Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a specific Generalized Eigenvalue Problem (GEVP) where i am primary not interested in solving this problem but concluding from a standard EVP the spectrum of the GEVP.

The Problem
Let $A$ be a $n\times x$ possibly complex matrix and $B$ a diagonal, real $n\times n$ matrix with maximal rank of $n-1$ (e.g. the matrix $B$ has at minimum 1 zero column and row).
Solving

$(B\lambda-A)\cdot v=0$

with $|v|=1$, so that we have $n+1$ equations for $n+1$ unknown is the GEVP. The GEVP can not be reformulated as EVP because $det(B)=0$ and therefore $B$ is not invertible.

As I said the goal is not just solving this problem (this could be done by solving $det(\lambda B I-A)=0$ to obtain the eigenvalues) but to conclude eigenvalues for the stated GEVP from the following, already solved, EVP (the $n$ eigenvalues $\mu_1\leq\mu_2\leq\dots\leq \mu_n$ of $A$ are known):

$(I\mu-A)\cdot w=0$.

What I have already learned
*As $A$ and $B$ in general do not commutate it is not possible to diagonalize $A$ and $B$ simultanously. Therefore the spectra will be different.
*If the EVP results in eigenvalues $\mu=0$, so there will be the same number of eigenvalues $\lambda=0$ in the GEVP. (Because in both cases $det(A)=0$ must be fullfilled and the geometric multiplicity comes from the dimension of $kern(A)$.)
*For every zero-row in $B$ the number of eigenvalues $\mu$ is one less then in the EVP. This is because the order of the characteristic polynom (CP) goes one down for every zero-row in $B$ compared to the order of the CP in the EVP.

Questions
*Can be said which eigenvalues (in addition to the zeros) of the EVP are also eigenvalues of the GEVP (the eigenvectors may not be the same in both cases, but the eigenvalues).
*Is there a pertubation theory? Can I somehow make a taylor series of the CP in the GEVP where the zeroth-term is the CP of the EVP?
*The number of eigenvalues in the GEVP is less than in the EVP, can be concluded which eigenvalues vanish?


In case anybody wants to know, where my question emerges from (this is not essential for my questions but possibly from general interest):

If one wants to conclude the stability of a fix point $x^*$ of ODEs one needs to solve the variational ODE $\dot{\delta x}(t)=D_xf(x^*)\delta x(t)$. Where $\delta x$ is a small pertubation away from the fix point: $\delta x(t)=x(t)-x^*$. Solving this with $\delta x(t)=\delta x_0 e^{\mu t}$ results in the EVP
$\mu\delta x_0 = D_x f(x^*) \delta x_0$.
Using $D_xf(x^*)=A$ and $w=\delta x_0$ results in the stated EVP.

If one has additional constraints in a implicit way
$g(x(t))=0$
the stability of a fix point in the ODE may change (e.g. the constraint acts in a unstable direction. The eigenvalue of $A$ in this direction is still greater zero (obviously the matrix $A$ does not change if constraints are imposed) but it is a "forbidden" direction as the corrisponding eigenvector is in a direction which is not allowed due to the constraint).
Taking the time derivative of $g$ results in $D_x g(x)\cdot \dot{x}(t)=0$. Inserting the pertubation away from the fix point results in
$(D_xg(x^*)+D_x(D_xg(x^*))\cdot \delta x)\cdot \dot{\delta x}(t)+\dots=0$
$D_xg(x^*)\cdot \dot{\delta x}(t)+O(\delta x^2)=0$
Inserting the exp-ansatz results in
$D_xg(x^*)\cdot \delta x_0\mu\approx0$
This means that the small pertubations need to be orthogonal to the gradient on the invariant manifold near the fix point (e.g. they are inside the invariant manifold).

One possible way to go to study the change of stability of the fix point when constraints are imposed, is to solve the EVP and then to check consistency with the last equation.
I want to include the constraint directly in the EVP, which leads to the GEVP by simply adding the last equation to the EVP (with $D_x g(x^*)=\hat{B}$ and $w=\delta x_0$):
$(\hat{B}\mu+I\mu-A)\cdot w=0$
and with $\hat{B}+I=B$
$(B\mu-A)\cdot w=0$
The criterion of "low rank $B$" comes from the generic constraints like $g(x_1,\dots,x_n)=x_0^1-x_1\rightarrow D_xg=(-1,0,\dots,0)\rightarrow B=\mbox{diag}(0,1,\dots,1)$.

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.