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Let us consider a block matrix of the form

$$A= \begin{bmatrix} -(k+\mu)I & B \\ kI & -(\gamma + \mu)I \end{bmatrix},$$

where $I$ is the $n\times n$ identity matrix, $\gamma, k$ and $\mu$ are positive constants and

$$B= \begin{bmatrix} a_1b_1& \ldots & a_1b_n \\ \vdots& \ddots& \vdots \\ a_nb_1& \ldots& a_nb_n \end{bmatrix}$$

with $a_i, b_i >0$ for all $i=1,\dots,n $.

Which are the eigenvalues of matrix $A$? Is there any way to evaluate them through the eigenvalues of the blocks of matrix $A$?

Thanks in advance.

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  • $\begingroup$ Hey :) It is good practice to include a bit of what you have tried in the question. Try to show where you got stuck. $\endgroup$ Jun 5, 2020 at 11:43
  • $\begingroup$ Sure, I'll edit it then, thanks for the advice! $\endgroup$
    – Vlntn
    Jun 5, 2020 at 12:53

1 Answer 1

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Given vectors $\rm{a}, \rm{b} \in \mathbb R^n$, let

$$\rm{M} := \begin{bmatrix} -(\kappa + \mu) \rm{I}_n & \rm{a}\rm{b}^\top\\ \kappa \,\rm{I}_n & -(\gamma + \mu) \rm{I}_n \end{bmatrix}$$

whose characteristic polynomial is

$$\begin{aligned} \det \left( s \rm{I}_{2n} - \rm{M} \right) &= \det \begin{bmatrix} (s + \kappa + \mu) \rm{I}_n & - \rm{a}\rm{b}^\top\\ - \kappa \,\rm{I}_n & (s + \gamma + \mu) \rm{I}_n \end{bmatrix}\\ &= \det \left( (s + \kappa + \mu) (s + \gamma + \mu) \, \rm{I}_n - \kappa \, \rm{a}\rm{b}^\top \right)\end{aligned}$$

because multiples of the identity matrix do commute. Let $q$ be the characteristic polynomial of rank-$1$ matrix $\kappa \, \rm{a}\rm{b}^\top$. Hence,

$$q (s) = s^{n-1} \left( s - \kappa \, \rm{b}^\top \rm{a} \right)$$

and, thus,

$$\det \left( s \rm{I}_{2n} - \rm{M} \right) = q \left( (s + \kappa + \mu) (s + \gamma + \mu) \right)$$

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    $\begingroup$ I think that the following is worth adding for the sake of completeness: the eigenvalues are the zeros of $\det(s I_{2n} - M)$, which in this case turn out to be $$ \lambda = -(\kappa + \mu), -(\gamma + \mu), $$ each with algebraic mulitplicity $n-1$, and the solutions to the quadratic equation $$ (\lambda + \kappa + \mu)(\lambda + \gamma + \mu) = \kappa b^Ta, $$ each with multiplicity $1$. $\endgroup$ Jun 5, 2020 at 13:45

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