\section{The Routh-Hurwitz Stability Criterion}
In this section we discuss the problem of determining if the zeros of an $n$th degree polynomial are located in the stable half-plane $\Re s <0$. Consider an $n$th degree polynomial
\[
p(s)=a_ns^n+a_{n-1}s^{n-1}+\cdots+a_1s+a_0.
\]
Associated with the polynomial $p(s)$ is the Hurwitz matrix
\[
\b H=\left[\begin{array}{ccccc}
a_{n-1} & a_{n-3} & a_{n-5} & a_{n-7} & \cdots\\
a_n & a_{n-2} & a_{n-4} & a_{n-6} & \cdots\\
0 & a_{n-1}& a_{n-3} & a_{n-5} & \cdots\\
0 & a_{n} & a_{n-2} & a_{n-4} & \cdots\\
\vdots&{}&{}&\vdots&\ddots
\end{array}\right]\in \mb{\Bb R}^{n \times n}.
\]
In our further discussion we assume, without loss of generality, that $a_n=1$. We denote the leading principal minors of the Hurwitz matrix by
\begin{eqnarray*}
H_1 &=& a_{n-1},\quad H_2=\det \left[\begin{array}{cc}
a_{n-1} & a_{n-3}\\
1 & a_{n-2}
\end{array}\right],\quad H_3=\det \left[\begin{array}{ccc}
a_{n-1} & a_{n-3}& a_{n-5}\\
1 & a_{n-2}& a_{n-4}\\
0 & a_{n-1} & a_{n-3}
\end{array}\right],\ldots \\
& & H_n=\det \b H.
\end{eqnarray*}
We next define
\[
b_1=H_1,\quad b_2=\frac{H_2}{H_1},\quad b_3=\frac{H_3}{H_2H_1},
\]
and
\[
b_k=\frac{H_{k-3}H_k}{H_{k-2}H_{k-1}}\quad \mb{for}\quad k=4,5,\ldots,n.
\]
Thus, for example, $b_4=\frac{H_1H_4}{H_2H_3}$. Let
\[
\b B=\left[\begin{array}{cccccccc}
0 & 1 & 0 &0&\cdots& 0 & 0&0\\
-b_n & 0 & 1 &0& \cdots& 0 & 0&0\\
0 & -b_{n-1}& 0 & 1 & \cdots & 0 & 0&0\\
\vdots& {} & {}& {}&{}&{}&{}&\vdots\\
0&0&0&0&\cdots&-b_3&0&1\\
0&0&0&0&\cdots&0&-b_2&-b_1
\end{array}\right]\in \mb{\Bb R}^{n \times n}.
\]
For example, for $n=4$ we have
\[
\b B=\left[\begin{array}{cccc}
0 & 1 & 0 & 0\\
-b_4 & 0 & 1 & 0\\
0 & -b_3 & 0 & 1\\
0 & 0 & -b_2 & -b_1
\end{array}\right].
\]
Let
\[
\b A=\left[\begin{array}{cccccc}
0 & 1 & 0 &\cdots & 0 & 0\\
0 & 0 & 1& \cdots & 0 & 0\\
\vdots& {}& {}& {}& {}& \vdots\\
0 & 0 & 0 & \cdots & 0 & 1\\
-a_0& -a_1 & -a_2&\cdots&-a_{n-2}&-a_{n-1}
\end{array}\right]\in \mb{\Bb R}^{n \times n}.
\]
\begin{lemma}
The matrices $\b B$ and $\b A$ are similar, that is, there is a nonsingular matrix $\b T\in \mb{\Bb R}^{n \times n}$ such that
\[
\b B=\b T^{-1}\b A\b T.
\]
\end{lemma}
{\bf Proof}\quad The similarity matrix is a lower triangular matrix of the form
\[
\b T=\left[\begin{array}{cccccccc}
1 & 0 & 0 & 0&0&0&\cdots & 0\\
0 & 1 & 0 & 0 & 0&0&\cdots & 0\\
t_{31}& 0 & 1&0&0&0&\cdots&0\\
0&t_{42}&0&1&0&0&\cdots&0\\
t_{51}&0&t_{53}&0&1&0&\cdots&0\\
0&t_{62}&0&t_{64}&0&1&\cdots&0\\
\vdots&{}&{}&{}&{}&{}&{}&\vdots
\end{array}\right].
\]
To determine the nonfixed coefficients $t_{ij}$ of the matrix $\b T$ we represent the equation $\b B=\b T^{-1}\b A\b T$ in the eqiuvalent form
\[
\b T\b B=\b A\b T,
\]
and compare both sides of it to determine the coefficients $t_{ij}$. We obtain, for example, $t_{31}=-b_n$, $t_{42}=-b_n-b_{n-1}$, etc.\\
\hspace*{5.5in} $\Box$

Let $\b P$ be an $n\times n$ diagonal matrix defined as
\[
\b P=\mb{diag}\left[\begin{array}{ccccc}
b_1b_2\cdots b_n, & b_1b_2\cdots b_{n-1}, & \ldots & b_1b_2, & b_1
\end{array}\right],
\]
and $\b Q$ be an $n\times n$ diagonal matrix defined as
\[
\b Q=2 b_1^2\left[\begin{array}{ccccc}
0 & 0 & \ldots & 0 & 1
\end{array}\right].
\]
We represent the matrix $\b Q$  as $\b Q=\b c^T\b c$, where
\[
\b c=\sqrt{2} b_1\left[\begin{array}{ccccc}
0 & 0 & \ldots & 0 & 1
\end{array}\right].
\]
Note that the pair $\left( \b B,\b c\right)$ is observable. The matrices $\b B$, $\b P$, and $\b Q$ satisfy the Lyapunov matrix equation, that is,
\[
\b B^T\b P+\b P\b B=-\b Q.
\]
By Theorem~\ref{thm:ver_lyap} on page~\pageref{thm:ver_lyap} applied to the above equation, the matrix $\b B$ is asymptotically stable if, and only if, the symmetric matrix $\b P$ is positive definite. The matrix $\b P$ is positive definite if, and only if, its diagonal elements are positive. Using the above results we can easily prove the following theorem:
\begin{thm}[Hurwitz]
The real polynomial $p(s)$ of degree $n$ with positive coefficients $a_i$, has its zeros in the open half plane if, and only if, the leading principal minors of the Hurwitz matrix are positive.
\end{thm}
{\bf Proof}\quad The polynomial $p(s)$ has its zeros in the open left plane if, and only if, the matrix $\b B$ is asymptotically stable. This in turn is true if, and only if, diagonal elements of $\b P$ are positive. We have
\[
b_1=H_1,\quad b_1b_2=H_2,
\]
and for $k=3,\ldots,n$,
\[
b_1b_2\cdots b_k=\frac{H_k}{H_{k-2}}.
\]
Thus, the diagonal elements of $\b P$ are all positive if, and only if, the leading principal minors of the Hurwitz matrix are positive, which completes the proof of the theorem.\\
\hspace*{5.5in} $\Box$