how to make symmetric matrix with vector of its non repeated elements?

suppose we have non repeated elements of a matrix in a vector . I need to make a symmetric matrix by using this vector in matlab.

for example if we have :

v= [ 1 4 5 6 9 0]

the answer must be the matrix:

1 4 5

4 6 9

5 9 0

[Luca Ferro]

could you better define the rules of the reshape?

Answer 1

[Jan]

v = [1 4 5 6 9 0];

d = numel(v);
n = (sqrt(8 * d + 1) - 1) / 2;  % Number of rows and colums
A = zeros(n, n);
A(tril(true(n, n))) = v;
A(triu(true(n, n))) = v   % [EDITED, faster than transposing the mask]

A = 3×3

1 4 6 4 5 9 5 9 0

Or with a loop (assumably faster for large arrays):

k = 0;
A = zeros(n, n);
for i2 = 1:n
    k         = k + 1;     % Diagonal element
    A(i2, i2) = v(k);    
    for i1 = i2 + 1:n      % Below and above diagonal
        k         = k + 1;
        A(i1, i2) = v(k);
        A(i2, i1) = v(k);
    end
end
A

A = 3×3

1 4 5 4 6 9 5 9 0

[Jan]

Some speed comparisons:

tic
for k = 1:1e3
    n = (k * (k+1)) / 2;
    A = ToSymMatrix_vec(1:n);
end
toc

Elapsed time is 1.327950 seconds.

tic
for k = 1:1e3
    n = (k * (k+1)) / 2;
    A = ToSymMatrix_loop(1:n);
end
toc

Elapsed time is 0.839996 seconds.

function A = ToSymMatrix_vec(v)
d      = numel(v);
n      = (sqrt(8 * d + 1) - 1) / 2;  % Number of rows and colums
A      = zeros(n, n);
A(tril(true(n, n))) = v;
A(triu(true(n, n))) = v;
end

function A = ToSymMatrix_loop(v)
d = numel(v);
n = (sqrt(8 * d + 1) - 1) / 2;  % Number of rows and colums
k = 0;
A = zeros(n, n);
for i2 = 1:n
    k         = k + 1;     % Diagonal element
    A(i2, i2) = v(k);    
    for i1 = i2 + 1:n      % Below and above diagonal
        k         = k + 1;
        A(i1, i2) = v(k);
        A(i2, i1) = v(k);
    end
end
end

[Isabelle Davies]

Heads up: The first method does not return the correct symmetric matrix A, as the elements in row 2, column 2 and row 1, column 3 have been switched.

[Stephen23]

@Isabelle Davies: you are right.

It is because the order of the elements in v are not suitable for allocating into the upper triangle.

One solution:

v = [1,4,5,6,9,0];

n = (sqrt(8*numel(v)+1)-1)/2; % must be whole
A = zeros(n,n);
A(tril(~A)) = v;
A = A+triu(A.',1)

A = 3×3

1 4 5 4 6 9 5 9 0

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Answer 2

[Pratheek]

The first line of the code is to input for the size of the matrix that you want to generate.

% ask the user for the size of the matrix
n = input('Enter the size of the square matrix: ');

% get the input vector
v = [1 4 5 6 9 0]; % or prompt the user to enter the vector as well

% create the symmetric matrix
M = zeros(n);
k = 1;
for i = 1:n
    for j = (i+1):n
        M(i,j) = v(k);
        M(j,i) = v(k);
        M(i,i) = v(randi(length(v)));
        k = k + 1;
    end
end

% display the resulting matrix
disp(M);

[Jan]

Why do you set the diagonal elements randomly and repeatedly in each iteration of the inner loop?