☞
**((users:au:index_e Author))**
☞
**((:dets:resultant Русская версия))**
!!⊗!! The page is under reconstruction; repair works: 18.12.2015 --- ??.??.????
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To understand the stuff of the present page, it would be desirable to remind some prerequisites in polynomial theory and determinant formalism.
==Resultant==
~~TOC~~
Denote by $ \mathbb A_{} $ any of the sets $ \mathbb Z, \mathbb Q, \mathbb R_{} $ or $ \mathbb C_{} $.
===Resultant in the Sylvester form==
For the polynomials $ f(x)=a_{0}x^n+a_1x^{n-1}+\dots+a_n $ and $ g(x)=b_{0}x^m+b_1x^{m-1}+\dots+b_m $ при $ a_{0}\ne 0, b_0\ne 0 $ consider the square matrix of the order $ m+n_{} $
$$
M=\left(\begin{array}{cccccccccc}
a_0&a_1&a_2&\ldots&\ldots&a_n&0&\dots &0 &0\\
0&a_0&a_1&\ldots&\ldots&a_{n-1}&a_n&\dots&0 &0\\
\vdots&&\ddots&&&&&&\ddots\\
0&0&\ldots&a_0&\ldots&\ldots & & \ldots &a_{n-1} &a_n\\
0&0&\ldots&&b_0&b_1&\ldots& \ldots &b_{m-1}&b_m\\
0&0&\ldots&b_0&b_1&\ldots &&\ldots &b_m&0\\
\vdots&&&\ldots&&&& &&\vdots\\
0&b_0&\ldots&\ldots&&b_m&\ldots&&\ldots&0\\
b_0&\ldots&\ldots&&b_m&0&\ldots&&&0
\end{array}\right)
\begin{array}{l}
\left.\begin{array}{l}
\\ \\ \\ \\
\end{array}\right\} m
\\
\left. \begin{array}{l}
\\ \\ \\ \\ \\
\end{array}\right\} n
\end{array}
$$
(the entries above $ a_{n} $ and $ b_{0} $, and below $ a_{0} $ and $ b_{m} $ equal to zero). The expression
$$
{\mathcal R}(f,g)= (-1)^{n_{}(n-1)/2} \det M
$$
is called **the resultant**[[Resultant (//Lat.//) --- reflective; the notion was introduced by Laplace in 1776.]] (**in the Sylvester form**) of the polynomials $ f_{} $ and $ g_{} $ .
In the majority of textbooks, the definition of the resultant is presented in a slightly different manner: the rows of the coefficients of the polynomial $ g_{}(x) $ are reordered in such a way that form the staircase ri sing up from the lower rigt corner. Thus, for instance,
$$
{\mathcal R}(a_0x^3+a_1x^{2}+a_2x+a_3,\ b_0x^3+b_1x^{2}+b_2x+b_3)=
$$
$$
=
\left|\begin{array}{cccccc}
a_0&a_1&a_2&a_3 & &\\
& a_0&a_1&a_2&a_3 & \\
& & a_0&a_1&a_2&a_3 \\
b_0&b_1&b_2&b_3 & &\\
& b_0&b_1&b_2&b_3 & \\
& & b_0&b_1&b_2&b_3
\end{array}\right| \ .
$$
Such a modification gives one an economy in the sign $ (-1)^{n_{}(n-1)/2} $ involved into the definition. Nevertheless, due to some reasons soon to be explained
☟
((#subresultants_and_the_greatest_common_divisor BELOW)), it is more convenient for me to utilize the above presented form of the matrix $ M_{} $.
!!Ex!! **Example.**
$$
{\mathcal R}(a_0x^2+a_1x+a_2,\ b_0x^2+b_1x+b_2)
=(a_0b_2-a_2b_0)^2-(a_0b_1-a_1b_0)(a_1b_2-a_2b_1)
$$
!!Th!! **Theorem.** //Denote the zeros of polynomial// $ f_{}(x) $ //by// $ \lambda_{1},\dots,\lambda_n $ ,// and the zeros of polynomial //$ g_{}(x) $ //by[[For the both cases, in accordance with their multiplicities.]] //$ \mu_{1},\dots,\mu_m $. //One has//
$$
{\mathcal R}(f,g)= a_0^m b_0^n \prod_{k=1}^m \prod_{j=1}^n (\lambda_j - \mu_k) \ .
$$
!!=>!!
$$
{\mathcal R}(f,g)= a_0^m \prod_{j=1}^n g(\lambda_j)= (-1)^{mn} b_0^n \prod_{k=1}^m f(\mu_k) \ .
$$
!!=>!! $ {\mathcal R}(f,g)=0_{} $ if and only if the polynomials $ f_{}(x) $ and $ g_{}(x) $ possess a common zero.
