Say $X \in \mathbb{R}^{m\times m}$,

Is it possible to have a constraint on $X$, such that all the eigenvalues has negative or zero real part?

What I conjecture

The following $X$ has only negative or zero real part:

$X = \frac{A - A^\top}{2} - \textrm{diag}(\gamma_1^2,\ldots,\gamma_m^2)$

for $A$ and $\gamma_i$ defined on the $\mathbb{R}$.

  • $\begingroup$ Your "parametric" method of generation looks very good. You have surely based on the fact that a skew symmetric matrix, alias antisymmetric matrix (that can always be given the form $(A-A^T)/2$) has all its eigenvalues of the form $ia$ when $m$ is even and $ia$ plus an additional $0$ when $m$ is odd. But I have no proof for it. Recurrence on $m$ ? $\endgroup$ – Jean Marie Feb 19 at 9:22
  • $\begingroup$ @JeanMarie Thank you. The proof may be think about $x^\top A x = x^\top A^\top x$, so $x^\top (A-A^\top) x = 0$, so the rest is simply a non-positive diagonal matrix, it takes everything as non-positive, then you might wonder: hey it is not a symmetric matrix! Indeed, luckily, it is proved to be non-positive real part here: math.stackexchange.com/questions/83134/… $\endgroup$ – ArtificiallyIntelligence Feb 19 at 16:44
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    $\begingroup$ One last thing : if a matrix is such that all its eigenvalues have their real part <0, it hasn't necessarily the form you have given : for example $\begin{pmatrix}-3&-1\\3&0\end{pmatrix}$ has eigenvalues $(-3\pm i\sqrt{3})/2$ without being of this form. $\endgroup$ – Jean Marie Feb 19 at 21:19
  • $\begingroup$ Thanks for the URL you have given. $\endgroup$ – Jean Marie Feb 19 at 21:20
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    $\begingroup$ @JeanMarie Yes. I agree, this parameterization is only a sufficient condition, or you can it a hack. I am a engineer and just want to hardwire a stable matrix in my optimization process. $\endgroup$ – ArtificiallyIntelligence Feb 19 at 22:15

Do you know the concept of companion matrix ?

To a polynomial

$$p(x)=x^n+c_{n-1}x^{n-1}+ \dots+c_1x+c_0 $$ one can associate a matrix, called its companion matrix, whose eigenvalues are precisely the roots of this polynomial ; this matrix is : $$ \begin{pmatrix} 0&1&0& \dots & 0\\ 0&0&1& \dots & 0\\ & \vdots & & \ddots &\\ 0&0&\ddots & \dots & 1\\ -c_0&-c_1&-c_2& \dots & -c_{n-1}\\ \end{pmatrix}$$

and the work is done.

Check it with a polynomial whose roots have a negative real part, such as


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    $\begingroup$ Thanks! But I want a unconstrained parameterization of the family of matrix with non-positive matrix. Is that even possible? The process you mentioned need to determine the combination of roots being conjugate pair OR single. $\endgroup$ – ArtificiallyIntelligence Feb 15 at 21:07

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