Calculation of Determinant of a two-variable Polynomial Matrix in
Complex Basis
DIMITRIOS VARSAMIS, ANGELIKI KAMILALI
Department of Computer, Informatics and Telecommunications Engineering
International Hellenic University - Serres Campus
Terma Magnisias, Serres - 62124
GREECE
Abstract: - In this paper, an innovation in evaluation part of Evaluation Interpolation technique (E-I) is presented.
The variables of this technique have been defined in Complex domain. The use of complex numbers instead of real
ones in evaluation part, provides us with the opportunity of using conjugate properties. With these properties, the
number of evaluations in evaluation part is being decreased. The interpolation part is common for both methods.
Key-Words: - Complex Basis, Polynomial Matrix, Evaluation - Interpolation, Two-variable polynomials
Received: July 29, 2021. Revised: June 26, 2022. Accepted: July 29, 2022. Published: September 23, 2022.
1 Introduction
A lot of solutions among Control Theory problems
consists of polynomials. Some of them are calcula-
tion of the determinant of a polynomial matrix [1],
[2], calculation of the greatest common divisor among
polynomials [3], computation of the generalised in-
verse of a polynomial matrix [4], [5], [6], polyno-
mial matrices as solutions of diophantine equations
[7], transfer function computation for multidimen-
sional systems [8].The need for solving Control The-
ory problems has led to the development of several
numerical computational methods. Evaluation - In-
terpolation technique is one of these methods, and is
used for solving problems whose solution is a polyno-
mial. It consists of two parts. In the first part (evalua-
tion part), a set of fixed required points is defined and
evaluated, and the second part (interpolation part) is
for finding the unique polynomial that passes through
all these points. The main advantage of this technique
is the use of numerical analysis instead of analytical
solutions, which makes it easier for algorithms to han-
dle.
The word ’interpolation’ has been introduced by J.
Wallis in early 1655. Nevertheless polynomial inter-
polation in several variables is a relatively new topic
that has been raised in the second half of the last cen-
tury. Multivariate polynomial interpolation is a basic
subject both in Approximation Theory and in Numer-
ical Analysis, as it has many applications in mathe-
matical problems [9], [10]. That’s why it has been
the concern of many scientists. Some of the most
known multivariate polynomial interpolation meth-
ods are: the use of Vandermonde matrix in addition
with LU factorization, Lagrange interpolation, Her-
mite - Birkhoff interpolation, Newton interpolation,
Discrete Fourier Transform (DFT) and Fast Fourier
Transform algorithms (FFT) [2] [11] [12].
This work is based on Newton two-variable poly-
nomial interpolation method. First of all a set of fixed
points is defined (evaluation part), and then with the
interpolation method that has been just referred, the
exact polynomial that approximates these points is be-
ing constructed. We study the special case, where the
upper bounds of the degree of each variable is known,
so we use a rectangular basis. In addition we define
the required set of points in Complex domain. Taking
into account conjugate properties, we prove that the
number of required points can be reduced even up to
the half. We perform this consideration through ex-
amples such as the calculation of the determinant of
a two variable polynomial matrix both in Real and in
Complex domain. As a conclusion, the number of re-
quired operations for this process is reduced.
2 Calculation of Determinant in Real
basis
The following algorithm is based on Determinant Cal-
culation with Evaluation-Interpolation technique [2].
2.1 A general algorithm
Let the two-variable polynomial matrix A∈
R[x, y]n×n
A(x, y) =
a1,1(x, y)· · · a1,n(x, y)
.
.
.....
.
.
an,1(x, y)· · · an,n(x, y)
Step 1:(Calculation of matrice’s degree in each vari-
able)
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Degree matrices of xand yvariable respectively, are
Dx=
degxa1,1(x, y)· · · degxa1,n(x, y)
.
.
.....
.
.
degxan,1(x, y)· · · degxan,n(x, y)
and
Dy=
degya1,1(x, y)· · · degya1,n(x, y)
.
.
.....
.
.
degyan,1(x, y)· · · degyan,n(x, y)
where degxai,j (x, y), i = 1,· · · , n, j = 1,· · · , n
is the degree of ai,j (x, y)element of Amatrix
with respect to xvariable and degyai,j (x, y), i =
1,· · · , n, j = 1,· · · , n is the degree of ai,j (x, y)
element of Amatrix with respect to yvariable.
