Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.