Consider the polynomial $$P(x)=x^n+a_{n-1} x^{n-1}+\dots+a_1 x+a_0 $$ with real coefficients $\{a_i\}$. It is called stable if all its roots have negative real part. The Routh-Hurwitz stability criterion gives a characterization of stable polynomials. However, I'm looking for a simpler way to show the following non-characterizing property

If $P(x)$ is stable, then all of its coefficients are positive.

I've tried using Vieta's formulas, as well the Laplace transform, but couldn't do it. Thanks!


A (monic) polynomial with real coefficients can be written as a product of

  • linear factors $x - a$, corresponding to a real zero of $P$, and
  • quadratic factors, corresponding to a pair of complex-conjugate zeroes $a - ib$, $a+ib$ of $P$.

Each quadratic factor is of the form $$ (x - a - ib)(x - a + ib) = (x-a)^2 + b^2 $$

with $b \ne 0$. It follows that if all zeros of $P$ have negative real part then it can be written as a product of monic polynomials with positive coefficients.


Assuming real coefficients . . .

Assume all the roots have negative real part.

For the real roots $r$, if any, the corresponding factor $(x-r)$ has positive coefficients.

For the non-real roots, pair them up as $r,\bar{r}$, and note that the quadratic polynomial $$(x-r)(x-\bar{r})$$ has positive coefficients.

Then just multiply the quadratic factors and the real linear factors.


  • $\begingroup$ Why can $P$ not have a real root? $\endgroup$ – Martin R Oct 8 '17 at 9:59
  • $\begingroup$ Oops ...${}{}{}{}$ $\endgroup$ – quasi Oct 8 '17 at 10:00
  • $\begingroup$ @Martin R: Thanks for the alert. Fixed. $\endgroup$ – quasi Oct 8 '17 at 10:04

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