Let dcx
s,drx
s,dcy
sand dry
sthe sum of the maximum el-
ement of each column and row of the matrices above.
dcx
j=max{degxai,j (x, y)|i= 1,· · · , n},
for j = 1,· · · , n
dcx
s=
n
j=1
dcx
j
drx
i=max{degxai,j (x, y)|j= 1,· · · , n},
for i = 1,· · · , n
drx
s=
n
i=1
drx
i
dcy
j=max{degyai,j (x, y)|i= 1,· · · , n},
for j = 1,· · · , n
dcy
s=
n
j=1
dcy
j
dry
i=max{degyai,j (x, y)|j= 1,· · · , n},
for i = 1,· · · , n
dry
s=
n
i=1
dry
i
Thus, according to [2] we have
k1=min{dcx
s, drx
s}
k2=min{dcy
s, dry
s}
Therefore, the number of required points is N=
(k1+ 1)(k2+ 1).
Step 2: (Evaluation Part)
Step 2a: Determination of the set of Nrandom fixed
points
S(k1,k2)={(xi, yj)|i= 0, . . . , k1, j = 0, . . . , k2}
Step 2b: Evaluation of all points of S(k1,k2)set in
the numerical matrix Aand computation of their de-
terminant det(A(xi, yj)) where i= 0, . . . , k1, j =
0, . . . , k2. The interpolation set is determined as fol-
lows:
{(A(xi, yj),det(A(xi, yj))) |i= 0,· · · , k1j=
0,· · · , k2}
Step 3: (Interpolation Part)
Step 3a: Construction of the zero order table (k= 0)
of required points (interpolation set).
Step 3b: Computation of the k-th order table of di-
vided differences, where k= 1,· · · , n and n=
max{k1, k2}according to [12]
Step 3c: Construction of the Newton polynomial
p(x, y) = XT·P·Y
where
X=
1
(x−x0)
(x−x0)·(x−x1)
.
.
.
(x−x0)·(x−x1)· · · · · (x−xk1−1)
Pis the n-th order table of divided differences and
Y=
1
(y−y0)
(y−y0)·(y−y1)
.
.
.
(y−y0)·(y−y1)· · · · · (y−yk2−1)
The calculated Newton polynomial is the determi-
nant of the two-variable polynomial matrix, namely,
p(x, y) = det(A(x, y)).
3 Calculation of Determinant in
Complex basis
In this case we are following the same procedure, just
the way we did it above. The only difference is that
we use interpolation points of the complex domain,
which allow us to reduce their number due to the con-
jugate properties of the complex domain.
Proposition 1. Let the two-variable polynomial
p(x, y) =
k1
n=0
k2
m=0
cn,mxnym
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where k1is the degree of x variable and k2is the de-
gree of y variable.
Let the point (ai, b), where iis the imaginary unit, and
a, b ∈Z, then holds
p(ai, b) = p(ai, b)(1)
Proof. We compute the value of p(ai, b)
p(ai, b) =
k1
n=0
k2
m=0
cn,m(ai)nbm
=
k1
n=0
k2
m=0
cn,manbmin
=
k1
n=0
k2
m=0
cn,manbmfor n= 4k
k1
n=0
k2
m=0
cn,manbmifor n= 4k+ 1
k1
n=0
k2
m=0
cn,manbm(−1) for n= 4k+ 2
k1
n=0
k2
m=0
cn,manbm(−i)for n= 4k+ 3
=
k1
n=0
k2
m=0
cn,manbmfor n= 4k
k1
n=0
k2
m=0
cn,manbmifor n= 4k+ 1
−k1
n=0
k2
m=0
cn,manbmfor n= 4k+ 2
−k1
n=0
k2
m=0
cn,manbmifor n= 4k+ 3
where k∈N.
Similarly, we compute the value of p(ai, b)
p(ai, b) = p(ai, b) = p(−ai, b)
=
k1
n=0
k2
m=0
cn,m(−ai)nbm
=
k1
n=0
k2
m=0
(−1)ncn,manbmin
=
k1
n=0
k2
m=0
(+1)cn,manbmfor n= 4k
k1
n=0
k2
m=0
(−1)cn,manbmifor n= 4k+ 1
k1
n=0
k2
m=0
(+1)cn,manbm(−1) for n= 4k+ 2
k1
n=0
k2
m=0
(−1)cn,manbm(−i)for n= 4k+ 3
=
k1
n=0
k2
m=0
cn,manbmfor n= 4k
−k1
n=0
k2
m=0
cn,manbmifor n= 4k+ 1
−k1
n=0
k2
m=0
cn,manbmfor n= 4k+ 2
k1
n=0
k2
m=0
cn,manbmifor n= 4k+ 3
where k∈N.
