Let us consider an invertible matrix $\mathbf{A}\in GL\_d(\mathbb{R})$ such that all its diagonal entries $\mathbf{A}\_{ii}=-1 \; \forall \, i$.
My question is the following:


Does it always exists a diagonal matrix $\mathbf{\Lambda}\in \mathbb{R}^d) $ such that 


$\sigma(\mathbf{\Lambda A}) \subset \{ z\in\mathbb{C} \, | \, Re(z)<0 \,\} $ ?


In other words, does exists a diagonal matrix $\mathbf{\Lambda}$ such that the characteristic polynomial of the matrix $\mathbf{\Lambda A}$ is Hurwitz, so the corresponding dynamical system $ \dot{\mathbf{X}}= (\mathbf{\Lambda A})\mathbf{X} $ is stable? 


In the case when the system induced by $\mathbf{A}$ is already stable then there exists a trivial solution $\mathbf{\Lambda}=\mathbf{I}$. Also if we allow $\mathbf{\Lambda}$ to live in a bigger space, for example general linear group $GL\_d(\mathbb{R})$, the problem becomes trivial by choosing $\mathbf{\Lambda}=-\mathbf{A}^{-1}$. But if we only allow it to be diagonal the question becomes more saddle. 


I would be very grateful if somebody could point me to some reference concerning this question, or giving to me some hint.



Stability of a linear system and spectrum of the product of two matrices
The answer is no, in general. Of course it is true for $d=1,2$. But already in the case $d=3$ there are counterexamples. One such counterexample is the matrix


$$
\mathbf{A} = \begin{bmatrix} -1 & 1 & 1\\ 1 & -1 & 1 \\ 1 & 1 & -1\end{bmatrix}
$$


All the 2x2 principal minors of this matrix are zero, so that the coefficient of $s$ of the characteristic polynomial $\det(sI-\mathbf{A})$ is zero. This you cannot change by scaling of the rows. So that also for all matrices $\mathbf{\Lambda A}$ the coefficient of $s$ of the characteristic polynomial is zero. Therefore these matrices are not Hurwitz.


The question has been treated in the paper


M. E. Fisher and A. T. Fuller, On the stabilization of
matrices and the convergence of linear iterative processes, Proc.
Cambridge Philos. Soc. 54:417-425 (1958).
The paper may be downloaded [here](https://1988c0fbd715ddc21e4eab393a153f2278df184c.googledrive.com/host/0B8_joYJa2eNzfkNIQ2ZiY1hxS0FlMTI4ZC1IYnRUZ0ttd0NrQlYxcEdMZnFSMnpYZ3ltbUE/v54/4/S0305004100002978.pdf).


Theorem 1 of the paper is actually a far reaching positive result: If there is a sequence of *nonzero* principal minors of the matrix $A$, i.e. $M\_d, M\_{d-1},\ldots,M\_1$ where each minor $M\_{k-1}$ is a principal $(k-1)\times (k-1)$-minor of $M\_k$, then it can be done.
On page 3 of the paper you find a necessary condition, namely that you need at least one nonzero principal minor of every order, which is what is used in the counterexample.


