Y.S.Han Cyclic codes 8 Proof:A linear combination of g(),xg(a),...g(), v(x)=aog(x)+alzg(x)+...+ak-1xk-1g(z) =(a0+a1x+·+ak-1x-1)g(x), is a polynomial of degree n-1 or less and is a multiple of g(x). There are a total of 2%such polynomial and they form an (n,k) linear code. Let v(x)=vo +v12+...+Un-1x"-1 be a code polynomial in this code.We have xv(c)=0x+u1x2+…+n-1x” =vn-1(x”+1)+(n-1+0x+…+vn-2xn-1) =vn-1(xn+1)+v1(c). Since both xu(x)and x"+1 are divisible by g(a),v(1)must be divisible by g().Hence,v(1)(x)is a code polynomial and the code generated by g()is a cyclic code. School of Electrical Engineering Intelligentization,Dongguan University of Technology Y. S. Han Cyclic codes 8 Proof: A linear combination of g(x), xg(x), . . . , xk−1 g(x), v(x) = a0 g(x) + a1 xg(x) + · · · + ak−1 x k−1 g(x) = (a0 + a1 x + · · · + ak−1 x k−1 )g(x), is a polynomial of degree n − 1 or less and is a multiple of g(x). There are a total of 2 k such polynomial and they form an (n, k) linear code. Let v(x) = v0 + v1 x + · · · + vn−1 x n−1 be a code polynomial in this code. We have xv(x) = v0 x + v1 x 2 + · · · + vn−1 x n = vn−1(x n + 1) + (vn−1 + v0 x + · · · + vn−2 x n−1 ) = vn−1 (x n + 1) + v (1)(x). Since both xv(x) and x n + 1 are divisible by g(x), v (1) must be divisible by g(x). Hence, v (1)(x) is a code polynomial and the code generated by g(x) is a cyclic code. School of Electrical Engineering & Intelligentization, Dongguan University of Technology