I know that the eigenvectors produced by eig(A) have 2-norm 1. But what about the vectors produced in the generalized eigenvalue problem eig(A,B)? A natural conjecture is that such a vector v should satisfy v'Bv=1. When B is the identity matrix, then v'Bv is exactly the square of the 2-norm. I ran the following test for various matrices A and B:
[p,d]=eig(A,B);
v=p(:,1);
v'*B*v
I always choose B to be diagonal. I noticed that v'Bv is not always 1. However, it is indeed 1 when A is symmetric. Does anyone know the rule for the way that Matlab normalizes the generalized eigenvectors? I can't find it in the document.
According to the documentation (emphasis mine):
The form and normalization of V depends on the combination of input arguments:
[...]
[V,D] = eig(A,B) and [V,D] = eig(A,B,algorithm) returns V as a matrix whose columns are the generalized right eigenvectors that satisfy A*V = B*V*D. The 2-norm of each eigenvector is not necessarily 1. In this case, D contains the generalized eigenvalues of the pair, (A,B), along the main diagonal.
When eig uses the 'chol' algorithm with symmetric (Hermitian) A and symmetric (Hermitian) positive definite B, it normalizes the eigenvectors in V so that the B-norm of each is 1.
This means that, unless you are using the 'chol' algorithm, V is not normalized.
If I get you correctly, you are looking for a way to generalize a vector then given a vector you can divide it by its norm to obtain a secondary vector whose norm is 1.
If you are looking for the mathematical background, then Eigendecomposition of a matrix contains a good introduction.
Related
When I get the eigenvalues of the diagonal of a PCA transformed image, I always get 1, whatever the image. What's the reason behind this?
I used the following code.
coeff = pca(pmap);
disp(coeff);
[V,L]=eig (coeff'*coeff);
Lamda = diag(L);
disp(Lamda);
The coeff which pca outputs are already eigenvectors, which are all orthogonal. They are even orthonormal, since MATLAB normalises them. Relative weight is in the explained output parameter of pca.
So transpose(coeff)*coeff gives you the identity matrix, which just contains ones and the eigenvectors of the identity matrix are, obviously, all just 1 in a single dimension.
The reason is thus because that's how linear algebra works.
In Matlab, I'm trying to solve for the energies and eigenstates of a Hamiltonian matrix which has a highly degenerate set of eigenvectors. The matrix is a 55x55 hermitian matrix, and when I call either eig or schur to do the diagonalization I find that some (but not all) of the eigenvectors are the "wrong" linear combinations within each degenerate subspace. What I mean by "wrong" is that there are additional constraints in the problem. In this case, there is a good quantum number, M, which I want to preserve by not allowing states with different M values to be mixed--- but that mixing is exactly what I see when I run the code. Is there a way to tell Matlab to diagonalize the matrix while simultaneously maintaining the eigenvectors of another operator?
you can use diag to diagonalize a matrix and [eig_vect,eig_val] = eig(A) to give you eigenvectors.
I don't know matlab well enough to know whether there is a routine for this this, but here's how to do it algorithmically:
First diagonalise H, as you do now. Then for each degenerate eigen-space V, diagonalise the restriction of C to V, and use this diagonalisation to compute simulaneous diagonalisations of C and H
In more detail:
I assume you have an operator C that commutes with your Hamiltonian H. If V is the eigen-space of H for a particular (degenerate) eigen value, and you have a basis x[1] .. x[n] of V , then for each i, Cx[i] must be in V, and so we can expand Cx[i] in terms of the x[], and get a matrix representation C^ of the restriction of C to V, that is we compute
C^[k,j] = <x[k]|C*x[j]> k,j =1 .. n
Diagonalise the matrix C^, getting
C^ = U*D*U*
Then for each row (r1,..rn) of U* we can form
chi = Sum{ j | r[j]*x[j]}
A little algebra shows that this is an eigenvector of C, and also of H
When using eig function in Matlab, it seems that this function has already normalize the values of the eigenvalues. Do we need to write some lines of code to normalize the eigenvalues after using the eig function.
The function eig in MATLAB normalizes the eigenvectors (not the eigenvalues).
See the following from the documentation:
[V,D] = eig(A) returns matrix V, whose columns are the right
eigenvectors of A such that AV = VD. The eigenvectors in V are
normalized so that the 2-norm of each is 1.
Eigenvectors can vary by a scalar, so a computation algorithm has to choose a particular scaled value of an eigenvector to show you. eig chooses 2-norm = 1. Just look at the eigenvector definition to see why: AV=VD. V shows up on both sides, so you can multiple V by anything without affecting the equation.
Eigenvalues do not vary. Look again at AV=VD. D is only on one side, so it can't be scaled.
Does anybody know how the sign of the singular vectors resulting from Matlab's svd function is determined?
Let:
B = U*S*V'
be a valid svd decomposition of a real or complex 2-by-2 matrix B, then:
B = (U*c)*S *(V*c)'
is also valid, where c is a matrix that changes the sign of one or both singular vectors:
c = diag([1 -1]), diag([-1 1]) or diag([-1 -1]).
I want to know how Matlab's svd algorithm determines the sign of the singular vectors in U and V.
Matlab uses LAPACK's DGESVD implementation for singular value decomposition, which doesn't take into account direction of the resulting vectors. In applications, when SVD is performed, decomposed data is processed and then data is reconstructed back signs make no difference. They became only important, when decomposed data is being analyzed.
One might apply sign correction algorithm after performing SVD with Matlab. But I believe sign correction depends on actual meaning of the data.
In the paper you provided direction is chosen to be the same as the most of the data points. This won't work for data with symmetrical distribution as theoretical direction is zero and sample direction will be random resulting in high numerical instability.
If the goal is just to have numerical stability of the solution, then it would be enough to choose some vector and change all SVD vectors to lie in the same half-space with it.
I am using LAPACK's ssteqr function to calculate eigenvalues/eigenvectors. The documentation for ssteqr says that the eigenvalues are sorted "in ascending order". Is it reasonable to assume that the list of eigenvectors is also sorted in ascending order?
Yes, it is reasonable to assume that the eigenvectors are ordered so that the i-th eigenvector corresponds to the i-th eigenvalue.
Nevertheless, if I were you, I would check for each eigenvalue the result of the multiplication of the eigenvector by the matrix. This way you are sure that you interpret the output right, and you see explicitly the accuracy of the calculations.
An old question, this, but I struggled with this recently, so am adding this for current and future readers.
The basic answer is that, yes, the eigenvectors are sorted such that the ith eigenvector corresponds to the ith eigenvalue. However, note that the eigenvectors thus obtained may not be the actual eigenvectors you want. This is so because of the following.
Since the ?steqr functions work only on tridiagonal matrices, one typically uses LAPACK's ?sytrd functions to first transform one's original symmetric matrix, call it M, to a tridiagonal form, call it T, such that M = QTQT where Q is an orthogonal matrix (and QT denotes its transpose). One then applies the ?steqr function on this tridiagonal matrix T to find its eigenvalues and eigenvectors. Now the eigenvalues thus obtained (of T) are exactly the same as the eigenvalues of M, so if one only wants the eigenvalues one can stop here. But if one is interested in the eigenvectors, like the OP, then one needs to bear in mind that the eigenvectors of T and M are different. To find the eigenvectors of the original matrix M, one needs to left-multiply the obtained eigenvectors of T by Q. This is very easily done by using the LAPACK functions orgtr or ormtr. See here for a clear explanation: https://software.intel.com/en-us/mkl-developer-reference-fortran-sytrd.