!!Ex!! **Example.** Find all the values for the parameter $ {\color{Red} \alpha } $ for which the polynomials
$$ x^{3}+{\color{Red} \alpha }\,x+1 \ \mbox{ and } \ x^{2}+{\color{Red} \alpha }\,x+1 $$
possess a common zero.
**Solution.** Compute the determinant with the aid of elementary row operations
$$
{\mathcal R}(x^3+{\color{Red} \alpha }\,x+1,\ x^2+{\color{Red} \alpha }\,x+1)= (-1)^{3\cdot 2/2}\left| \begin{array}{ccccc}
1 & 0 & {\color{Red} \alpha } & 1 & 0 \\
0 & 1 & 0 & {\color{Red} \alpha } & 1 \\
0 & 0 & 1 & {\color{Red} \alpha } & 1 \\
0 & 1 & {\color{Red} \alpha } & 1 & 0 \\
1 & {\color{Red} \alpha } & 1 & 0 & 0
\end{array}
\right| =-
\left| \begin{array}{ccccc}
1 & 0 & {\color{Red} \alpha } & 1 & 0 \\
0 & 1 & 0 & {\color{Red} \alpha } & 1 \\
0 & 0 & 1 & {\color{Red} \alpha } & 1 \\
0 & 0 & {\color{Red} \alpha } & 1-{\color{Red} \alpha } & -1 \\
0 & {\color{Red} \alpha } & 1-{\color{Red} \alpha } & -1 & 0
\end{array}
\right|=
$$
(the first row is subtracted from the last one, while the second row from the last-but-one), expand the resulting determinant by the last column:
$$
=-\left| \begin{array}{cccc}
1 & 0 & {\color{Red} \alpha } & 1 \\
0 & 1 & {\color{Red} \alpha } & 1 \\
0 & {\color{Red} \alpha } & 1-{\color{Red} \alpha } & -1 \\
{\color{Red} \alpha } & 1-{\color{Red} \alpha } & -1 & 0
\end{array}
\right|=-
\left| \begin{array}{cccc}
1 & 0 & {\color{Red} \alpha } & 1 \\
0 & 1 & {\color{Red} \alpha } & 1 \\
0 & {\color{Red} \alpha } & 1-{\color{Red} \alpha } & -1 \\
0 & 1-{\color{Red} \alpha } & -1-{\color{Red} \alpha }^2 & -{\color{Red} \alpha }
\end{array}
\right|=
$$
(the first row multiplied by $ {\color{Red} \alpha } $ is subtracted from the last one), expand the determinant by the first column:
$$
=-\left| \begin{array}{ccc}
1 & {\color{Red} \alpha } & 1 \\
{\color{Red} \alpha } & 1-{\color{Red} \alpha } & -1 \\
1-{\color{Red} \alpha } & -1-{\color{Red} \alpha }^2 & -{\color{Red} \alpha }
\end{array}
\right|=
-\left| \begin{array}{ccc}
1 & {\color{Red} \alpha } & 1 \\
{\color{Red} \alpha }+1 & 1 & 0 \\
1 & -1 & 0
\end{array}
\right|=
$$
(the first row is added to the second one, the first row multiplied by $ {\color{Red} \alpha } $ is added to the third one),
expand the determinant by the last column:
$$
=-\left| \begin{array}{cc}
{\color{Red} \alpha }+1 & 1 \\
1 & -1
\end{array}
\right|=-(-({\color{Red} \alpha }+1)-1)={\color{Red} \alpha }+2 \ .
$$
**Answer.** $ {\color{Red} \alpha }=-2_{} $.
**Check.** $ x^{3}-2\,x+1\equiv (x-1)(x^2+x-1),\ x^2-2\,x+1\equiv (x-1)(x+1) $.
!!?!! Find all the values for the parameter $ {\color{Red} \alpha } $ for which the polynomials
$$ x^{3}+{\color{Red} \alpha }+\,x^2-14 \quad and \quad x^{3}+{\color{Red} \alpha }+\,x-14 $$
possess a multiple zero.
!!§!! For the particular case $ g_{}(x)\equiv f^{\prime}(x) $, the resultant is transformed into ((detse:discrime#discriminant discriminant)).
!!!!! The main application of the resultant
☞
((:detse:resultante#elmination_of_variables_in_a_system_of_polynomial_equations
elimination of variables in a system of polynomial equations )).
