Suppose $A,E$ are Hermitian $(n \times n)$-matrices and $E$ is of low rank. There are well known results bounding the difference in spectra of $A$ and $A+E$. For example the Eigenvalue Interlacing Theorem states \begin{align*} & \lambda_{j-\operatorname{rank}(E)}(A) \geq \lambda_{j}(A+E) \geq \lambda_{j+\operatorname{rank}(E)}(A) \end{align*} for $\lambda_1(M) \geq ... \geq \lambda_n(M)$, whenever the eigenvalues with the corresponding indices exist.

Question: Can we do something similar when $A$ is not Hermitian, but still has real eigenvalues? Let $A,E$ be (not necessarily Hermitian) $(n \times n)$-matrices such that the eigenvalues of $A$ and $A+E$ are all real. Is there a link or bound on the difference between the eigenvalues of $A$ and those of $A+E$ for low rank $E$?

[If it helps, we may assume the eigenvalues of $A$ to be positive, instead of just real.]

The standard counter example for low-rank perturbation having large effects on the eigenvalues of non-Hermitian matrices is $\tilde{A} = (\delta_{k-1,l})_{k,l \leq n}$ and $\tilde{E}_{k,l}=\begin{cases} 1 & \text{ for } k=1, \, l=n \\ 0 & \text{ else} \end{cases}$, since $\tilde{A}$ has only the eigenvalue $0$ with multiplicity $n$ and $(\tilde{A}+\tilde{E})$ has eigenvalues $(e^{i2\pi \frac{j}{n}})_{j \leq n}$. A similar example, where all eigenvalues are in $\mathbb{R}$ would answer my question in the negative.

Any help is much appreciated!

Suppose $A,E$ are Hermitian $(n \times n)$-matrices and $E$ is of low rank. There are well known results bounding the difference in spectra of $A$ and $A+E$. For example the Eigenvalue Interlacing Theorem states \begin{align*} & \lambda_{j-\operatorname{rank}(E)}(A) \geq \lambda_{j}(A+E) \geq \lambda_{j+\operatorname{rank}(E)}(A) \end{align*} for $\lambda_1(M) \geq ... \geq \lambda_n(M)$, whenever the eigenvalues with the corresponding indices exist.

Question: Can we do something similar when $A$ is not Hermitian, but still has real eigenvalues? Let $A,E$ be (not necessarily Hermitian) $(n \times n)$-matrices such that the eigenvalues of $A$ and $A+E$ are all real. Is there a link or bound on the difference between the eigenvalues of $A$ and those of $A+E$ for low rank $E$?

[If it helps, we may assume the eigenvalues of $A$ to be positive, instead of just real.]

The standard counter example for low-rank perturbation having large effects on the eigenvalues of non-Hermitian matrices is $\tilde{A} = (\delta_{k-1,l})_{k,l \leq n}$ and $\tilde{E}_{k,l}=\begin{cases} 1 & \text{ for } k=1, \, l=n \\ 0 & \text{ else} \end{cases}$, since $\tilde{A}$ has only the eigenvalue $0$ with multiplicity $n$ and $(\tilde{A}+\tilde{E})$ has eigenvalues $(e^{i2\pi \frac{j}{n}})_{j \leq n}$. A similar example, where all eigenvalues are in $\mathbb{R}$ would answer my question in the negative.

Any help is much appreciated!

Thank you Ilya Bogdanov for the answer, it is exactly what was needed. Please allow me too simplify/elaborate in case anyone else has trouble reading your answer.

Take any polynomial $p(x) = x^n + a_{n-1}x^{n-1} + ... + a_0$ whose roots $\lambda_1,...,\lambda_n$ are all real. For $A = (\delta_{k-1,l})_{k,l \leq n}$ and the rank one matrix $E$ defined by $E_{k,n} = -a_{k-1}$ and $E_{k,l}=0$ for $l<n$. The matrix $A$ again has only the eigenvalue $0$ with multiplicity $n$ and the matrix $A+E$ will have the roots $\lambda_1,...,\lambda_n$ as eigenvalues, since its characteristic polynomial will be $p$.

This is thus a counter example, since adding a low rank matrix to a non-Hermitian matrix can change the eigenvalues arbitrarily.

Suppose $A,E$ are Hermitian $(n \times n)$-matrices and $E$ is of low rank. There are well known results bounding the difference in spectra of $A$ and $A+E$. For example the Eigenvalue Interlacing Theorem states \begin{align*} & \lambda_{j-\operatorname{rank}(E)}(A) \geq \lambda_{j}(A+E) \geq \lambda_{j+\operatorname{rank}(E)}(A) \end{align*} for $\lambda_1(M) \geq ... \geq \lambda_n(M)$, whenever the eigenvalues with the corresponding indices exist.

Question: Can we do something similar when $A$ is not Hermitian, but still has real eigenvalues? Let $A,E$ be (not necessarily Hermitian) $(n \times n)$-matrices such that the eigenvalues of $A$ and $A+E$ are all real. Is there a link or bound on the difference between the eigenvalues of $A$ and those of $A+E$ for low rank $E$?

[If it helps, we may assume the eigenvalues of $A$ to be positive, instead of just real.]

The standard counter example for low-rank perturbation having large effects on the eigenvalues of non-Hermitian matrices is $\tilde{A} = (\delta_{k+1,l})_{k,l \leq n}$ and $\tilde{E}_{k,l}=\begin{cases} 1 & \text{ for } k=1, \, l=n \\ 0 & \text{ else} \end{cases}$, since $\tilde{A}$ has only the eigenvalue $0$ with multiplicity $n$ and $(\tilde{A}+\tilde{E})$ has eigenvalues $(e^{i2\pi \frac{j}{n}})_{j \leq n}$. A similar example, where all eigenvalues are in $\mathbb{R}$ would answer my question in the negative.

Any help is much appreciated!

Suppose $A,E$ are Hermitian $(n \times n)$-matrices and $E$ is of low rank. There are well known results bounding the difference in spectra of $A$ and $A+E$. For example the Eigenvalue Interlacing Theorem states \begin{align*} & \lambda_{j-\operatorname{rank}(E)}(A) \geq \lambda_{j}(A+E) \geq \lambda_{j+\operatorname{rank}(E)}(A) \end{align*} for $\lambda_1(M) \geq ... \geq \lambda_n(M)$, whenever the eigenvalues with the corresponding indices exist.

Question: Can we do something similar when $A$ is not Hermitian, but still has real eigenvalues? Let $A,E$ be (not necessarily Hermitian) $(n \times n)$-matrices such that the eigenvalues of $A$ and $A+E$ are all real. Is there a link or bound on the difference between the eigenvalues of $A$ and those of $A+E$ for low rank $E$?

[If it helps, we may assume the eigenvalues of $A$ to be positive, instead of just real.]

The standard counter example for low-rank perturbation having large effects on the eigenvalues of non-Hermitian matrices is $\tilde{A} = (\delta_{k-1,l})_{k,l \leq n}$ and $\tilde{E}_{k,l}=\begin{cases} 1 & \text{ for } k=1, \, l=n \\ 0 & \text{ else} \end{cases}$, since $\tilde{A}$ has only the eigenvalue $0$ with multiplicity $n$ and $(\tilde{A}+\tilde{E})$ has eigenvalues $(e^{i2\pi \frac{j}{n}})_{j \leq n}$. A similar example, where all eigenvalues are in $\mathbb{R}$ would answer my question in the negative.

Any help is much appreciated!