How to create a symmetric random matrix?
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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
2 Comments
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.
Accepted Answer
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'.
3 Comments
Alaa
on 3 Sep 2023
What if the matrix has definite numbers? (i,e., numbers are not randomly generated)
How can one develop a symmetric matrix (given that the diagonals will be zeros)
More Answers (4)
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.
2 Comments
Irfan Ahmed
on 11 Apr 2020
I think this should not be element-wise multiplication, instead, it should be A*A'
Antong Cheng
on 3 May 2022
The problem is that this only generates positive semi-definite matrices, so it's unfit for certain scenarios.
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.
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
2 Comments
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