Preserving Hurwitz stability of block matrices with same eigenvalues of each block

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 $$\tilde A$$.

• Is it $I_n\otimes Q$ or $Q\otimes I_n$? Are you also sure of your first inclusion?
– KBS
May 12 at 20:45
• 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 $n$. May 13 at 22:09
• Yes but it is not block diagonal.
– KBS
2 days ago
• you are right. I am sorry, I meant block matrices yesterday
• 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...
– KBS
yesterday