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Given the parameter dependent matrix $A=\begin{pmatrix} 0 & I\\ A_1 & kA_2\end{pmatrix}$, with $A_1, A_2\in \mathbb{R}^{n\times n}$ and $k\in\mathbb{R}>0$, is there a way to display the eigenvalues of $A$ as a function of $A_1, A_2, k$? Or to give an estimation of the eigenvalues' location? What can be said about the corresponding eigenvectors of $A$ dependent on $k$?

We can suppose that $A_1,A_2$ have full rank and all eigenvalues are negative or have a negative real part. They may have repeated eigenvalues.

Clearly if $\lambda(A)$ is an eigenvalue, for $k=0$ we have $\lambda(A)=\pm \sqrt{\lambda(-A_1)}$, and for $k\rightarrow \infty$ we have $n$ eigenvalues $\lambda(A)=0$ and $n$ eigenvalues $\lambda(A)=\lambda(kA_2)$. Furthermore we have $k\mathrm{tr}(B)=\sum_{i=1}^{2n} \lambda_i(A)$. Assuming that $\mathrm{tr}(B)<0$ this would mean that at least some eigenvalues of $A$ have negative real parts of magnitude growing with $k$.

I'm interested in what happens for $0\ll k\ll\infty$. My intuition is that with growing $k$ the eigenvalues of $A$ will move into the left half-plane, before tending to $-\infty$ (as all eigenvalues of $A_2$ have negative real parts) or $0$ but so far I haven't been able to find a suitable proof method for this.

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It is not a complete answer. Assume $\begin{pmatrix} u\\ v \end{pmatrix}$ is an eigenvector with the corrersponding eigenvalue $\lambda$. Then we have $$ \begin{equation} v =\lambda u,\\ A_1 u+kA_2 v =\lambda v. \end{equation} $$ Substituting the first equation into the second one, we get $$ (A_1+kA_2\lambda)u=\lambda^2u. $$ Case 1) $\lambda=0$. Then $A_1u=0$. So $\lambda=0$ is eigenvalue and the eigenspace is the null-space of $A_1$. Case 2) $\lambda\neq0$. Then $$ (A_1+kA_2\lambda-\lambda^2 I)u=0. $$ So you have to find $\lambda$ such that $$ \det(A_1+kA_2\lambda-\lambda^2 I)=0. $$ If we assume that $\lambda$ tends to $\infty$ on the complex plane then calculating the coefficients of $\lambda^{2n}$ and $\lambda^{2n-1}$ we obtain an asymptotic formula for $\lambda$. It is an idea only but it may give some information and further intuition.

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