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Given a real square matrix $A \in \Re^{n \times n}$, let's say its eigenvalues have a lovely property: $Re(\lambda^A_i) < 0$. We also have another diagonal real square matrix $K \in \Re^{n \times n}$ whose entries are all positive. Now, if we define $B = KA$, what is the lower bound of element $k_{ii}$ that guarantees all $Re(\lambda^{B}_{i}) < 0$ too?

If $K = kI$, we don't have any problem. If $A$ is a diagonal matrix, we again do not have any problem. This seems to be a problem when both are not true.

motivation

Given a continuous-time system $\tau \frac{dh}{dt} = W_{hh}h + W_{hi}x$, I was thinking of adding a gain matrix $K$ which can be left-multiplied to the right side of the equation. With that, $K_{ii}$ can properly scale all the input to the ith node, say $W_{ij}$.

However, if $K$ is not chosen properly, the system $\tau \frac{dh}{dt} = KW_{hh}h + KW_{hi}x$ can become unstable, as the real part of the eigenvalues of $KW_{hh}$ can become positive even when $Re(\lambda^{W_{hh}}_i) < 0$.

EDIT 1

Constraining on $K$ is an interesting idea. As Robert suggested, can we think of a way to constrain $K$ to keep the real part negative?

One another boring example is $A$ being block-diagonal and $K$ having a unique gain factor $k_i$ for each block. Though, this seems to be a bit too special case.

lostintimespace
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    Even in the case where $A$ is symmetric and definite, there still isn't a nice relationship https://math.stackexchange.com/questions/1846769/eigenvalues-of-product-of-a-matrix-with-a-diagonal-matrix – whpowell96 Feb 29 '24 at 01:59
  • As stated, there is no such lower bound. Note that the eigenvalues of $t K A$ are $t$ times the eigenvalues of $K A$. So if you have one matrix $K$ for which the eigenvalues of $B$ all have negative real part, suitable scaling gives you another one where the lowest eigenvalue is as small or as large as you want. Perhaps something could be done if you normalized, say by fixing $\text{Tr}(K)$. – Robert Israel Feb 29 '24 at 04:10

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