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Let us assume that $$A=-\text{diag}(d_1,\ldots,d_m) +Q \text{ is Hurwitz},$$ where $d_i>0, \forall i\in \{1,\ldots,m\}$ and $Q\in \mathbb R^{m \times m}$ is a possibly full matrix.

Can we deduce that (or under which conditions?) $$\tilde A=-\text{blkdiag}(D_1,D_2,\ldots D_m)+ Q \otimes I_n \text{ is Hurwitz},$$ where $D_1\in \mathbb R^{n\times n}$ is full matrix, while $D_2, \ldots D_m\in \mathbb R^{n\times n}$ are diagonal. Assume also that $\min( \mathrm{eigenspectrum}(D_i))=d_i$ and all eigenvalues of $D_i$ are positive, for all $i\in \{1,\ldots,m\}$.

In other words, both matrices $A$ and $\tilde A$ consist of diagonal matrices which share common eigenvalues. Note that $$\mathrm{eigenspectrum}(\text{blkdiag}(d_1,\ldots,d_m)) \subset \mathrm{eigenspectrum}(\text{diag}(D_1,D_2,\ldots D_m))$$ and $$\mathrm{eigenspectrum}(Q)\subset \mathrm{eigenspectrum}(Q \otimes I_n).$$ Intuitively, it seems that some properties should be preserved between both matrices $A$ and $\tilde A$.

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  • $\begingroup$ Is it $I_n\otimes Q$ or $Q\otimes I_n$? Are you also sure of your first inclusion? $\endgroup$
    – KBS
    May 12 at 20:45
  • $\begingroup$ I made a mistake in the first inclusion. I rewrote it. It is $Q\otimes I_n$. The eigenvalues of the later are the same as the eigenvalues of $Q$ with multiplicity $n$. $\endgroup$
    – Con_T
    May 13 at 22:09
  • $\begingroup$ Yes but it is not block diagonal. $\endgroup$
    – KBS
    2 days ago
  • $\begingroup$ you are right. I am sorry, I meant block matrices $\endgroup$
    – Con_T
    yesterday
  • $\begingroup$ Then that will not work. The block diagonal structure $I\otimes Q$ is much nicer in this respect. The fact you will have off-diagonal block entries in $\tilde A$ will perturb the eigenvalues of the whole matrix. Moreover, eigenvalues have terrible additive properties. I am also not even factoring in the fact that $D_1$ is not diagonal... $\endgroup$
    – KBS
    yesterday

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