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2
votes
Accepted
Distribution of the spectrum of a perturbed matrix
Since $P(G)\propto{\rm exp}\bigl(-\frac{1}{2}{\rm Tr}\,GG^{\rm T}\bigr)$ is invariant under orthogonal transformations, if $A$ is real Hermitian you can work in a basis where $A$ is diagonal, with the …
3
votes
Proof (or reference) about $\lambda_i(A+\epsilon e_je_j^*) = \lambda_i(A) + \epsilon |v_{i,j...
This is first order perturbation theory: a perturbation $\delta A$ to a Hermitian matrix $A$ gives to first order a correction $\delta \lambda$ to an eigenvalue $\lambda$ (with corresponding eigenvec …
4
votes
Lipschitz continuity of eigenvalues and eigenvectors of Hermitian matrices
Here is a counterexample: consider the real symmetric matrix
$$A=\begin{pmatrix}
\cos\phi&\sin\phi\\
\sin\phi&-\cos\phi
\end{pmatrix}.$$
The eigenvectors $v_\pm$ for the eigenvalues $\pm 1$ are
$$v_\p …
0
votes
Analytic perturbation of the eigenvalues/eigenvectors of non-Hermitian matrix
The functions $\lambda(x)$ and $w(x)$ obey a set of coupled first-order differential equations in $x$, derived in Structure of trajectories of complex matrix eigenvalues (Bohigas, De Carvalho, and Pat …
3
votes
Perturbation theory of eigenvalues - Effects of degeneracy/ multiplicity
Kato's theory has been extended to include degenerate eigenvalues by Hunziker and Pillet, Degenerate asymptotic perturbation theory (1983).
These beautiful results [of Kato and others] are not qui …
1
vote
Finding $\theta$ such that at least one eigenvalue of $A(\theta)$ is real
Since the eigenvalues of a real matrix $A(\theta)$ come in complex conjugate pairs, an eigenvalue on the real axis with multiplicity 1 cannot move off the real axis when $\theta$ is varied over a smal …
1
vote
Separating the spectrum of a Hermitian matrix
since you want $f$ to be the same for all $A$, let's first take the special case $A=0$; then the eigenvalues of $A+\epsilon G$ have an average spacing of order $\epsilon/\sqrt{n}$, but the fraction of …
2
votes
Accepted
Sum of positive self-adjoint operator and an imaginary "potential": literature request
Let me consider the case of the Schrödinger equation, $L=-\nabla^2+V(x)$. Then the operator $L+ikx$ has PT-symmetry, meaning that it is invariant under the combined action of inversion $x\mapsto-x$ (p …
3
votes
Asymptotics of integral representation of distribution
The Fourier transform $F_\epsilon(x,t)$ has a closed-form expression in terms of hypergeometric functions,
$$F_\epsilon(x,t)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{ikx}e^{i\epsilon^4 tk^4}\,dk=$$
$$\q …
3
votes
Local differentiability of eigenvalues and eigenvectors of a real symmetric matrix
Theorem (1.1) of Perturbation theory for normal operators is likely what you are looking for.
See also Differentiable perturbation of unbounded operators.
If you would order the eigenvalues by their m …
6
votes
How do I solve the following definite integral (preferably by an asymptotic method)?
Here is a log-log plot of
$$\delta I=I_{\text{appr}}-\int_{1}^{\infty} \frac{\sin^2 (\mu \sqrt{x^2 -1})}{(x+1)^{\frac{9}{2}} (x-1)^{\frac{3}{2}}} \,dx,$$
as a function of $\mu$, with
$$I_{\text{appr}} …
6
votes
Accepted
First derivative of $f(A) = \frac{1}{\lambda_{\min}(A)}$ for perturbed matrix
First order perturbation theory gives you
$$\frac{1}{\lambda_{\rm min}(A+\epsilon B)}=\frac{1}{\lambda_{\rm min}(A)}-\frac{\epsilon}{\lambda_{\rm min}(A)^2}\langle v_0|B|v_0\rangle+{\cal O}(\epsilon^2 …
3
votes
Accepted
Why is the second order correction to energy zero for a fully degenerate eigensystem?
If $H$ is fully degenerate, all eigenvalues are identical, it means that $H$ is proportional to the unit matrix. The zeroth order eigenstates can be chosen as any orthonormal basis. Degenerate perturb …