If there a relationship between a submatrix's eigenvalues and the matrix's eigenvalues.

Question

My question

Let $A$ be a $n \times n$ matrix. We denote the submatrix whose entries are $\{a_{ij}\}\in A$ where $i=2,...,n$ and $j=2,...,n$ as $A_{n-1}$.

Suppose we know the eigenvalues of $A_{n-1}$. Does there exist a relationship between the eigenvalues of the matrix $A$ and the eigenvalues of the submatrix $A_{n-1}$?

Context

I am trying to do linear analysis on a system of $n$ ODEs an equilibrium point. The Jacobin of the system has the following structure \begin{align} J&=\begin{pmatrix} -1 & a_{12} & \dots & a_{1n} \\ a_{21} & \\ \vdots & &A_{n-1}\\ a_{n1} & \end{pmatrix} \end{align} where $A_{n-1}$ is an $n-1 \times n-1$ matrix whose eigenvalues I know from another unrelated method. The entries $a_{12},...,a_{1n}\geq 0$ while entries $a_{21},...,a_{n1}\leq 0$, note these entries are functions ans vary in value based on the equilibrium point the Jacobian is evaluated at.

Notes

• I suspect that the answer to my question is that there is an upper bound for the eigenvalues of $A$. I haven't found anything that supports this suspicion.
• I provide additional context to assist those answering the question, you can feel free to ignore the context if you can answer the original question.
• If any clarification is needed feel free to ask.

Answer 1

There is an answer to your question for symmetric matrices : the eigenvalues $\lambda_k$ of $A_{n-1}$ and $\mu_j$ of $A_n$ are interleaved :

$$\mu_1 \leq \lambda_1 \leq \mu_2 \leq ... \leq \lambda_{n-1} \leq \mu_n$$

This is a theorem established by Cauchy in the 1830s.

See {Eigen values of a principal sub-matrix of a symmetric matrix}

For non-symmetric matrices, there is no general result like this one.

Answer 2

The only eigenvalue if $[0]$ is, of course, $0$. However, the eigenvalues of\begin{bmatrix}-1&x\\x&0\end{bmatrix}are $\dfrac{-1\pm\sqrt{1+4x^2}}2$, which can be arbitrarily large.