EDIT: I simplified the question:

Let $N$ be a self-adjoint matrix and consider a small perturbation of this matrix by another diagonal skew-symmetric rank one matrix $A=-i \langle e_1,\bullet \rangle e_1$.

If both $N$ and $A$ were self-adjoint, we'd have the nice Weyl inequalities linking the eigenvalues of the sum.

The perturbed matrix $M:=N+A$ is not assumed to have any nice structure besides the fact that we assume that $\Im(\sigma(M)) \le -\delta$ for some $\delta>0$. Here, $\Im$ is the real part and $\sigma$ the spectrum.

Moreover, we assume the following property:

There is a normalized eigenvector $v$ of $N$ with $\Vert A v \Vert \le \varepsilon$ for some $\varepsilon,$ i.e. one of the eigenvectors of $N$ is (almost) in the nullspace of $A.$

Also, I am equally interested in making the assumption that that there is an approximate eigenvector $v$ for $N$ such that

$$\Vert (N-\lambda)v \Vert \le \varepsilon \ \text{ and } Av=0.$$

The question:

Does all this give us any information about how large $\delta$ can be in terms of $\varepsilon$ or are these two parameters independent?

Put differently, can we conclude the existence of an eigenvalue of $M$ close to the real axis?

** Remark **: The big difficulty here is really to get the existence of eigenfunctions of $M$ supported (almost) away from $e_1.$ If anybody has an idea how to get this, please feel free to comment as well.

You may also consider the following example:

$$N= \left( \begin{array}{cccccc} 0 & 0 & 0 & 3 & -1 & 0 \\ 0 & 0 & 0 & -1 & 3 & -1 \\ 0 & 0 & 0 & 0 & -1 & 3 \\ 3 & -1 & 0 & 0 & 0 & 0 \\ -1 & 3 & -1 & 0 & 0 & 0 \\ 0 & -1 & 3 & 0 & 0 & 0 \\ \end{array} \right)$$ and $A$ as above then the eigenvalues are and as you can see, they are quite close to the real axis, i.e. much close to the real axis than $\Vert A \Vert=1$.

Let $N$ be a normal matrix.

Now I consider a perturbation of the matrix by another matrix $A.$

The perturbed matrix shall be called $M=N+A.$

Now assume there is a normalized vector $u$ such that $\Vert (N-i\lambda)u \Vert \le \varepsilon$

for some $\lambda \in \mathbb R.$

Since $N$ is normal this implies that $d(i\lambda,\sigma(N))\le \varepsilon.$

Moreover, assume that $\Re(\sigma(M)) \le -\delta$ for some $\delta>0.$

If we assume additionally that $Au=0.$ Does this give us any information about how large $\delta$ can be in terms of $\varepsilon$ or are they independent?

EDIT: I simplified the question:

Let $N$ be a self-adjoint matrix and consider a small perturbation of this matrix by another diagonal skew-symmetric rank one matrix $A=i \langle e_1,\bullet \rangle e_1$.

If both $N$ and $A$ were self-adjoint, we'd have the nice Weyl inequalities linking the eigenvalues of the sum.

The perturbed matrix $M:=N+A$ is not assumed to have any nice structure besides the fact that we assume that $\Im(\sigma(M)) \le -\delta$ for some $\delta>0$. Here, $\Im$ is the real part and $\sigma$ the spectrum.

Moreover, we assume the following property:

There is a normalized eigenvector $v$ of $N$ with $\Vert A v \Vert \le \varepsilon$ for some $\varepsilon,$ i.e. one of the eigenvectors of $N$ is (almost) in the nullspace of $A.$

Also, I am equally interested in making the assumption that that there is an approximate eigenvector $v$ for $N$ such that

$$\Vert (N-\lambda)v \Vert \le \varepsilon \ \text{ and } Av=0.$$

The question:

Does all this give us any information about how large $\delta$ can be in terms of $\varepsilon$ or are these two parameters independent?

Put differently, can we conclude the existence of an eigenvalue of $M$ close to the real axis?

** Remark **: The big difficulty here is really to get the existence of eigenfunctions of $M$ supported (almost) away from $e_1.$ If anybody has an idea how to get this, please feel free to comment as well.

EDIT: I simplified the question:

Let $N$ be a self-adjoint matrix and consider a small perturbation of this matrix by another diagonal skew-symmetric rank one matrix $A=-i \langle e_1,\bullet \rangle e_1$.

If both $N$ and $A$ were self-adjoint, we'd have the nice Weyl inequalities linking the eigenvalues of the sum.

