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The fact that the entries of the matrix are real does seem to help. The state of the art is the following. The spectrum is continuous functions of $\xi$. However, it is not always possible to label …

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=- …

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 …

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 …

Set $c:=\cos\frac1t$ and $s:=\sin\frac1t$. Choose a function $\phi$ that is flat at $t=0$ ($\phi$ and all its derivatives vanish at $t=0$). Then set $$A(t):=\begin{pmatrix} \phi c^2 & -\phi sc \\\\ -\ …

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 $ …

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 …

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 …

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 …

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 …

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 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 …

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 …

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 …