requirement. For example:
A = randn(4);
As = (A + A')/2;
eig(As)
ans =
-1.9167
-1.6044
-0.37354
2.1428
As is symmetric always. But there is no requirement that it is SPD. As you see, it had 3 negative eigenvalues in this simple example.
Even if rand had been used to generate the matrix, instead of randn, there would still be no assurance the result is positive definite. A counter-example for that took me only one try too.
A = rand(4);
eig((A + A')/2)
ans =
-0.32868
0.088791
0.32729
1.9184
The symmetric computation shown will insure only that the eigenvalues are real. Positive definite requires positivity of the eigenvalues.

Youssef Khmou on 30 Mar 2014
the random matrix is generated using the following :
N=500;
M=rand(N);
M=0.5*(M+M');
L=100; % magnitude
for n=1:N
M(n,n)=L*rand;
end

1 Comment

Muhammad Shafique on 30 Mar 2014
Youssef, thank you very much for taking the time to help. I'm wondering whether it is possible to get the output in whole numbers.

mike will on 22 Mar 2019
This is the solution:
A = rand(4, 4)
A_symmetric = tril(A) + triu(A', 1)
Where A will be a square matrix, and
tril(A)
returns lower triangular part of matrix A, and
triu(A', 1)
returns upper triangular part of matrix transpose(A).

1 Comment

John D'Errico on 22 Mar 2019
To be pedantic, Mike has shown ONE solution. Not THE solution. It has different numerical properties from the other solutions shown.