The perturbed matrix $M:=N+A$ is not assumed to have any nice structure besides the fact that we assume that $\Im(\sigma(M)) \le -\delta$ for some $\delta>0$. Here, $\Im$ is the real part and $\sigma$ the spectrum.

Moreover, we assume the following property:

There is a normalized eigenvector $v$ of $N$ with $\Vert A v \Vert \le \varepsilon$ for some $\varepsilon,$ i.e. one of the eigenvectors of $N$ is (almost) in the nullspace of $A.$

Also, I am equally interested in making the assumption that that there is an approximate eigenvector $v$ for $N$ such that

$$\Vert (N-\lambda)v \Vert \le \varepsilon \ \text{ and } Av=0.$$

The question:

Does all this give us any information about how large $\delta$ can be in terms of $\varepsilon$ or are these two parameters independent?

Put differently, can we conclude the existence of an eigenvalue of $M$ close to the real axis?

** Remark **: The big difficulty here is really to get the existence of eigenfunctions of $M$ supported (almost) away from $e_1.$ If anybody has an idea how to get this, please feel free to comment as well.

You may also consider the following example:

$$N= \left( \begin{array}{cccccc} 0 & 0 & 0 & 3 & -1 & 0 \\ 0 & 0 & 0 & -1 & 3 & -1 \\ 0 & 0 & 0 & 0 & -1 & 3 \\ 3 & -1 & 0 & 0 & 0 & 0 \\ -1 & 3 & -1 & 0 & 0 & 0 \\ 0 & -1 & 3 & 0 & 0 & 0 \\ \end{array} \right)$$ and $A$ as above then the eigenvalues are and as you can see, they are quite close to the real axis, i.e. much close to the real axis than $\Vert A \Vert=1$.

EDIT: I simplified the question:

Let $N$ be a self-adjoint matrix and consider a small perturbation of this matrix by another diagonal skew-symmetric rank one matrix $A=i \langle e_1,\bullet \rangle e_1$.

If both $N$ and $A$ were self-adjoint, we'd have the nice Weyl inequalities linking the eigenvalues of the sum.

The perturbed matrix $M:=N+A$ is not assumed to have any nice structure besides the fact that we assume that $\Im(\sigma(M)) \le -\delta$ for some $\delta>0$. Here, $\Im$ is the real part and $\sigma$ the spectrum.

Moreover, we assume the following property:

There is a normalized eigenvector $v$ of $N$ with $\Vert A v \Vert \le \varepsilon$ for some $\varepsilon,$ i.e. one of the eigenvectors of $N$ is (almost) in the nullspace of $A.$

Also, I am equally interested in making the assumption that that there is an approximate eigenvector $v$ for $N$ such that

$$\Vert (N-\lambda)v \Vert \le \varepsilon \ \text{ and } Av=0.$$

The question:

Does all this give us any information about how large $\delta$ can be in terms of $\varepsilon$ or are these two parameters independent?

Put differently, can we conclude the existence of an eigenvalue of $M$ close to the real axis?

EDIT: I simplified the question:

Let $N$ be a self-adjoint matrix and consider a small perturbation of this matrix by another diagonal skew-symmetric rank one matrix $A=i \langle e_1,\bullet \rangle e_1$.

If both $N$ and $A$ were self-adjoint, we'd have the nice Weyl inequalities linking the eigenvalues of the sum.

The perturbed matrix $M:=N+A$ is not assumed to have any nice structure besides the fact that we assume that $\Im(\sigma(M)) \le -\delta$ for some $\delta>0$. Here, $\Im$ is the real part and $\sigma$ the spectrum.

Moreover, we assume the following property:

There is a normalized eigenvector $v$ of $N$ with $\Vert A v \Vert \le \varepsilon$ for some $\varepsilon,$ i.e. one of the eigenvectors of $N$ is (almost) in the nullspace of $A.$

Also, I am equally interested in making the assumption that that there is an approximate eigenvector $v$ for $N$ such that

$$\Vert (N-\lambda)v \Vert \le \varepsilon \ \text{ and } Av=0.$$

The question:

Does all this give us any information about how large $\delta$ can be in terms of $\varepsilon$ or are these two parameters independent?

Put differently, can we conclude the existence of an eigenvalue of $M$ close to the real axis?

** Remark **: The big difficulty here is really to get the existence of eigenfunctions of $M$ supported (almost) away from $e_1.$ If anybody has an idea how to get this, please feel free to comment as well.