Bézout matrix

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In mathematics, a Bézout matrix (or Bézoutian) is a special square matrix associated to two polynomials. Such matrices are sometimes used to test the stability of a given polynomial.

1 Definition
2 Examples
3 Properties
4 Applications
5 References

[edit] Definition

Let f(z) and g(z) be two complex polynomials of degree at most n with coefficients (note that any coefficient could be zero):

$f(z)=\sum_{i=0}^n u_i z^i,\quad\quad g(z)=\sum_{i=0}^n v_i z^i.$

The Bézout matrix of order n associated to the polynomials f and g is

B n (f, g) = [b i j] i j

It is in $\mathbb{C}^{n\times n}$ and the entries of that matrix are such that if we note for each i,j=1,...,n, $m i j = min{i, n + 1 - j}$ , then:

$b_{ij}=\sum_{k=1}^{m_{ij}}u_{j+k-1}v_{i-k}-u_{i-k}v_{j+k-1}.$

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

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

[edit] Examples

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

$B_3(f,g)=\left[\begin{matrix}u_1v_0-u_0 v_1 & u_2 v_0-u_0 v_2 & u_3 v_0-u_0 v_3\\u_2 v_0-u_0 v_2 & u_2v_1-u_1v_2+u_3v_0-u_0v_3 & u_3 v_1-u_1v_3\\u_3v_0-u_0v_3 & u_3v_1-u_1v_3 & u_3v_2-u_2v_3\end{matrix}\right].$

Let $f (z) = 3 x 3 - x$ and $g (z) = 5 x 2 + 1$ be 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 (equal 4). The left zeros are due to the fact that for each i=0,...,n, either $u i$ or $v i$ is equal to zero.

[edit] Properties

$B n (f, g)$ is symmetric (as a matrix);
$B n (f, g) = - B n (g, f)$ ;
$B n (f, f) = 0$ ;
$B n (f, g)$ is bilinear in (f,g);
$B n (f, g)$ is in $\mathbb{R}^{n\times n}$ if f and g have real coefficients;
$B n (f, g)$ is nonsingular with $n = m a x (d e g (f), d e g (g))$ if and only if f and g have no common roots.

[edit] 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 note 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 iff $B n (p, q)$ 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.

[edit] References

D. Hinrichsen and A.J. Pritchard, Mathematical Systems Theory I: Modelling, State Space Analysis, Stability and Robustness, Springer-Verlag, Berlin-Heidelberg, 2005

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Categories: Polynomials | Matrices