As we can see, it holds p(ai, b) = p(ai, b)
Thus, because of the fact that the determinant of a
polynomial matrix is also a polynomial, it comes out
that
det(A(−ai, b)) = det(A(ai, b)) = det(A(ai, b))
(2)
where a,b ∈Z
3.1 A general algorithm
Let the two-variable polynomial matrix A∈
R[x, y]n×n
A(x, y) =
a1,1(x, y)· · · a1,n(x, y)
.
.
.....
.
.
an,1(x, y)· · · an,n(x, y)
Step 1:(Calculation of matrice’s degree in each vari-
able)
The Step 1 is the same with the algorithm in subsec-
tion 2.1. Then
k1=min{dcx
s, drx
s}
k2=min{dcy
s, dry
s}
Therefore, the number of required points is N=
(k1+ 1)(k2+ 1).
Step 2: (Evaluation Part)
Step 2a: Determination of the set of Nrandom fixed
points
S(k1,k2)={(xl1·i, yl2)|l1= 0, . . . , k1,
l2= 0, . . . , k2}
where iis the imaginary unit.
Step 2b: From 2 we have that
det(A(−ai, b)) = det(A(ai, b)) = det(A(ai, b))
Thus, evaluation is executed only for ⌈N
2⌉complex
points of S(k1,k2)set in the numerical matrix Aand the
computation of their determinant det(A(xl1·i, yl2))
follows. The determnants of the next N− ⌈ N
2⌉points
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of S(k1,k2)set occur as conjugate numbers of the pre-
vious one. The interpolation set is determined as fol-
lows:
{(A(xl1·i, yl2), det(A(xl1·i, yl2))) |l1= 0, . . . , k1,
l2= 0, . . . , k2}
and iis the imaginary unit
Step 3: (Interpolation Part)
Step 3a: Construction of the zero order table (k= 0)
of required points (interpolation set).
Step 3b: Computation of the k-th order table of di-
vided differences, where k= 1,· · · , n and n=
max{k1, k2}according to [12]
Step 3c: Construction of the Newton polynomial
p(x, y) = XT·P·Y
where
X=
1
(x−x0)
(x−x0)·(x−x1)
.
.
.
(x−x0)·(x−x1)· · · · · (x−xk1−1)
Pis the n-th order table of divided differences and
Y=
1
(y−y0)
(y−y0)·(y−y1)
.
.
.
(y−y0)·(y−y1)· · · · · (y−yk2−1)
The calculated Newton polynomial is the determi-
nant of the two-variable polynomial matrix, namely,
p(x, y) = det(A(x, y)).
4 Examples
4.1 Example in Real basis
Let the two-variable polynomial matrix
A(x, y) =
xy21 2xy
−xy −1 3x
2y−2x xy
Step 1:(Calculation of matrice’s degree in each vari-
able)
Degree matrices with respect to xand yvariable, are
Dx=101
101
011and Dy=201
100
101
We define dcx
s,drx
s,dcy
sand dry
sas the sum of the
maximum element of each column and row for xand
yvariable respectively.
dcx
1=max{1,1,0}= 1
dcx
2=max{0,0,1}= 1
dcx
3=max{1,1,1}= 1 ⇒
dcx
s=dcx
1+dcx
2+dcx
3= 3
and drx
1=max{1,0,1}= 1
drx
2=max{1,0,1}= 1
drx
3=max{0,1,1}= 1 ⇒
drx
s=drx
1+drx
2+drx
3= 3
and
dcy
1=max{2,1,1}= 2
dcy
2=max{0,0,0}= 0
dcy
3=max{1,0,1}= 1
⇒
dcy
s=dcy
1+dcy
2+dcy
3= 3
and
dry
1=max{2,0,1}= 2
dry
2=max{1,0,0}= 1
dry
3=max{1,0,1}= 1
⇒
dry
s=dry
1+dry
2+dry
3= 4
Thus, according [2] we have
k1=min{dcx
s, drx
s}=min{3,3}= 3
k2=min{dcy
s, dry
s}=min{3,4}= 3
Therefore, the number of required points is N=
(k1+ 1)(k2+ 1) = 16.
Step 2: (Evaluation Part)
Step 2a: We determine 16 random fixed points
S(3,3) ={(xi, yj)|xi=i, yj=jand
i= 0, . . . , 3, j = 0, . . . , 3}
Step 2b: For each of those points we evaluate the nu-
merical matrix A(xi, yi)and compute their determi-
nant det(A(xi, yi)). For example:
A(x0, y0) = A(0,0) = 010
0−1 0
000
and det(A(0,0)) = 0
A(x0, y1) = A(0,1) = 010
0−1 0
200
and det(A(0,1)) = 0
.
.
.
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A(x3, y3) = A(3,3) = 27 1 18
−9−1 9
6−6 9
and det(A(3,3)) = 2430
Step 3: (Interpolation Part)
Step 3a: We construct the zero order table (k= 0) of
required points (interpolation set), with the determi-
nants we computed in the previous step.
P(0) =
P(0)
0,0P(0)
0,1P(0)
0,2P(0)
0,3
P(0)
1,0P(0)
1,1P(0)
1,2P(0)
1,3
P(0)
2,0P(0)
2,1P(0)
2,2P(0)
2,3
P(0)
3,0P(0)
3,1P(0)
3,2P(0)
3,3
=
=
0 0 0 0
0 20 64 126
0 100 360 756
0 300 1128 2430
where P(0)
i,j =det(A(xi, yj)) for i= 0, . . . , 3, j =
0, . . . , 3.
Step 3b: For k= 1 to n, where n=max{k1, k2}=
max{3,3}= 3 we compute the k-th order table of
divided differences according to [12].
For k= 3 we have
P(3) =
0 0 0 0
0 20 12 −1
0 30 28 −1
0 10 10 0
Step 3c: The Newton polynomial is
p(x, y) = 10x3y2−x2y3+x2y2+ 4xy2+ 6xy
It occurs from the following
p(x, y) = XT·P·Y
where
X=
1
x
x(x−1)
x(x−1)(x−2)
P=
0 0 0 0
0 20 12 −1
0 30 28 −1
0 10 10 0
and
Y=
1
y
y(y−1)
y(y−1)(y−2)
The calculated Newton polynomial is the determi-
nant of the two-variable polynomial matrix, namely,
p(x, y) = det(A(x, y)).
4.2 Example in Complex basis
Construction of the determinant of a two-variable
polynomial matrix, using Newton interpolation tech-
nique.
Let the two-variable polynomial matrix
A(x, y) =
xy21 2xy
−xy −1 3x
2y−2x xy
Step 1: (Calculation of matrice’s degree in each vari-
able)
According to example in Real basis we have
k1=min{dcx
s, drx
s}=min{3,3}= 3
k2=min{dcy
s, dry
s}=min{3,4}= 3
Therefore, the number of required points is
N= (k1+ 1)(k2+ 1) = 16.
Step 2: (Evaluation Part)
Step 2a: We define the set
S(3,3) ={(xℓ1·i, yℓ2)|xℓ1={−3,−1,1,3},
yℓ2={−3,−1,1,3}and ℓ1= 0, . . . , 3, ℓ2= 0, . . . , 3}
where iis the imaginary unit.
Step 2b: (Evaluation part) From 2 we have that
det(A(−ai, b)) = det(A(ai, b)) = det(A(ai, b))
Thus, we evaluate only ⌈N
2⌉= 8 complex interpola-
tion points. Specifically, we determine the following
interpolation points (−3i, −3),(−3i, −1),(−3i, 1),
(−3i, 3),(−i, −3),(−i, −1),(−i, 1),(−i, 3), and
we compute the determinant of the matrix that occurs
from the evaluation of each one of those points.
