1
$\begingroup$

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 $\mathbb{R}$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
1
$\begingroup$

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

$$(x+1)((x+1)^2+1)((x+3)^2+1)$$

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.