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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.

David Goodmanson
on 5 Jan 2021

Edited: David Goodmanson
on 5 Jan 2021

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.

David Goodmanson
on 7 Jan 2021

For

[vA lamA] = eig(A)

[vB lamB] = eig(B)

then the columns of vA are the eigenvectors (evecs). Same for vB. You want to find if any column of A is the same as some column of B. All the evecs are normalized to 1 but unfortunately, an evec can be multiplied by a phase factor exp(i*theta) (which includes the phase factor -1) and still be an evec. No guarantee what eig produces, so an evec of A and an evec of B can differ by a phase factor. So you have a couple of choices.

[1] compare each evec eA of A with each evec eB of B, (n^2 cases), determine the phase factor between them, take out the phase factor, look at the difference eA-eB and decide if it's small enough. But unless both eA and eB are real, the phase factor issue is harder than it looks.

[2]

Use the fact that if eA and eB are almost equal, then since they are normalized their dot product eA'*eB will close to 1. Here eA' turns column vector to row vector; and row vector times column vector eB is the scalar dot product. So you need the transpose. Multiplying the matrix vA' by the matrix vB automatically finds all n^2 possible dot products of a column of A with a column of B and you can search the resulting matrix for values near 1.

As far as tolererances, you have to decide for yourself what is appropriate. In this case 1e-2 gives 1 equality, and as you point out, 1e-1 gives 4 equalites. If you use 1e-12 then nothing is equal to anything, and if you use 1e1 then everything equals everything else. Maybe your data is imprecise enough that there should not be any equalities. It's a judgment call.

I'm not sure what to make of null(FISH_sp*FISH_xc-FISH_xc*FISH_sp) having a nonempty null space.

Bruno Luong
on 7 Jan 2021

null(FISH_sp*FISH_xc-FISH_xc*FISH_sp)

returns the basis of subspace where the two linear operator commutes when restricted on the subspace.

As example I(n) and 2*I(n) (or any matrix)

commutes, the above NULL returns n vectors, yet they the operators not share any eigen vector.

Bruno Luong
on 4 Jan 2021

Edited: Bruno Luong
on 4 Jan 2021

K = null(A-B);

[W,D] = eig(K'*A*K);

X = K*W, % common eigen vectors

lambda = diag(D), % common vector

Bruno Luong
on 4 Jan 2021

Quote: " I just want Common eigen vectors" meaning

AX = BX

This is equivalent to

(A-B)*X = 0

Therefore

X belongs to span of NULL(A-B)

If NULL returns empty result then there is NO common eigen vector.

This also implies eigen values are common, this is a consequence of YOUR request, not because I add it as an extra requirement. If you don't understand it you do not understand the math logic.

Might be you redefine the meaning of word "common"? If you do then right I don't understand what you want.

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