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Let $N:=(M+M^T)/2$. besides the obvious equality $Tr(N)=Tr(M)$ which is an equality of the sums of eigenvalues, you have the following. Let $\lambda_\pm$ be the smallest/largest eigenvalues of $N$. Th …

This is true, as long as your domain depends smoothly upon one real parameter. Say that you are insterested in the $n$ first eigenvalues. Using a Lyapunov-Schmidt procedure, you may reduce to the situ …

The central question in this area was Kato's conjecture. From Wikipedia: Tosio Kato asked whether the square root of certain elliptic operators, defined via functional calculus, are analytic. The pr …

I think that the numerical range is an appropriate tool for your question. Your naive approximations $L_n$ of the shift operator are nilpotent. For such matrices $M$ (nilpotent of size $n$), the numer …

Selberg's Trace Formula, together with its avatars, gives strong information in a lot of topics: asymptotics of closed geodesics over manifolds of constant negative curvature, asymptotics of the numbe …

I don't have a general answer (I guess it is yes, there are uniformy bounded, at least when $A$ is a smooth bounded domain). At least, let me mention the case of the torus $\mathbb T^d=\mathbb R^d/\ma …

Let me extend, and correct, the argument expressed in the comment made by user378654. Let us start with a surface $S$ for which $\Delta$ admits a double eigenvalue $\lambda$. For instance, you may cho …

My guess is No. You do not need a one parameter Lie group of symmetry to have infinitely many double eigenvalues. Just one involution suffices. And one involution is not enough to make a symmetric spa …

First, let me rephrase your remark. Let $L=HU$ be the polar factorization of $L$ ($H$ hermitian positive definite, $U$ unitary). Then $\Sigma=LL^\ast=H^2$ tells you that the Hermitian part of $L$ is $ …

To begin with, the ellipticity condition is useless if you don't ask also that $$\sum_{i,j}a_{ij}\xi_i\xi_j\le M|\xi|^2$$ for some finite constant $M$. Now the resolvant estimate is used to define an …

An equivalent trick : Let $J:= \operatorname{diag}(1,i,-1,-i)$. Then $J^*AJ=iB$ where $B$ is real and skew-symmetric. Hence the spectrum of $iB$ (thus that of $A$) comes by pairs $\pm\lambda$.

If $A,A^T=\ell^p\rightarrow\ell^p$, then the adjoints $A^T,A$ map $\ell^{p'}$ into itself. By interpolation (Riesz-Thorin), they map $\ell^2$ into itself. It will be often the case that the spectrum o …

The claim is true with $\epsilon=\frac6{\pi^2}\,$. To see this, remark that by changing variable $x_i=y_i\sqrt i\,$, this is equivalent to proving that $$\epsilon\left(\left(\frac1{ij}\right)\right …

Here is a construction. It elaborates from perturbation analysis of eigenvalues. However it starts from the situation of a non-simple eigenvalue. So, let me start with the standard self-adjoint $L_0=- …

When $M$ is negatively curved, and especially when the curvature is constant, the distribution of the eigenvalues tells something about the distribution of lengths of closed geodesics. This is because …