University Problem Set #5: Solutions ECE 456
of Illinois Page 1 of 2 Fall 2002
1. Let X = (x0, x1, . . . , xm-1) be the representation of X ∈ GF(pm) w.r.t. the basis {δ0, δ1, . . . , δm-1},
that is, X = ∑ xiδi where xi ∈ GF(p) and δi ∈ GF(pm). Let {λ0, λ1, . . . , λm-1} denote the dual basis.
Let γ = (γ0, γ1, . . . , γm-1) be the representation of β with respect to the dual basis, that is β = ∑ γiλi.
Then, Tr(βX) = Tr([∑ γiλi][∑ xjδj] = ∑∑ γixj•Tr(λiδj)= ∑ γixi = γ•X by the linearity of the trace function
and the property of the dual bases. Thus, γ is the representation of β with respect to the dual of the basis
that we used to represent X as a m-vector X.
2.(a) 1 = 1
α= γ+γ3
α2=γ+γ2
α3=γ
α4= 1 +γ +γ3
α5=γ2+γ3
α6=γ2
α7= 1 +γ3
α8= 1 +γ+γ2
α9=γ3
α10 = 1 +γ2+γ3
α11 =γ+γ2+γ3
α12 = 1 +γ+γ2+γ3
α13 = 1 +γ2
α14 = 1 +γ
Hence we get that x0 + x1α + x2α2 + x3α3 = x0 + x1(γ + γ3) + x2(γ + γ2) + x3γ
= x0 + (x1 + x2 + x3)γ + x2γ2 + x1γ3. It follows that y0 = x0, y1 = x1 + x2 + x3, y 2 = x2, y3 = x1, and we
get that A =
1 0 0 0
0 1 0 1
0 1 1 0
0 1 0 0
(b) Given any basis {α0, α1, … , αm–1 } and the representation of X = ∑ xiαi in terms of the basis, write each
αi in terms of the dual basis to get the representation of X in the dual basis. That is, write αi = ∑ aijβj
where, as we have seen in class, aij = Tr(αjαi). Hence, if we define the matrix A as [Tr(αiαj)], then
^
X = [^
x0, ^
x1, … , ^
xm–1] = [x0, x1, … , xm–1 ]A = XA is the representation of X with respect to the dual
basis. In our specific case we have that m = 4, and αi = αi. Since Tr(1) = Tr(α) = Tr(α2) = Tr(α4) =
Tr(α5) = 0, and Tr(α3) = Tr(α6) = 1, we get the matrix as A = [Tr(αi+j)] =
0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 1
. It can be used to
transform from the polynomial basis to the dual basis. The dual basis {β0, β1, β2, β3} itself is represented
by the unit vectors [1, 0, 0, 0], [0, 1, 0, 0], … with respect to itself, and we can obtain these elements in
terms of the original (polynomial) basis by noting that ^
X = XA implies that βi is the i-th row of A–1.
This is readily found by adding row 1 to row 4, and then interchanging rows 1 and 4, 2 and 3. The same
transformations applied to the identity matrix give A–1 =
1 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0
. Thus, β0 = 1 + α3 = α14, β1 = α2,
β2 = α, and β3 = 1. See also the bottom of page 243 in the text.
(c) {α,α2,α4,α8} cannot be a basis since α + α2 + α4 + α8 = Tr(α) = 0 and thus the elements are not linearly
independent. The other two sets are readily verified as being linearly independent.
(d) X•Y = (x0 + x1α + x2α2 + x3α3)•(y0 + y1α + y2α2 + y3α3)
= x0y0 + α(x0y1+x1y0) + α2(x0y2+x1y1+x2y0) + α3(x0y3+x1y2+x2y1+x3y0) + α4(x1y3+x2y2+x3y1)
+ α5(x2y3+x3y2) + α6x3y3
= (x0y0+x1y3+x2y2+x3y1) + α(x0y1+x1y0+x1y3+x2y2+x3y1+x2y3+x3y2)
+ α2(x0y2+x1y1+x2y0+x2y3+x3y2+x3y3) + α3(x0y3+x1y2+x2y1+x3y0+x3y3). The spaghetti
diagram can be sketched from this information. Obviously, 16 AND gates and 18 XOR gates are required
(though there may be some savings in XORs since, for example, x2y3+x3y2 occurs in both α and α2).
(e) First, let us assume (as in class) that X is represented with respect to the dual basis, that is, X = ∑xiβi, and
that Y is represented with respect to the polynomial basis, that is, Y = ∑~
yjαj and that Z = X•Y = ∑zkβk is
