• The Routh–Hurwitz theorem provides an algorithm for determining if a given polynomial is Hurwitz stable, which is implemented in the Routh–Hurwitz and Liénard–Chipart tests.
• To test if a given polynomial P (of degree d) is Schur stable, it suffices to apply this theorem to the transformed polynomial

${\displaystyle Q(z)=(z-1)^{d}P\left({{z+1} \over {z-1}}\right)}$ 

obtained after the Möbius transformation ${\displaystyle z\mapsto {{z+1} \over {z-1}}}$  which maps the left half-plane to the open unit disc: P is Schur stable if and only if Q is Hurwitz stable and ${\displaystyle P(1)\neq 0}$ . For higher degree polynomials the extra computation involved in this mapping can be avoided by testing the Schur stability by the Schur-Cohn test, the Jury test or the Bistritz test.

• Necessary condition: a Hurwitz stable polynomial (with real coefficients) has coefficients of the same sign (either all positive or all negative).
• Sufficient condition: a polynomial ${\displaystyle f(z)=a_{0}+a_{1}z+\cdots +a_{n}z^{n}}$  with (real) coefficients such that

${\displaystyle a_{n}>a_{n-1}>\cdots >a_{0}>0,}$ 

is Schur stable.

• Product rule: Two polynomials f and g are stable (of the same type) if and only if the product fg is stable.
• Hadamard product: The Hadamard (coefficient-wise) product of two Hurwitz stable polynomials is again Hurwitz stable.^[1]

• ${\displaystyle 4z^{3}+3z^{2}+2z+1}$  is Schur stable because it satisfies the sufficient condition;
• ${\displaystyle z^{10}}$  is Schur stable (because all its roots equal 0) but it does not satisfy the sufficient condition;
• ${\displaystyle z^{2}-z-2}$  is not Hurwitz stable (its roots are −1 and 2) because it violates the necessary condition;
• ${\displaystyle z^{2}+3z+2}$  is Hurwitz stable (its roots are −1 and −2).
• The polynomial ${\displaystyle z^{4}+z^{3}+z^{2}+z+1}$  (with positive coefficients) is neither Hurwitz stable nor Schur stable. Its roots are the four primitive fifth roots of unity

${\displaystyle z_{k}=\cos \left({{2\pi k} \over 5}\right)+i\sin \left({{2\pi k} \over 5}\right),\,k=1,\ldots ,4\,.}$ 

Note here that

${\displaystyle \cos({{2\pi }/5})={{{\sqrt {5}}-1} \over 4}>0.}$ 

It is a "boundary case" for Schur stability because its roots lie on the unit circle. The example also shows that the necessary (positivity) conditions stated above for Hurwitz stability are not sufficient.

Just as stable polynomials are crucial for assessing the stability of systems described by polynomials, stability matrices play a vital role in evaluating the stability of systems represented by matrices.

Hurwitz matrix edit

A square matrix A is called a Hurwitz matrix if every eigenvalue of A has strictly negative real part.

Schur matrix edit

Schur matrices is an analogue of the Hurwitz matrices for discrete-time systems. A matrix A is a Schur (stable) matrix if its eigenvalues are located in the open unit disk in the complex plane.

• Kharitonov region
• Stability criterion
• Stability radius

• Mathworld page