MATLAB: Find the common eigenvectors and eigenvalues between 2 matrices

diagonlisationeigenvalueeigenvectorinversematrixmatrix manipulation

Hello,
I am looking for finding or rather building common eigenvectors matrix X between 2 matrices A and B such as :
   AX=aX with "a" the diagonal matrix corresponding to the eigenvalues
    
   BX=bX with "b" the diagonal matrix corresponding to the eigenvalues
I took a look in a similar post topic https://stackoverflow.com/questions/56584609/finding-the-common-eigenvectors-of-two-matrices but had not managed to conclude, i.e having valid results when I build the final wanted endomorphism F defined by : F = P D P^-1
I have also read the wikipedia topic on Wikipedia and this interesting paper https://core.ac.uk/download/pdf/82814554.pdf but couldn't have to extract methods pretty easy to implement.
Particularly, I am interested by the 
eig(A,B)
Matlab function.
I tried to use it like this :
    % Search for common build eigen vectors between FISH_sp and FISH_xc
    [V,D] = eig(FISH_sp,FISH_xc);
    % Diagonalize the matrix (B^-1 A) to compute Lambda since we have AX=Lambda B X
    [eigenv, eigen_final] = eig(inv(FISH_xc)*FISH_sp);
    % Compute the final endomorphism : F = P D P^-1
    FISH_final = V*eye(7).*eigen_final*inv(V)
But the matrix `FISH_final` don't give good results since I can do other computations from this matrix FISH_final (this is actually a Fisher matrix) and the results of these computations are not valid.
So surely, I must have done an error in my code snippet above.
If someone could help me to build these common eigenvectors and finding also the eigenvalues associated, this would be fine to tell it, I am a little lost between all the potential methods that exist to carry it out.

Best Answer

Hi petit,
Eigenvectors calculated by Matlab are normalized, but neither (a) the the overall phase of each one or (b) the order ot the eigenvalues and the corresponding columns of the eigenvectors are guaranted to be anything in particular. But if AX = aX and BX = bX, then for
[vA lambdaA] = eig(A)
[vB lambdaB] = eig(B)
there will be a case where a column of Va and a column vB differ by only a phase factor. So in the matrix product vA'*vB there be an entry of absolute value 1, the phase factor.
n = 6;
% set up matrics A and B with two eigenvectors in common
v = 2*rand(n,n)-1 +i*(2*rand(n,n)-1);   % eigenvalues
w = 2*rand(n,n)-1 +i*(2*rand(n,n)-1);
lamv = rand(n,1);                       % eigenvectors
lamw = rand(n,1);
% two common eigenvectors
w(:,2) = v(:,3);
w(:,4) = v(:,5);
A = (v*diag(lamv))/v;
B = (w*diag(lamw))/w;
% given A and B, find the common eigenvectors
[vA lamA] = eig(A);
[vB lamB] = eig(B);
vAvB = vA'*vB;
[j k] = find(abs(abs(vAvB)-1)<1e-12)
% show that the jth A eigenvector and kth B eigenvector are proportional 
vA(:,j(1))./vB(:,k(1))
vA(:,j(2))./vB(:,k(2))
This method works as long as for the eigenvectors in question, the eigenvalues are distinct. If there are repeated eigenvalues, then if A has eigenvectors x and y, B might have eigenvectors that are linear combinartions of x and y. Then the job gets a lot harder.
You can also use the fact that
(A-B)X = (a-b)X
to look for cases where an eigenvalue of (A-B) equals (a-b), where a is one of the eigenvalues of A and B is one of the eigenvalues of B. However, this method is likely to be more prone to false positives than is the first method.
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