det(A(x0i, y0)) = det(A(−3i, −3))
=det(−27i1 18i
−9i−1−9i
−6 6i9i)
=−324 + 2376i
det(A(x0i, y1)) = det(A(−3i, −1)) =
=det(−3i1 6i
−3i−1−9i
−2 6i3i)
=−18 + 276i
det(A(x0i, y2)) = det(A(−3i, 1)) =
=det(−3i1−6i
3i−1−9i
2 6i−3i)
= 240i
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det(A(x0i, y3)) = det(A(−3i, 3)) =
=det(−27i1−18i
9i−1−9i
6 6i−9i)
= 162 + 2268i
det(A(x1i, y0)) = det(A(−i, −3)) =
=det(−9i1 6i
−3i−1−3i
−6 2i3i)
=−36 + 72i
det(A(x1i, y1)) = det(A(−i, −1)) =
=det(−i1 2i
−i−1−3i
−2 2i i )
=−2 + 12i
det(A(x1i, y2)) = det(A(−i, 1)) =
=det(−i1−2i
i−1−3i
2 2i−i)
= 0
det(A(x1i, y3)) = det(A(−i, 3)) =
=det(−9i1−6i
3i−1−3i
6 2i−3i)
= 18 + 36i
For the rest interpolation points, according to,
det(A(−ai, b)) = det(A(ai, b)) = det(A(ai, b))
we have
det(A(x3i, y0)) = det(A(3i, −3))
=det(A(−3i, −3))
=det(A(−3i, −3))
=−324 −2376i
det(A(x3i, y1)) = det(A(3i, −1))
=det(A(−3i, −1))
=det(A(−3i, −1))
=−18 −276i
det(A(x3i, y2)) = det(A(3i, 1))
=det(A(−3i, 1))
=det(A(−3i, 1))
=−240i
det(A(x3i, y3)) = det(A(3i, 3))
=det(A(−3i, 3))
=det(A(−3i, 3))
= 162 −2268i
det(A(x2i, y0)) = det(A(i, −3))
=det(A(−i, −3))
=det(A(−i, −3))
=−36 −72i
det(A(x2i, y1)) = det(A(i, −1))
=det(A(−i, −1))
=det(A(−i, −1))
=−2−12i
det(A(x2i, y2)) = det(A(i, 1))
=det(A(−i, 1))
=det(A(−i, 1))
= 0
det(A(x2i, y3)) = det(A(i, 3))
=det(A(−i, 3))
=det(A(−i, 3))
= 18 −36i
Step 3: (Interpolation Part)
Step 3a: We construct the zero order table (k= 0) of
initial values (interpolation set).
P(0) =
−324 + 2376i−18 + 276i240i162 + 2268i
−36 + 72i−2 + 12i0−2 + 12i
−36 −72i −2−12i 0 −2−12i
−324 −2376i −18 −276i −240i 162 −2268i
where P(0)
ℓ1,ℓ2=det(A(xℓ1·i, yℓ2)) for ℓ1=
0, . . . , 3, ℓ2= 0, . . . , 3. The values in bold are in-
dicating the conjugate numbers that we do not need
to calculate.
Step 3b: For k= 1 to n, where n=max{k1, k2}=
max{3,3}= 3 we compute the k-th order table of di-
vided differences according to [12].
For k= 3 we have
P(3) =
−324 + 2376i153 −1050i−36 + 258i9
−1152 −144i510 + 68i−126 −16i4i
36 −270i−17 + 120i4−30i−1
90 −40 10 0
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Step 3c: The Newton polynomial is
p(x, y) = 10x3y2−x2y3+x2y2+ 4xy2+ 6xy
which is the determinant of the matrix A(x, y). It oc-
curs from the following
p(x, y) = XT·P·Y
where
X=
1
x+ 3i
(x+ 3i)(x+i)
(x+ 3i)(x+i)(x−i)
, P =
−324 + 2376i153 −1050i−36 + 258i9
−1152 −144i510 + 68i−126 −16i4i
36 −270i−17 + 120i4−30i−1
90 −40 10 0
and
Y=
1
y+ 3
(y+ 3)(y+ 1)
(y+ 3)(y+ 1)(y−1)
5 Conclusion
An optimized technique of calculating the determi-
nant of a two - variable polynomial matrix has been
proposed. This technique is based on Newton Eval-
uation - Interpolation method in rectangular basis, in
complex domain.
As we can see from the above, step 1 as well as
steps 3a, 3b and 3c, are the same regardless of the do-
main (Real or Complex) we choose every time. The
novelty of the proposed technique is that if we choose
conjugate numbers as required points, we can reduce
numerical operations in evaluation part even up to the
half.
References:
[1] Henrion, D. and Sebek M., Improved polyno-
mial matrix determinant computation. Circuits
and Systems I: Fundamental Theory and Ap-
plications, IEEE Transactions , Vol.46, No.10,
1999, pp. 1307-1308.
[2] Varsamis D., Calculation of Determinant of a
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WSEAS TRANSACTIONS on SYSTEMS and CONTROL
DOI: 10.37394/23203.2022.17.44
Dimitrios Varsamis, Angeliki Kamilali
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WSEAS TRANSACTIONS on SYSTEMS and CONTROL
DOI: 10.37394/23203.2022.17.44
Dimitrios Varsamis, Angeliki Kamilali
E-ISSN: 2224-2856
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