What matrix has only negative or zero real part for all the eigenvalues?

Question

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 $\mathbb{R}$.

Answer 1

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

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