Find the set of eigenvectors of a 4x4 matrix elements whose matrix elements have some VARIABLE PARAMETERS.

Below is how I defined the matrix. It is a 4x4 matrix with variables B and N on the off diagonal. Since each of these variable takes certain values in a given range, my goal is to read and display the eigenvectors for each of the values in the range of the variables.

    delta=45;
    %create a 4x4 zeros matrix
    mat=zeros(4,4);
    %set values for corresponding entries of the matrix
    mat(1,1)= delta;
    mat(1,2)= 0 ; 
    mat(1,3)=0;
    mat(2,1)=0 ;
    mat(2,2)= delta; 
    mat(2,4)=0;
    mat(3,1)=0; 
    mat(3,3)= -delta; 
    mat(3,4)= 0 ;
    mat(4,2)=0; 
    mat(4,3)=0 ; 
    mat(4,4)= -delta;
    
   %define the off diagonal elements of the matrix
    for N = 0:1:5
        for B = 0:0.001:10
            mat(1,4)=sqrt(750*B*N);
            mat(2,3)=sqrt(750*B*N);
            mat(3,2)=sqrt(750*B*N);
            mat(4,1)=sqrt(750*B*N);
            
        end
    end

[John D'Errico]

Please don't ask this same question any more times.

[Muhsin]

My bad!

I realized I could edit the original question after I did it.

Answer 1

[Jon]

If I understand what you are trying to do I think this should do what you want.

The resulting values of the eigenvector for each value of the parameter B*N will be stored in the columns of matrix V

    delta=45;

    %set values for corresponding diagonal entries of the matrix
    mat = diag([delta,delta,-delta,-delta]);

   % define vectors of variable parameters
   N = 0:1:5;
   B = 0:0.001:10;
 

   % matrix is parameterized by product B*N, so just compute for unique
   % values of this parameter
   BN = unique(N'*B);
   %define the off diagonal elements of the matrix and compute the
   %eigenvectors
   numParam = numel(BN); % number of parameter values to be evaluated
   V = zeros(4,numParam); % preallocate array to hold eigenvectors
   for k = 1:numParam
       a = sqrt(750*BN(k));
       mat(1,4)=a;
       mat(2,3)=a;
       mat(3,2)=a;
       mat(4,1)=a;
       V(:,k) = eig(mat);
   end

[Muhsin]

V(:,k) = eig(mat);

The ouput of the code isn't right. I think part of the issues has to do with this line. I want to believe it doesn't return the right eigenvectors. Doesn't "eig(mat)" return the eigenvalues instead??

[Jon]

Yes, good catch. The syntax seems a little strange with this, e = eig(A) returns a vector of eigen values, and so does [e] = eig(A), but [V,~]=eig(A) returns the matrix of eigenvectors. So even though in both cases we look at the first returned argument, the definition changes. So you must use the two argument calling syntax to get the eigenvectors.

This also means you need to store a matrix of Eigenvectors for each value of B*N, here is some modified code where I store the matrices of eigenvectors in a 3d array (each page is for a value of B*N).

So if you wanted look at the eigenvectors corresponding to BN(3) you would look at V(:,:,3)

    delta=45;

    %set values for corresponding diagonal entries of the matrix
    mat = diag([delta,delta,-delta,-delta])

   % define vectors of variable parameters
   N = 0:1:5;
   B = 0:0.001:10;
 

   % matrix is parameterized by product B*N, so just compute for unique
   % values of this parameter
   BN = unique(N'*B);
   %define the off diagonal elements of the matrix and compute the
   %eigenvectors
   numParam = numel(BN); % number of parameter values to be evaluated
   V = zeros(4,4,numParam); % preallocate array to hold eigenvectors
   for k = 1:numParam
       a = sqrt(750*BN(k));
       mat(1,4)=a;
       mat(2,3)=a;
       mat(3,2)=a;
       mat(4,1)=a;
       [V(:,:,k),~] = eig(mat);
   end

[Muhsin]

Thanks.

[V(:,:,k),~] = eig(mat); is a significant improvement. It, returns a set of eigenvectors but they are the same eigenvectors repeated. Some of the values of the eigenvectors are controlled by B*N, hence the eigenvectors shouldn't be exactly the same.

Answer 2

[Torsten]

I set a = sqrt(750*B*N) in the below code. So for every combination of B and N, it gives you the eigenvalues (diagonal of D) and eigenvectors (columns of V) of your matrix "mat".

syms delta a real
mat = delta*sym(eye(4));
mat(3,3) = -mat(3,3);
mat(4,4) = -mat(4,4);
mat(1,4) = a;
mat(4,1) = a;
mat(3,2) = a;
mat(2,3) = a;
[V,D] = eig(mat)

V = 

D = 

simplify(mat*V-V*D)

ans = 

[Muhsin]

B and N are range of values and not fixed. The problem here is to loop to the possible BN combinations and find the eigenvectors for each combination. Your code here doesn't show how.

[Torsten]

The eigenvectors don't depend on BN combinations as you can see.

They are fixed as V does not depend on a.

I wrote that the columns of V are the eigenvectors, didn't I ?

[Jon]

@TorstenIt seems that when I explicitly loop through the BN combinations (code in my answer) the eigenvectors the eigenvectors do seem to depend upon "a" (the value of B*N). Did I do something wrong or miss something?

[Torsten]

I misread

mat = diag([delta,delta,delta,delta]);

instead of

mat = diag([delta,delta,-delta,-delta]);

But nevertheless, after incorporating the changes in the symbolic computation (see above), the results are still easy to implement for the numerical case.

Answer 3

[Walter Roberson]

I showed you in https://www.mathworks.com/matlabcentral/answers/1848378-how-to-read-the-eigenvectors-of-a-4x4-matrix-elements-for-a-given-range-of-variable-elements-b-and-n#answer_1096903 how to loop over the elements of B and record the values properly.