===Properties==
1.
If $ \deg f(x) = n_{} $ and $ \deg g(x)=m_{} $, then
$$ {\mathcal R}(f,g)=(-1)^{nm}{\mathcal R}(g,f) \ . $$
2.
$$
{\mathcal R}\left(f_1\cdot f_2,\, g\right)={\mathcal R}(f_1,\,g)\cdot {\mathcal R}(f_2,\, g)
$$
3.
$${\mathcal R}(Af(x)+Bg(x),Cf(x)+Dg(x))=(AD-BC)^n{\mathcal R}\left(f(x),g(x)\right)$$
The last equality is valid under an additional assumption that
$$ \deg f= n\ge m =\deg g \ge 1 \quad and \quad \deg (Af(x)+Bg(x))=\deg (Cf(x)+Dg(x))= n \ .$$
4.
If $ f(x)=a_{0}x^n + a_1x^{n-1} + \dots + a_n $ and $ g(x)=b_{0}x^m+b_1x^{m-1}+\dots+b_m $,
and $ a_{0}\ne0,a_n\ne 0,b_0\ne 0,b_m\ne 0 $ then
$$ {\mathcal R}(f,g) = (-1)^{mn}{\mathcal R}(a_0+a_1x+\dots+a_nx^n,\ b_0+b_1x+\dots+b_mx^m) \ .$$
===Resultant as a polynomial function of the coefficients==
By definition, resultant is a polynomial over $ \mathbb Z $ with respect to the coefficients of polynomials $ f_{}(x) $ and $ g_{}(x) $:
$$ {\mathcal R}(a_0 x^n+ \ldots + a_{n}, b_0x^m + \ldots + b_m)\equiv R(a_0,\dots,a_n,b_0,\dots,b_m) \in {\mathbb Z}[a_0,\dots,a_n,b_0,\dots,b_m]
\ .$$
Its degree equals $ mn_{} $. Treated with respect to $ a_0,\dots,a_{n} $, the resultant is a homogeneous polynomial of the degree $ m_{} $; treated with respect to $ b_{0},\dots,b_m $
the resultant is a homogeneous polynomial of the degree $ n_{} $.
!!Th!! **Theorem.** //If polynomials// $ f_{}(x) $ and $ g_{}(x) $ //possess a unique common root// $ \lambda_{} $ //then//
$$ \lambda^j : \lambda^k = \frac{\partial R}{\partial a_{n-j-\ell}} :
\frac{\partial R}{\partial a_{n-k-\ell}} = \frac{\partial R}{\partial b_{m-j-q}} :
\frac{\partial R}{\partial b_{m-k-q}} $$
//for any// $ \{j,k,\ell,q_{} \}\subset \{0,1,2,\dots \} $.
**Proof.** Consider the case $ j=0, k=1 $. Differentiate the equality for the resultant
$$
{\mathcal R}(f,g)= a_0^m \prod_{j=1}^n g(\lambda_j)
$$
with respect to $ b_{q} $. One gets:
$$
\frac{\partial R}{\partial b_{q}}= \sum_{j=1}^n \frac{\partial g(\lambda_j)}{\partial b_{q}} \cdot \prod_{j=1 \atop j\ne q}^n g(\lambda_j) =
\sum_{j=1}^n \lambda_j^{m-q} \prod_{j=1 \atop j\ne q}^n g(\lambda_j) \ .
$$
If polynomials $ f_{}(x) $ and $ g_{}(x) $ possess a unique common root $ \lambda_1 $ then the last equality is transformed into
$$
\frac{\partial R}{\partial b_{q}}=\lambda_1^{m-q} \prod_{j=2}^n g(\lambda_j)
$$
and the product $ \prod $ does not vanish. Since the equality is valid for any specializations of $ q_{} $, one has
$$
\frac{\partial R}{\partial b_{q-1}} \Bigg/ \frac{\partial R}{\partial b_{q}}= \lambda_1 \ ,
$$
and the other equality from the theorem is proved similarly.
