2
$\begingroup$

I'm reading about resultant theory from Peter Stiller's notes (Link to pdf). The author mentions that, given two univariate polynomials $$f(x)=a_r x^r+ ...+a_1 x+ a_0, \qquad g(x)=b_s x^s+ ...+b_1 x+ b_0,$$ there are common roots iff their resultant $R_{r,s}(f,g)$ is zero, where the resultant is defined as the determinant of the Sylvester matrix of the two polynomials.

For example, for $r=s=2$, the resultant reads $$R_{2,2}(f,g) = a_0^2 b_2^2 + a_0 a_2 b_1^2 - a_0 a_1 b_1 b_2 + a_1^2 b_0 b_2 - a_1 a_2 b_0 b_1 + a_2^2 b_0^2 - 2 a_0 a_2 b_0 b_2.$$ I'm trying to get a better understanding of where this expression comes from.

I understand that the resultant is what should come out imposing the two polynomials to have common roots. However, if I simply multiply $f(x)$ by $b_0/a_0$ (assuming $a_0\neq 0$) and remove the corresponding common term in both polynomials, I get a first order polynomial in $x$. Similarly if I remove the $x^2$ term from the two polynomials.

From these first-order polynomials, I can easily deduce the solutions to the system. How would one combine these solutions to obtain the expression for the resultant?


Consider two quadratic polynomials $a(x)=a_0 + a_1 x + a_2 x^2$ and $b(x)=b_0 + b_1 x + b_2 x^2$. Performing the Euclidean algorithm we get, $$a(x) = q_0 b(x) + (r_0 + r_1 x), \qquad q_0 = \frac{a_2}{b_2}, \\ r_0 = a_0 - q_0 b_0 = a_0 - \frac{a_2}{b_2} b_0, \qquad r_1 = a_1 - q_0 b_1 = a_1 - \frac{a_2}{b_2}b_1.$$ Dividing $b(x)$ by $r(x)$ we then get $$b(x) = (q_0' + q_1' x) (r_0 + r_1 x) + r_0', \qquad q_1' = \frac{b_2}{r_1} = \frac{b_2^2}{a_1 b_2 - a_2 b_1}, \\ q_0' = \frac{1}{r_1}\left[ b_1 - q_1' r_0 \right] = \frac{1}{r_1}\left[ b_1 - q_1' r_0 \right], \qquad r_0' = b_0 - q_0' r_0.$$

If $a$ and $b$ have a common root, then we need $r_0'=0$, as each remainder must be a multiple of the GCD of the polynomials. Thus we get the condition $$r_0' = b_0 - q_0' r_0 = 0.$$ Replacing the known expressions for $q_0'$ and $r_0$ we do indeed get a condition on the coefficients $a_i, b_i$, but this hardly seems like the best approach to the problem. I don't see how the Sylvester matrix is supposed to arise, and it doesn't seem prone to easy generalisation.

$\endgroup$
3
  • 1
    $\begingroup$ You can compute the common factor using the Euclidean algorithm (for polynomial division), which is in effect what you have started to do. Carefully done, this simply gives the same result. $\endgroup$ Commented Apr 27, 2021 at 16:03
  • $\begingroup$ @MarkBennet continuing with the Euclidean algorithm I do get conditions on the coefficients, see the edit, but this hardly seems like the best approach. How would I generalise this way of proceeding for more general case? And I also don't see where the Sylvester matrix emerges $\endgroup$
    – glS
    Commented Apr 28, 2021 at 16:33
  • $\begingroup$ I suggested the Euclidean algorithm as the way to proceed from your starting point. The point of the Sylvester matrix is that it is easily written down and easily generalised. However that does not necessarily mean that the determinant is simple to determine. Both the Euclidean algorithm and the Sylvester matrix involve systematic computation. The Euclidean algorithm could be trick if there were two roots in common (with higher degree) - it doesn't seem readily adapted to degenerate cases. $\endgroup$ Commented Apr 28, 2021 at 19:01

1 Answer 1

2
$\begingroup$

I'm trying to get a better understanding of where this expression [the resultant] comes from

For the case of two quadratics, a common root means the following equations are satisfied for at least one value of $x$. The system includes the two original equations, plus the same equations multiplied by $x$.

$$ \begin{cases} a_2 x^3 + & a_1 x^2 + & a_0 x && = 0 \\ & a_2 x^2 + & a_1 x + & a_0 & = 0 \\ b_2 x^3 + & b_1 x^2 + & b_0 x && = 0 \\ & b_2 x^2 + & b_1 x + & b_0 & = 0 \\ \end{cases} $$

Consider this as a system of linear equations in $x_0 = 1, x_1 = x, x_2 = x^2, x_3 = x^3$.

$$ \begin{cases} a_2 \cdot x_3 & +\; a_1 \cdot x_2 & +\; a_0 \cdot x_1 + & \,0 \cdot x_0 & = 0 \\ \,0 \cdot x_3 & +\; a_2 \cdot x_2 & +\; a_1 \cdot x_1 + & a_0 \cdot x_0 & = 0 \\ b_2 \cdot x_3 & +\; b_1 \cdot x_2 & +\; b_0 \cdot x_1 + & \,0 \cdot x_0 & = 0 \\ \,0 \cdot x_3 & +\; b_2 \cdot x_2 & +\; b_1 \cdot x_1 + & b_0 \cdot x_0 & = 0 \\ \end{cases} $$

This is a linear homogeneous system which has the non-trivial solution $(1, x, x^2, x^3)$ so its matrix must be singular, which happens to be precisely the Sylvester matrix. This amounts to the following determinant being zero.

$$ \left| \begin{matrix} \;a_2 \;&\; a_1 \;&\; a_0 \;&\; 0 \\ \;0 \;&\; a_2 \;&\; a_1 \;&\; a_0 \\ \;b_2 \;&\; b_1 \;&\; b_0 \;&\; 0 \\ \;0 \;&\; b_2 \;&\; b_1 \;&\; b_0 \\ \end{matrix} \right| \;\;=\;\; 0 $$

Expanding the determinant results in the expression posted for $R_{2,2}(f,g)$.

In the general case of polynomials $f$ of degree $r$ and $g$ of degree $s$, multiplying the equation $f(x) = 0$ by $x, x^2, \dots, x^{s-1}$ and $g(x) = 0$ by $x, x^2, \dots, x^{r-1}$ gives a system of $r+s$ equations with a similar Sylvester matrix.

Performing the Euclidean algorithm we get [...]

Looking at it from another angle, the first step of the euclidean division eliminates the $x^2$ term between equations, then the second step eliminates the $x$ term. In the end, this is solving the same equations by successive elimination, instead of determinants, but it becomes more laborious to do for higher degrees.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .