Lecture 13:
Review of vector spaces V over fields F .
Emphasis in Math 221: F = R
Emphasis in Math 342: Finite field, F = GF (q). Recall that
GF (q) is the unique field of size q and q must be a prime power:
q = pk , where p is prime and k is a positive integer. Recall that
GF (p) = Zp. Recall that GF (4) 6= Z4 and is instead given by the
addition and multiplication tables on page 36.
Defn: A vector space over a field F is a nonempty set V with
vector addition and scalar multiplication operations that satisfy the
nine axioms on pp. 41-42 of the text.
The first five axioms can be summarized to say that (V, +) is an
abelian group. The other axioms are:
(vi) (Closure of scalar multiplication): for a ∈ F, u ∈ V , we have
au ∈ V
(vii) (Distributivity): for a, b ∈ F, u, v ∈ V , we have a(u + v) =
au + av and (a + b)u = au + bv.
(viii) (Compatibility of multiplication in the field and scalar multiplication): for a, b ∈ F, u ∈ V , we have (ab)u = a(bu)
(ix) (Identity) for u ∈ V , we have 1u = u where 1 is the multiplicative identity of F
For any field F and positive integer n, F n is a vector space over
F ; here, vector addition is coordinate-wise addition:
(u1, . . . , un) + (v1, . . . , vn) = (u1 + v1, . . . , un + vn)
and scalar multiplication is coordinate-wise scalar multiplication:
a(u1, . . . , un) = (au1, . . . , aun)
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For F = R, F n = Rn, n-dimensional Euclidean space.
We focus on the vector space V (n, q) := GF (q)n. Note that
|V (n, q)| = q n and so is finite (in contrast to Rn).
Example: V (2, 3) = Z23 = {(x, y) : x, y ∈ Z3}
(1, 2) + (2, 2) = (0, 1), 2(1, 2) = (2, 1).
Defn: A subset W of a vector space V is a subspace if it is a
vector space, in its own right, using the same vector addition and
scalar multiplication as in V (and over the same field).
The following characterization of subspaces is similar to the characterization of subgroups that we gave earlier.
Theorem 4.1 A nonempty subset W of a vector space V is a subspace iff it is closed under vector addition and scalar multiplication.
Proof: “only if:” obvious.
“if:” Axioms (vii) - (ix) for W are inherited from V .
Axiom (vi) holds by assumption.
So, it remains to show that (V, +) is an abelian group. For this,
we know by our earlier characterization of subgroups that it suffices
to prove that V is closed under 1) vector addition and 2) additive
inverses. We are assuming 1). And 2) follows from the fact that for
any u ∈ W , (−1)v = −v: We leave this fact as an exercise in HW3.
Example: {(x, y, 0) : x ∈ Z3} and {(x, x, x) : x ∈ Z3} are
subspaces of V (3, 3). To see this, apply Theorem 4.1.
Defn: Let V be a vector space, u1, . . . , uk ∈ V, a1, . . . , ak ∈ F .
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An expression of the form
k
X
aiui
i=1
is called a linear combination of {u1, . . . , uk }.
Defn: Let V be a vector space and S ⊆ V . The span of S,
denoted hSi, is the set of all linear combinations of elements of S.
And S is called a spanning set for hSi.
It follows from Theorem 4.1 that hSi is a subspace of V . In fact,
these are the only subspaces of V .
In the following examples, we consider subsets S ⊂ V (3, 3).
Example 1: Let S = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}.
Claim: hSi = V (3, 3).
Proof: Clearly LHS ⊆ RHS. For the reverse inclusion: given
(x, y, z) ∈ V (3, 3), we can write
(x, y, z) = x(1, 0, 0) + y(0, 1, 0) + z(0, 0, 1)
Example 2: Let S = {(1, 0, 0), (0, 1, 0)}.
Claim: hSi = {(x, y, 0) : x, y ∈ Z3}.
Proof: Clearly LHS ⊆ RHS. For the reverse inclusion: given
(x, y, 0), we can write
(x, y, 0) = x(1, 0, 0) + y(0, 1, 0)
Example 3: Let S = {(1, 0, 0), (2, 1, 0)}.
Claim hSi = {(x, y, 0) : x, y ∈ Z3}.
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Proof: Clearly LHS ⊆ RHS. For the reverse inclusion: given
(x, y, 0), we can write
(x, y, 0) = (x − 2y)(1, 0, 0) + y(2, 1, 0)
Note that any subspace of V (n, q) has a finite spanning set, namely
itself.
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Lecture 14:
Recall definitions of vector space over a field F , subspace, span
of a subset S: hSi, which is the set of all linear combinations of
elements of S.
W := hSi is a subspace and we say that S is a spanning set for
(or spans) W .
Our main interest is the vector space V (n, q) := GF (q)n, which
