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I have a real square matrix $A$ (not necessarily symmetric) with all its principal submatrices (including $A$) having eigenvalues with a negative real part.

On the other hand, I have a matrix $D$ which is diagonal and has only non-positive elements in its diagonal. I think that this implies that $A+D$ must also have eigenvalues with a negative real part. Does anyone know how to prove this, or if this is even true?

I have read some things about this in this link:

Principal minors of sum of a matrix and a diagonal matrix

That implies that the odd (resp. even) minors of $A+D$ will be negative (resp. positive). However, this doesn't prove my assertion.

Until now, I have only found the fact that $A+D$ cannot have a real positive eigenvalue, but could it have a complex-conjugate pair of eigenvalues with a positive real part?

EDIT: It is worth noticing that the condition on all principal submatrices to be stable is a requirement for the claim to be (possibly) true. In fact, if we only consider that $A$ is stable but we allow that some of its submatrices have eigenvalues with a positive real part, then the statement clearly does not hold. For instance, take $$A=\begin{pmatrix} 1 & -3 \\ 1 & -2 \end{pmatrix}.$$

This matrix is clearly stable. Nevertheless, if we consider the diagonal matrix

$$D=\begin{pmatrix} 0 & 0 \\ 0 & -3 \end{pmatrix},$$

then

$$A+D=\begin{pmatrix} 1 & -3 \\ 1 & -5 \end{pmatrix},$$

which is not a stable matrix. Notice that, in this case, $A$ has one principal submatrix with a positive real eigenvalue.

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By going over Nonnegative Matrices in the Mathematical Sciences by Abraham Berman & Robert J. Plemmons, if we add that $A \in \{ B = (b_{ij}) \in \mathbb{R}^{n \times n}: b_{ij}>=0 \text{ for } i \neq j \}$, then this is an equivalence. It might hold for a larger class of matrices, but I could not find any proof that holds for the broader case.

Your description that "I have a real square matrix $A$ (not necessarily symmetric) with all its principal submatrices (including $A$) having eigenvalues with a negative real part." Leads to $-A$ having all principal minor being positive (A1 in the book).

If you go over [Berman & Plemmons, Ch.6 M-matrices] you will find that the description above means that $-A$ is a non-singular M-matrix. Which is also equivalent to existing a positive diagonal matrix $P$ such that $ -AP-PA^\top \succ 0 $ (H24 in the book.)

From there, given any $D$ which is diagonal and has only non-positive elements, since both $-D$ and $P$ matrices are non-negative diagonal matrices then $-PD \succeq0$, we have that $$ (A+D)P+P(A+D)^\top = AP+PA^\top + 2DP \prec 2DP \preceq 0.$$ Therefore $A+D$ is Hurwitz stable.

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  • $\begingroup$ Thank you for your reply. I think there is only one problem with it. I don't know whether this is generalizable but it seems that all the 50 conditions stated in that book are valid for -A \in Z^{n\times n}, which is not the general case... $\endgroup$ Commented Aug 11, 2022 at 0:20
  • $\begingroup$ Oh, I missed that. I will add that as a condition for the equivalence to hold and leave it as it is, as I do not think I will come across any larger matrix class for which that has been proven. $\endgroup$
    – jDAQ
    Commented Aug 11, 2022 at 0:48

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