Search Results

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 …

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 …

The characteristic polynomial is even in both $X$ and $t$ : $P_t(X)=Q(X^2,t^2)$ where $$Q(Y,s)=(Y-s)^2-(2|a|^2+|b|^2+|c|^2)(Y-s)-4|a|^2s+|a^2-b\bar c|^2.$$ The variation of $s\mapsto Y(s)$ is given by …

Miscellaneous results. If $A$ is strictly upper triangular, then $x\cdot\nabla$ consists only is terms $x_j\partial_k$ with $j<k$. The action of $L$ over homogenous polynomials of degree $d$ is descr …

Your matrix $M_\mu$ is symplectic: $M_\mu^T\Omega M_\mu=\Omega$ where $$\Omega=\begin{pmatrix} 0_2 & Y \\ -Y & 0_2 \end{pmatrix},\qquad Y=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$$ Then every pro …

The solutions of $(L-\lambda)u=0$ are the functions $u(x)=e^{i\lambda x}v(x)$, where $v$ satisfies $Lv=0$. The periodicity amounts to $e^{i\lambda}v(1)=v(0)$. Thus your problem does admit infinitely m …

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=- …

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 …

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 $\pm\lambda$.

As mentionned by other answers, simple eigenvalues are $C^\infty$, while non-simple ones are not. Let me add however two important properties which you can find in Kato's book Perturbation theory of l …

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 …

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 …

The most elementary case of Weyl's inequality says that, if $\lambda_i(S)$ denote the $i$th eigenvaue of the Hermitian matrix $S$ (increasing order), then $\lambda_i(S)+\lambda_1(T)\le\lambda_i(S+T)\l …

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 …

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 …