is a vector space over the field F = GF (q).
Defn: Let V be a vector space over a field F . A subset {u1, . . . , uk } ⊂
V is linearly dependent if there exists a1, . . . , ak ∈ F, not all 0, s.t.
k
X
aiui = 0
(1)
i=1
Equivalently, there exists i s.t. ui is a linear combination of {uj :
j 6= i, 1 ≤ j ≤ k}.
Defn: A subset {u1, . . . , uk } ⊂ V is linearly independent if it is
not linearly dependent.
Equivalently, if a1, . . . , ak ∈ F, s.t.
k
X
aiui = 0
(2)
i=1
then each ai = 0.
Facts:
1. if u is a scalar multiple of v, then {u, v} is linearly dependent.
2. any set that contains the zero vector is linearly dependent.
3. if u 6= 0, then {u} is linearly independent:
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Proof: if au = 0 and a 6= 0, then a has a mult. inverse,
a−1(au) = a0 and so
u = 0, a contradiction.
So, the only solution to au = 0 is a = 0. Example: {(1, 0, . . . , 0), (0, 1, . . . , 0), . . . , (0, 0, . . . , 0, 1)} is linearly
independent over an field.
Example: In V (3, 3) is {(1, 2, 1), (0, 1, 2), (1, 0, 1)} linearly independent? We must see if there is a non-trivial solution to
a1(1, 2, 1) + a2(0, 1, 2) + a3(1, 0, 1) = (0, 0, 0)
(3)
This is equivalent to the scalar equations (over Z3):
a1 + a3 = 0, 2a1 + a2 = 0, a1 + 2a2 + a3 = 0
We can solve this in an ad hoc way:
From the first equation, we see that a3 = −a1.
From the second equation, we see that a2 = −2a1.
Plugging into the third equation, we see that
a1 + 2(−2a1) − a1 = 0 and so
Since 2 · 2 = 1, this implies that −a1 = 0 and so a1 = 0 and so
a2 = a3 = 0.
So, this set of vectors is linearly independent. We will develop a concrete algorithm for checking linear independence.
Defn: A basis for a vector space V is a subset S ⊂ V such that
S spans V and S is linearly independent.
Example 1: {(1, 0, . . . , 0), (0, 1, . . . , 0), . . . , (0, 0, . . . , 0, 1)} is a
basis for V (n, q).
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Example 2: {(1, 0, 0), (0, 1, 0)} is a basis for {(x, y, 0) : x, y ∈ Z3}.
Example 3: {(1, 1, 1)} is a basis for {(x, x, x) : x ∈ Z3}.
Theorem 4.2: Every subspace of V (n, q) has a basis. In fact, any
spanning set of a subspace of V (n, q) has a subset which is a basis.
The proof of this theorem inductively extracts a basis from a finite
spanning set. Read and understand the proof in the text.
Given a finite spanning set for a subspace W of V (n, q), we will
develop a concrete algorithm for finding a basis of W .
Theorem 4.3 Let W be a subspace of V (n, q) and let {u1, . . . , uk }
be a basis for W . Then
(i) every element of W can be expressed uniquely as a linear combination of {u1, . . . , uk }.
(ii) |W | = q k .
Proof:
(i): Since {u1, . . . , uk } is a basis for W , it spans W and so every element of W can be expressed as a linear combination of {u1, . . . , uk }.
For the proof of uniqueness, suppose that for some x ∈ W , there
exist a1, . . . , ak , b1, . . . , bk ∈ F s.t.
x=
k
X
aiui
i=1
and
x=
k
X
biui
i=1
Pk
i
Then,
i=1 (ai − bi )u = 0 and so by linear independence, each
ai − bi = 0 and so each ai = bi.
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(ii) This follows from (i) and the fact that there are q k distinct
P
linear combinations ki=1 aiui of the basis elements. It follows from part (ii) of the theorem above that any two bases
of a subspace of V (n, q) have the same size. The common size is
called the dimension of the subspace.
Example 1: {(1, 0, . . . , 0), (0, 1, . . . , 0), . . . , (0, 0, . . . , 0, 1)} is a
basis for V (n, q). Its dimension is n and |V (n, q)| = q n.
Example 2: {(1, 0, 0), (0, 1, 0)} is a basis for W := {(x, y, 0) :
x, y ∈ Z3}. Its dimension is 2 and |W | = 32 = 9.
Example 3: {(1, 1, 1)} is a basis for W := {(x, x, x) : x ∈ Z3}.
Its dimension is 1 |W | = 3.
Remark: Theorem 4.2 and Theorem 4.3(i) can be generalized to
any vector space that has a finite spanning set, such as F n for any
field F (finite or infinite). In fact, these are precisely the finitedimensional vector spaces and any subspace of a finite-dimensional
vector space is finite-dimensional.
The algorithms for checking linear independence and for finding a
basis depend on matrices. Let A be an m × n matrix (rectangular)
with entries in a ring F
Can multiply such matrices, compute determinants of square matrices, etc.
Examples: F = Z4






2 1 3
1 2
1 2
A =  1 1 2  , B =  0 2  AB =  3 2 
0 2 1
1 3
1 3
In order to do the things that we need to do, we will now
require that F be a field. In fact, we assume that F = GF (q).
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Write

R1

 R
A= 2
 ...
Rm





Defn: Elementary row operations on a matrix:
1. Exchange a pair of rows: Ri ↔ Rj
2. Multiply a row by a non-zero scalar: aRi replaces Ri
3. Add a scalar multiple of one row to a different row: aRi + Rj
replaces Rj
Note: any permutation of rows can be realized by a sequence of
row exchanges.
Defn: The row space of an m × n matrix A is the span of its rows.
Denoted R(A).
Defn: A matrix A is row equivalent to a matrix B if B can be
obtained from A by a sequence of elementary row operations.
Note: row equivalence is symmetric: each row operation is invertible (implicit in proof above); so, A is row equivalent to B iff A is
row equivalent to B.
Theorem 1: If B is row equivalent to A, then R(B) = R(A).
Proof postponed.
Defn: A leading entry of a matrix is the left-most non-zero entry
in a row. Sometimes called a pivot entry or pivot position.
Note: a row has a leading entry iff it is nonzero row (i.e., has at
least one nonzero entry).
Defn: A matrix is in Row Echelon form (REF) if
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1. The leading entry in each non-zero row is strictly to the right of
the leading entry in the preceding row.
2. All the non-zero rows (if any) are at the bottom.
It follows that in an REF, all entries in the column below a leading
entry are zero.
L = a leading entry (the different appearances of L below may be
different nonzero elements)

0
 0


 0

 0

 ...
0
...
...
...
...
...
...
0
0
0
0
...
0
L
0
0
0
...
0
∗
0
0
0
...
0
...
...
...
...
...
...
∗
0
0
0
...
0
a
L
0
0
...
0
∗
∗
0
0
...
0
...
...
...
...
...
...
∗
∗
0
0
...
0
b
c
L
0
...
0
∗
∗
∗
0
...
0
...
...
...
...
...
...

∗
∗ 


∗ 

0 

... 
0
Theorem 2: Every matrix is row equivalent to an REF.
Proof postponed.
Note: we say that a matrix B is an REF of a matrix A if B is in
REF and A and B are row equivalent.
Example over GF (3) = Z3: Let


2 1 0
A=1 1 2
0 2 2
Construct an REF B for A:
Step 1: multiply the first row by

2 1
0 2
0 2
10
-2 and add to the second row:

0
2
2
Step 2: subtract the second row

2 1
0 2
0 0
from the third row:

0
2=B
0
Theorem 3: Let W = hSi where S is a finite subset of V (n, q).
Let A be a matrix whose rows are the elements of S. Let B be a
REF of A. Then the nonzero rows of B from a basis for W .
Corollary: Let S be a finite subset of V (n, q). Let A be a matrix
whose rows are the elements of S. Let B be a REF of A. Then S is
linearly independent iff B has no zero rows.
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