♦
===Subresultants and the greatest common divisor==
Consider the matrix $ M_{} $ from the ((#resultant_in_the_sylvester_form definition of the resultant)) and delete its first and last columns as well as the first and the last rows. We get a square matrix of the order $ m_{}+n-2 $. Its determinant
$$
{\mathcal R}^{(1)}(f,g)=
\left|\begin{array}{cccccccc}
a_0&a_1&\ldots&\ldots&a_{n-1}&a_n&\dots&0 \\
\vdots&\ddots&&&&&\ddots\\
0&\ldots&a_0&\ldots&\ldots & & \ldots &a_{n-1} \\
0&\ldots&&b_0&b_1&\ldots& \ldots &b_{m-1}\\
0&\ldots&b_0&b_1&\ldots &&\ldots &b_m\\
\vdots&&\ldots&&&& &\vdots\\
b_0&\ldots&\ldots&&b_m&\ldots&&0
\end{array}\right|
\begin{array}{l}
\left.\begin{array}{l}
\\ \\ \\
\end{array}\right\} m-1
\\
\left. \begin{array}{l}
\\ \\ \\ \\
\end{array}\right\} n-1
\end{array}
$$
is called the **first subresultant** of the resultant $ {\mathcal R}(f_{},g) $.
!!Th!! **Theorem.** //For the polynomials// $ f_{}(x) $ //and// $ g_{}(x) $ //to possess a unique common root, it is necessary and sufficient that the following conditions be satisfied://
$${\mathcal R}(f,g)=0,\ {\mathcal R}^{(1)}(f,g)\ne 0\ .$$
!!=>!! Under the conditions of the previous theorem, the single common root of $ f_{}(x) $ and $ g_{}(x) $ can be expressed as a rational function of the coefficients of these polynomials:
$$
\lambda=-{\det M_1^{(1)}\over {\mathcal R}^{(1)}(f,g)} .
$$
Here the $ M_1^{(1)} $ denotes the matrix obtained from $ M_{} $ by deleting its first and last rows as well as its first and its __last but one__ column.
!!Ex!! **Example.** Under the conditions of the theorem, the single common root of the polynomials
$$
f(x)=a_0x^5+a_1x^4+a_2x^3+a_3x^2+a_4x+a_5 \quad \mbox{ and } \quad g(x)=b_0x^3+b_1x^2+b_2x+b_3
$$
($ a_{0} \ne 0, b_0 \ne 0 $) can be expressed by the formula:
$$ \lambda =-
\left|
\begin{array}{cccccc}
a_0&a_1&a_2&a_3&a_4&0\\
0&a_0&a_1&a_2&a_3&a_5\\
0&0&0&b_0&b_1&b_3\\
0&0&b_0&b_1&b_2&0\\
0&b_0&b_1&b_2&b_3&0\\
b_0&b_1&b_2&b_3&0&0
\end{array}\right| \Bigg/
\left|\begin{array}{cccccc}
a_0&a_1&a_2&a_3&a_4&a_5\\
0&a_0&a_1&a_2&a_3&a_4\\
0&0&0&b_0&b_1&b_2\\
0&0&b_0&b_1&b_2&b_3\\
0&b_0&b_1&b_2&b_3&0\\
b_0&b_1&b_2&b_3&0&0
\end{array} \right| \ .
$$
♦
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The determinant of the matrix $ M_{k} $ obtained from the matrix $ M_{} $ by deleting its first $ k_{} $ columns and its $ k_{} $ last columns, its first
$ k_{} $ rows and its $ k_{} $ last rows, is called the $ {\mathbf k} $**th subresultant** of the resultant
of the polynomials $ f_{} $ and $ g_{} $; we will denote it as $ {\mathcal R}^{(k)}(f,g) $. For simplicity, we term by the
**zero subresultant** the determinant of the matrix $ M_{} $:
$$ {\mathcal R}^{(0)}(f,g)= \det M =(-1)^{n(n-1)/2}{\mathcal R}(f,g) . $$
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!!Ex!! **Example.** For the polynomials $ f(x)=a_{0}x^5+a_1x^4+a_2x^3+a_3x^2+a_4x+a_5 $ and $ g(x)=b_{0}x^3+ b_1x^2+b_2x+b_3 $ one has:
$$
{\mathcal R}^{(2)}(f,\ g)=
\left|\begin{array}{cccc}
a_0&a_1&a_2&a_3\\
0&0&b_0&b_1\\
0&b_0&b_1&b_2\\
b_0&b_1&b_2&b_3
\end{array} \right| ;\
{\mathcal R}^{(3)}(f,\ g)
=\left|\begin{array}{cc}
0&b_0\\
b_0&b_1
\end{array}\right| = -b_0^2 .
