Bézout matrix

In mathematics, a Bézout matrix (or Bézoutian or Bezoutiant) is a special square matrix associated with two polynomials, introduced by James Joseph Sylvester in 1853 and Arthur Cayley in 1857 and named after Étienne Bézout.^[1]^[2] Bézoutian may also refer to the determinant of this matrix, which is equal to the resultant of the two polynomials. Bézout matrices are sometimes used to test the stability of a given polynomial.

Definition

Let $f(z)$ and $g(z)$ be two complex polynomials of degree at most n,

f(z)=\sum _{i=0}^{n}u_{i}z^{i},\qquad g(z)=\sum _{i=0}^{n}v_{i}z^{i}.

(Note that any coefficient $u_{i}$ or $v_{i}$ could be zero.) The Bézout matrix of order n associated with the polynomials f and g is

B_{n}(f,g)=\left(b_{ij}\right)_{i,j=0,\dots ,n-1}

where the entries $b_{ij}$ result from the identity

{\frac {f(x)g(y)-f(y)g(x)}{x-y}}=\sum _{i,j=0}^{n-1}b_{ij}\,x^{i}\,y^{j}.

It is an n × n complex matrix, and its entries are such that if we let $\ell _{ij}=\max\{0,i-j\}$ and $m_{ij}=\min\{i,n-1-j\}$ for each $i,j=0,\dots ,n-1$ , then:

b_{ij}=\sum _{k=\ell _{ij}}^{m_{ij}}(u_{j+k+1}v_{i-k}-u_{i-k}v_{j+k+1}).

To each Bézout matrix, one can associate the following bilinear form, called the Bézoutian:

\operatorname {Bez} :\mathbb {C} ^{n}\times \mathbb {C} ^{n}\to \mathbb {C} :(x,y)\mapsto \operatorname {Bez} (x,y)=x^{*}B_{n}(f,g)\,y.

Examples

For n = 3, we have for any polynomials f and g of degree (at most) 3:

B_{3}(f,g)=\left[{\begin{matrix}u_{1}v_{0}-u_{0}v_{1}&u_{2}v_{0}-u_{0}v_{2}&u_{3}v_{0}-u_{0}v_{3}\\u_{2}v_{0}-u_{0}v_{2}&u_{2}v_{1}-u_{1}v_{2}+u_{3}v_{0}-u_{0}v_{3}&u_{3}v_{1}-u_{1}v_{3}\\u_{3}v_{0}-u_{0}v_{3}&u_{3}v_{1}-u_{1}v_{3}&u_{3}v_{2}-u_{2}v_{3}\end{matrix}}\right]\!.

Let $f(x)=3x^{3}-x$ and $g(x)=5x^{2}+1$ be the two polynomials. Then:

B_{4}(f,g)=\left[{\begin{matrix}-1&0&3&0\\0&8&0&0\\3&0&15&0\\0&0&0&0\end{matrix}}\right]\!.

The last row and column are all zero as f and g have degree strictly less than n (which is 4). The other zero entries are because for each $i=0,\dots ,n$ , either $u_{i}$ or $v_{i}$ is zero.

Properties

$B_{n}(f,g)$ is symmetric (as a matrix);
$B_{n}(f,g)=-B_{n}(g,f)$ ;
$B_{n}(f,f)=0$ ;
$(f,g)\mapsto B_{n}(f,g)$ is a bilinear function;
$B_{n}(f,g)$ is a real matrix if f and g have real coefficients;
$B_{n}(f,g)$ is nonsingular with $n=\max(\deg(f),\deg(g))$ if and only if f and g have no common roots.
$B_{n}(f,g)$ with $n=\max(\deg(f),\deg(g))$ has determinant which is the resultant of f and g.

Applications

An important application of Bézout matrices can be found in control theory. To see this, let f(z) be a complex polynomial of degree n and denote by q and p the real polynomials such that f(iy) = q(y) + ip(y) (where y is real). We also denote r for the rank and σ for the signature of $B_{n}(p,q)$ . Then, we have the following statements:

f(z) has n − r roots in common with its conjugate;
the left r roots of f(z) are located in such a way that:
- (r + σ)/2 of them lie in the open left half-plane, and
- (r − σ)/2 lie in the open right half-plane;
f is Hurwitz stable if and only if $B_{n}(p,q)$ is positive definite.

The third statement gives a necessary and sufficient condition concerning stability. Besides, the first statement exhibits some similarities with a result concerning Sylvester matrices while the second one can be related to Routh–Hurwitz theorem.