Eigenvalues of symmetric matrix with skew-symmetric matrix perturbation If $A$ is diagonalizable, using the Bauer-Fike theorem, for any eigenvalue $λ$ of $A$, there exists an eigenvalue $μ$ of $A+E$ such that $|\lambda-\mu|\leq\|E\|_2$ (the vector induced norm).
Here I have the doubt that if we have more properties on $A$ and $E$:

*

*$A$ is symmetric and $E$ is skew-symmetric,

*The eigenvalues of $A$ are completely separated from each other, i.e.: there is a certain distance between each two eigenvalues so that with sufficiently small $\|E\|_2$, the perturbed eigenvalues are still different from each other.

Do we have $|\lambda-\mu|\leq c\|E\|_2^2$ for some constant $c$? Is there any strict proof or a counterexample?
I have only a rough idea that it may be true, because if we denote $\lambda(\epsilon)$ as the eigenvalue of $A+\epsilon E$, then:
$$ \lambda(-\epsilon)=\lambda(A-\epsilon E)=\lambda(A^T-\epsilon E^T)=\lambda(A+\epsilon E)=\lambda(\epsilon) $$
This means the function $\lambda(\epsilon)$ is an even function, so the Taylor expansion may be
$$ \lambda(\epsilon)=\lambda(0)+\lambda''(0)\epsilon^2/2+O(\epsilon^4) $$
Or equivalently,
$$ \lambda(A+\epsilon E)-\lambda(A)=\lambda''(0)\epsilon^2/2+O(\epsilon^4) $$
 A: Assume that $A\in M_n$ is real symmetric and has $n$ simple eigenvalues $\lambda_1>\cdots> \lambda_n$. Thus there is $\alpha>0$ s.t. if $||E||_2<\alpha$, then $A+ E$ has $n$ simple real eigenvalues $\lambda_1(E)>\cdots> \lambda_n(E)$. Moreover any function $\lambda_i:E\rightarrow \lambda_i(E)$ is real analytic. Put $\det(A+E-\lambda I)=\chi(\lambda,E)$ ; it is a polynomial in the $(E_{i,j})$ that has not any term of degree $1$. Thus $\dfrac{\partial \chi}{\partial E}(\lambda,0)=0$. For every skew-symmetric $H$,     $\lambda_i'(E)(H)=\dfrac{-\dfrac{\partial \chi}{\partial E}(\lambda_i,E)(H)}{\dfrac{\partial \chi}{\partial \lambda}(\lambda_i,E)}$ 
and $\lambda_i''(0)(H,H)=\dfrac{-\dfrac{\partial^2 \chi}{\partial E^2}(\lambda_i,0)(H,H)}{\dfrac{\partial \chi}{\partial \lambda}(\lambda_i,0)}$. 
According to Taylor formula, $\lambda_i(E)-\lambda_i\sim  \dfrac{-1/2\dfrac{\partial^2 \chi}{\partial E^2}(\lambda_i,0)(E,E)}{\dfrac{\partial \chi}{\partial \lambda}(\lambda_i,0)}$.
Finally there is $\beta<\alpha$ s.t. if $||E||_2<\beta$, then for every $i$, $|\lambda_i(E)-\lambda_i|\leq c||E||_2^2$ where $c=\sup_i \dfrac{||\dfrac{\partial^2 \chi}{\partial E^2}(\lambda_i,0)||_2}{|\dfrac{\partial \chi}{\partial \lambda}(\lambda_i,0)|}$.
