Bézout matrix

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

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

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

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

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

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

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

${\displaystyle b_{ij}=\sum _{k=0}^{m_{ij}}(u_{j+k+1}v_{i-k}-u_{i-k}v_{j+k+1}).}$

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

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

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

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

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

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

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 ${\displaystyle i=0,\dots ,n}$, either ${\displaystyle u_{i}}$ or ${\displaystyle v_{i}}$ is zero.

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

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 ${\displaystyle B_{n}(p,q)}$. 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 ${\displaystyle B_{n}(p,q)}$ 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.

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• Pan, Victor; Bini, Dario (1994). Polynomial and matrix computations. Basel, Switzerland: Birkhäuser. ISBN 0-8176-3786-9.
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