Math 342
Homework Assignment #4 (Due Thursday, March 10)
1. Show that row equivalence preserves linear independence (i.e., if A and B are row
equivalent, then the rows of A are linearly independent iff the rows of B are linearly
independent).
Solution:
Since row equivalence is a symmetric operation (i.e., A is row equivalent to B iff B is
row equivalent to A), it suffices to show that if A and B are row equivalent and the
rows of A are linearly independent, then so are the rows of B.
For this, we just need to check each of the three types of elementary row operations.
1. An exchange of rows does not affect the set of rows and so clearly linear independence
of the rows is preserved.
Let R1 , . . . , Rm denote the rows of A
2. Suppose B is obtained by replacing Rj by aRj for some a 6= 0 and j.Then to say
that a linear combination, with coefficients ai , of the rows B equals 0 is to say
(
m
X
ak Rk ) + aj (aRj ) = 0
k=1,k6=j
Since the rows of A are linearly independent, we have that for all k 6= j, ak = 0 and
aj a = 0; since a 6= 0 and the coefficients are in a field, aj = 0 as well. Thus, the rows
of B are linearly independent.
3. Suppose that B is obtained by replacing Rj by aRi + Rj where i 6= j. If
X
ak Rk ) + aj (aRi + Rj ) = 0
(
k=1,k6=j
then for all k 6= i, ak = 0 and ai + aj a = 0; since aj = 0, we have ai = 0 as well. Thus,
the rows of B are linearly independent.
2. (We did the following problem in class for the special case p = 2).
Let p be prime and H be a subset of V (n, p). Show that the following are equivalent
(a) H is closed under vector addition
(b) H is a subgroup of V (n, p).
(c) H is a subspace of V (n, p).
Solution:
1 ⇒ 2: It suffices to show that H is closed under vector addition and inverses.
We are assuming that H is closed under vector addition. Since GF (p) = Zp , for all
a ∈ GF (p), a + (p − 1)a = pa = 0. Thus, for all x ∈ V (n, p), x + (p − 1)x = 0. Thus,
1
−x = (p − 1)x. But (p − 1)x is the sum of p − 1 copies of x and so −x = (p − 1)x ∈ H.
Thus, H is closed under inverses.
2 ⇒ 3: It suffices to show that H is closed under vector addition and scalar multiplication. Since H is a subgroup, it is closed under vector addition. For any scalar
a = 0, . . . , p − 1, and any x ∈ H, ax is the sum of a copies of x and so the scalar
multiple ax ∈ H.
3 ⇒ 1: Obvious since any subspace is a vector space and is thus closed under vector
addition.
3. Consider the matrix


aa1b
G =  0ba1 
a10a
(a) Verify that G is a generator matrix for a linear code C over GF (4).
(b) What is the dimension and what is the size of C?
(c) Encode each of the following message vectors to a codeword of C.
i. a1b
ii. a01
Solution:
a. This is equivalent to showing that the rows of G are linearly independent (the linear
code would then be the span of the rows). We can do this by reducing to REF, but
let’s do it in a more ad hoc way: the vector equation:
c1 (aa1b) + c2 (0ba1) + c3 (a10a) = 0000
is equivalent to the four scalar equations:
(c1 + c3 )a = 0, c1 a + c2 b + c3 = 0, c1 + c2 a = 0, c1 b + c2 + c3 a = 0
From the first and third equaitons, we get c3 = −c1 = c1 , c2 = −c1 a−1 = c1 a−1 .
Plugging into the 2nd equation, we get c1 a−c1 a−1 b+c1 = 0, and so 0 = c1 (a+a−1 b+1) =
c1 (a + b2 + 1) = c1 and so c1 = c2 = c3 = 0, and so the rows are linearly indepedent
(we did not even need to use the last equation).
b. dim(C) = 3 and |C| = 43 = 64.
c.
(a1b)G = a(aa1b) + (0ba1) + b(a10a) = (a2 + ba, a2 + b + b, a + a, ab + 1 + ba) = (ab01)
(a01)G = a(aa1b) + (a10a) = (a2 + a, a2 + 1, a, ab + a) = (1aab)
4. Find a generator matrix in standard form for the linear span (or an equivalent code)
of each of the following sets.
2
(a) {100111, 010110, 101010, 111100, 011011, 001101}, q = 2.
(b) {11212, 10110, 12011, 11202, 22121}, q = 3.
a. Apply elementary row operations

1
 0

 1

 1

 0
0
as follows:
























to the matrix:
0
1
0
1
1
0
0
0
1
1
1
1
1
1
0
1
0
1
1
1
1
0
1
0
1
0
0
0
1
1

1
0
0
0
0
0
0
1
0
1
1
0
0
0
1
1
1
1
1
1
1
0
0
1
1
1
0
1
1
0
1
0
1
1
1
1

1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
1
0
1
1
1
1

1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
1
1
1
0
0
0
1
1
0
0
0
0
1
0
1
0
0
0





























Delete zero rows to get generator matrix in standard form:


1 0 0 1 1 1
 0 1 0 1 1 0 
0 0 1 1 0 1
b. Apply elementary row operations to:






1
1
1
1
2
1
0
2
1
2
3
2
1
0
2
1
1
1
1
0
2
2
0
1
2
1










































1
0
0
0
0
1
2
1
0
0
2
2
1
0
0
1
0
0
2
0
2
1
2
0
0

1
0
0
0
0
1
2
0
0
0
2
2
0
0
0
1
0
0
2
0
2
1
0
0
0

1
0
0
0
0
1
2
0
0
0
2
2
0
0
0
1
0
2
0
0
2
1
0
0
0

1
0
0
0
0
1
1
0
0
0
2
1
0
0
0
1
0
1
0
0
2
2
0
0
0

1
0
0
0
0
0
1
0
0
0
1
1
0
0
0
1
0
1
0
0
0
2
0
0
0

1
0
0
0
0
0
1
0
0
0
1
1
0
0
0
0
0
1
0
0
0
2
0
0
0































Delete zero rows to get generator matrix:


1 0 1 0 0
 0 1 1 0 2 
0 0 0 1 0
Permute columns to get generator matrix in standard form:
4


1 0 0 1 0
 0 1 0 1 2 
0 0 1 0 0
5