CC BY
33
UDK 517.43
Radon transform on the plane over a finite ring: a
new inversion formula 1
© E. V. Vodolazhskaya, S. V. Kuzovkin
Derzhavin Tambov State University, Tambov, Russia
The Radon transform R on the plane over a finite ring K assigns to a function f on K sums of its values on lines. We write a new inversion formula for a field and the ring of cosets modulo p2.
Keywords: Radon transform, finite fields, rings of residue classes
In our previous talk [1] we described the image of the Radon transform and wrote an inversion formula for the plane over a finite ring when this ring is either a field or the ring of cosets modulo p2 with prime p. In this paper we present an inversion formula of essentially other structure for the same cases.
Let K be a finite ring with q elements, K2 = K x K the plane over K. A line on the plane K2 is the set t of points z = (x,y) E K2 satisfying the equation
ax + by = c,
where a,b,c E K and a and b are not zero divisors simultaneously. Let H denote the set of all lines.
For a finite set X, let L(X) denote the linear space of C-valued functions on X. Its dimension is equal to the number of elements of X.
The Radon transform R assigns to any function f E L(K2) the function Rf E L(H) whose value at a line t is equal to an "integral" of the function f over this line, i.e.
(Rf )(t) = Y f (z).
zel
The transform R is a linear operator L(K2) ^ L(H).
1 Supported by the Russian Foundation for Basic Research (RFBR): grant 09-01-00325-a, Sci. Progr. "Development of Scientific Potential of Higher School": project 1.1.2/9191, Fed. Object Progr. 14.740.11.0349 and Templan 1.5.07
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For points z,w E K2, let us denote by T(z,w) the number of lines passing through these points, (for w = z it is the number of lines passing through the point z). We obtain a kernel T(z,w), or a matrix T of order q2. Let us define an operator
zel
Theorem 1.1 Let us assume that the matrix T is invertible, denote by S the reciprocal 'matrix: S = T-1, let S(z,w) be its elements. Then
Therefore, the matrix T can be written as T = qE + / where E is the identity matrix, / denotes the matrix whose entries are equal to 1, Using formula (3), where a = q, ft = 1, r = q2, we obtain
Kq
formula
Now let K be the ring of cosets modulo p2, p is prime. Let D be the set of zero divisors: 0,p, 2p,..., (p — 1)p. Denote D2 = D x D. Then
Therefore, the matrix T can be written as matrix (4), where m = r = p2, the matrix C has the form (2) with a = p2 + p — 1, ft = p — 1, such that we have a = p3 in formula (5), Using (6), we get
M : L(H) ^ L(K2) bv
(M(F))(z) = Y, F(t).
f (z)= S (z,w)(M (Rf))(w).
(1)
weK 2
Proof, We have
(M(Rf))(z) =Y, T(z,w)f(w),
weK2
therefore
s(z,w)(M(Rf))(w) = S(z,w)H T(w,u)f(u)
weK2
weK2
ueK2
f(z).
□
Kq
f (z) = J+njfa2+q—i){M {Rf ))z — T,(m (Rf ))(w)}
J- \J- /
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K p2 p
inversion formula
f(z) =
1
5 + p4 — p3 + p — 1)(M (Rf ))(z)
p6(p + 1)
(p3 — p +1) Y (M(Rf))(w) — Y (M(Rf))(w)}-
t?ez+D2,w=z
wez+D2
In proofs of Theorems 2 and 3 we use formulae for some reciprocal matrices. These formulae can be checked immediately,
Cr
C = aE + ft/.
Then
C-1 = - E-
ft
a a (a + rft)
/.
(2)
(3)
Let A be the following mr x mr matrix written as a block matrix with blocks of order r:
C + / / . . . /
/ C + / ... /
(4)
\ / / ... C + / y
C
C/ = /C = a/,
(5)
a being a number. Then the reciprocal matrix C 1 satisfies the eondition /C
-1
a-1/
A
1
( C-1 + 11/ 11/ ...
^/ C-1 +1/ ...
V i/
where 1 = —{a(a + mr)}-1.
l/
l/ \ l/
C-1 +1/
(6)
References
1, E, V, Vodolazhskava, Radon transform on a finite plane. International Workshop "Polynomial Computer Algebra April 4-7, 2008, St.-Peterburg, Russia, 78-80.
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