$$
♦
!!Th!! **Theorem.** //Polynomials// $ f_{}(x) $ //and// $ g_{}(x) $ //possess the greatest common divisor of the degree// $ k>0 $, //it is necessary and sufficient the fulfillment of the following conditions//
$${\mathcal R}(f,g)=0,\ {\mathcal R}^{(1)}(f,g)= 0,\ \dots , {\mathcal R}^{(k-1)}(f,g)= 0, {\mathcal R}^{(k)}(f,g)\ne 0.$$
!!=>!! Under the conditions of the previous theorem, the greatest common divisor for the polynomials
$ f_{}(x) $ and $ g_{}(x) $ can be represented in the form
$$
\operatorname{GCD} (f(x),g(x)) \equiv x^k{\mathcal R}^{(k)}(f,g) +x^{k-1} \det M_k^{(1)} +\ldots +\det M_k^{(k)} .
$$
Here $ M_k^{(j)} $ stands for the matrix obtained from the subresultant $ {\mathcal R}^{(k)}(f,g) $ by the replacement of its last column with
$$
[a_{m+n-2k+j-1}, a_{m+n-2k+j-2},\dots, a_{n-k+j},b_{m-k+j},b_{m-k+j+1},\dots,b_{m+n-2k+j-1}]^{\top}
$$
(we set here $ a_{K}=0 $ for $ K>n_{} $ and $ b_{L}=0 $ for $ L>m_{} $).
!!Ex!! **Example.** Find the greatest common divisor for the polynomials
$$ f(x)= x^6-x^5+3\,x^{3}-2\,x^2+1 \quad and \quad g(x)=x^{5}+x^3+x^2+2\,x+1 \ . $$
**Solution.** Omitting the intermediate computations, we present the final result:
$$
{\mathcal R}(f,g)=0,\ {\mathcal R}^{(1)}(f,g)= 0,\ {\mathcal R}^{(2)}(f,g)= 0, \ {\mathcal R}^{(3)}(f,g)=
\left|
\begin{array}{rrrrr}
1 & -1 & 0 & 3 & -2 \\
0 & 1 & -1 & 0 & 3 \\
0 & 0 & 1 & 0 & 1 \\
0 & 1 & 0 & 1 & 1 \\
1 & 0 & 1 & 1 & 2
\end{array}
\right|
=-7 \ne 0
$$
Therefore, $ \operatorname{GCD} (f(x),g(x)) $ is of the degree $ 3_{} $. For its computation, compose the determinant by replacing the last
column of the subresultant $ {\mathcal R}^{(3)} $:
$$
\operatorname{GCD} (f(x),g(x)) =\left|
\begin{array}{rrrrr}
1 & -1 & 0 & 3 & -2\,x^3+x \\
0 & 1 & -1 & 0 & 3\,x^3-2\,x^2+1 \\
0 & 0 & 1 & 0 & x^3+x^2+2\,x+1 \\
0 & 1 & 0 & 1 & x^3+2\,x^2+x \\
1 & 0 & 1 & 1 & 2\,x^3+x^2
\end{array}
\right|=-7\,x^3 +7\,x^2-7\,x-7 \ .
$$
**Answer.** $ x^3-x^{2}+x+1 $.
!!§!! For the particular case $ g(x)\equiv f^{\prime}(x) $, the subresultants transform into ((detse:discrime#subdiscriminants subdiscriminants)).
===Resultant in the Kronecker form==
For the polynomials
$$
f(x)=a_0x^n+a_1x^{n-1}+\ldots+a_n\quad and \quad g(x)=b_0x^m+b_1x^{m-1}+\ldots+b_m
$$
($ a_{0}\ne 0 $) find first the expansion of the fraction
$ g_{}(x)/f(x) $ in Laurent series in negative powers of $ x_{} $.
For the case $ \deg g < \deg f $, represent this expansion with undetermined coefficients
$$
\frac{g(x)}{f(x)}=\frac{c_0}{x}+\frac{c_1}{x^2}+\dots+\frac{c_j}{x^{j+1}}+\dots \ ,
$$
multiply then this by
$ f_{}(x) $, and equate the coefficients of equal powers of $ x_{} $ in both sides. For the case $ m=n-1 $
these formulas take the form
$$
\begin{array}{ll}
c_0&=b_{0}/a_0,\\
c_1& =(-c_{0}a_1+b_1)/a_0,\\
c_2& =(-c_{0}a_2-c_1a_1+b_2)/a_0, \\
\dots & \dots \\
c_{n-1}&=(-c_0a_{n-1}-c_1a_{n-2}-\dots-c_{n-2}a_1+b_{n-1})/a_0,\\
c_{K+n}&=(-c_{K}a_{n}-c_{K+1}a_{n-1}-\dots-c_{K+n-1}a_1)/a_0 \quad if \quad K \in \{0,1,2,\dots \} \ .
\end{array}
$$
If $ m
☞