2 The Number of Subspaces of F

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faculteit Wiskunde en Natuurwetenschappen

On the Schur Product of Vector Spaces over Finite Fields

Bachelor Thesis Mathematics

July 2012

Student: Christiaan Koster First Supervisor: Prof. dr. J. Top

Second Supervisor: Prof. dr. H. L. Trentelman

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Abstract

In this thesis we consider the vector space of all n-tuples over finite fields. We will investigate the Schur product, also known as entry-wise multiplication, of its linear subspaces, which is equal to the span of the Schur product over all pairs of vectors of a subspace. The number of d-dimensional linear subspaces will be counted, certain properties of the Schur product will be shown and the Schur product of cyclic codes will also be investigated. The main goal is to find good bounds on the number of subspaces for which the Schur product is equal to the total space. This will then be compared to the total number of subspaces as either the dimension of the vector space or the number of elements of the base field tend to infinity.

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1 Introduction

In this modern day and age we live in an information-driven society. Private information about people or companies is very valuable, while the people themselves do not wish this information to be known to others. One could think of PIN-codes or credit card numbers, but also the economic situation of a person and very sensitive data such as nuclear missile codes. While the information must not be known to certain others, it also has to be able to be used. The information in such cases can be regarded as a secret. The problem is to find out how a secret from a certain party can be used in combination with other secrets from other parties, without the secret being known to any of the other parties. An important tool used in solving this problem is known as secret-sharing.

In this thesis we consider vector spaces of n-tuples over finite fields and investigate the Schur product, also known as entry-wise multiplication, of linear subspaces of the vector spaces.

The Schur product is a product that is used in secret sharing. The main goal of this thesis is to find the number of subspaces, or upper and lower bounds, for which the Schur product is equal to the whole vector space and compare this to the total number of linear subspaces.

In order to do this we first count the exact number of linear subspaces. Next we investigate some basic properties of the Schur product and then we handle the main question by trying various methods. It will be done by a coding theoretic approach and we also consider how the Schur product acts on cyclic codes.

1.1 Preliminaries

Let K be a finite field, there is the following theorem on the classification of finite fields, found (in Dutch) in [3]:

Theorem 1.

1. Let K be a finite field. Then there is a prime p and an integer m ≥ 1 with #K = p^m. 2. Conversely, for each prime p and integer m ≥ 1 there is a finite field with p^m elements,

and this field is uniquely determined up to isomorphisms.

The field with q = p^m elements will be denoted by Fq. If p is a prime, then Fp ∼= Z/pZ and we write Fp = {0, 1, . . . , p − 1} where a ∈ Fp is identified with a + pZ. For convenience we just write a instead of a + pZ. In this thesis, the vector space Fⁿq over Fq will be considered, for any integer n, with standard vector addition and scalar multiplication. Define

e1= (1, 0, . . . , 0) e2= (0, 1, . . . , 0)

...

e_n= (0, . . . , 0, 1), then the set {e₁, . . . , e_n} forms the standard basis of Fⁿq.

Definition 1. A set V ⊂ Fⁿq is called a linear subspace of Fⁿq if for all v, w ∈ V and α ∈ Fq

it holds that:

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1. v + w ∈ V 2. αv ∈ V .

Definition 2. The collection of all d-dimensional linear subspaces of Fⁿq is denoted by G(d, n)(Fq), 1 ≤ d ≤ n.

The set G(d, n) is also known as the Grassmannian of Fⁿq. It has a lot of useful properties, but these will not be treated in this thesis. Here it will just be used as easy notation to pick a d-dimensional linear subspace of Fⁿq.

Definition 3. The general linear group GL(n, Fq) is defined as the group of all n×n invertible matrices over Fq.

Definition 4. For any n ∈ N let S = {1, 2, . . . , n}. The symmetric group of S consisting of all bijections of S onto itself is denoted by Sn. An element σ ∈ Sn is called a permutation.

In this thesis a subspace V ∈ G(d, n)(Fq) will be often be written as the span of d basis vectors v₁, . . . , v_d. Each v_i can be written as v_i = a_i,1e_σ(1)+ · · · + a_i,ne_σ(n) for a permutation σ ∈ S_n and where a_i,j ∈ Fq for i = 1, . . . , d, j = 1, . . . , n. Since v₁ 6= 0 we can assume that a1,1 6= 0 by choosing σ appropriately and since V = Span{v₁, . . . , vd} we can take a_i,i = 1.

Then subtracting a_k,1v₁ from v_k for each k = 2, . . . , d, a new basis for V is {v₁, w₂, . . . , w_d} where each w_j can be written as w_j = b_j,2e_σ(2)+ · · · + b_j,ne_σ(n). Since v₁ and w₂ are linearly independent, we can assume bj,2 6= 0 and even b_j,2 = 1. Now subtract bm,2w2 from wm for each m = 3, . . . , d. This can be continued repeatedly until we find a basis for V of the form

{u_i= e_η(i)+ ãi,i+1e_η(i+1)+ · · · + ãi,ne_η(n) : ãi,j ∈ Fq for i = 1, . . . d, j = i + 1, . . . , n}

for a η ∈ S_n. For any i > 1 we can now subtract ˜a_l,iu_i from u_lfor each l = 1, . . . , i − 1. Doing this for each i > 1 and replacing u_j with u_j− ˜a_j,iu_i whenever j < i we find a basis for V of the form

{e_{τ (i)}+ c_i,d+1e_{τ (d+1)}+ · · · + c_i,ne_{τ (n)}: c_i,j ∈ Fq for i = 1, . . . d, j = d + 1, . . . , n} (1) for a τ ∈ S_n. For τ fixed, this basis is unique.

1.2 Coding Theory

In this thesis we will also look at linear subspaces of Fⁿq from a coding theory perspective. All definitions and theorems in this part can be found (in Dutch) in [2]. Since we only consider linear subspaces of Fⁿq, the following definition of codes will be used.

Definition 5. Let K be a field. A code of length n is a linear subspace C of Kⁿ. An element c ∈ C is called a word of the code.

Definition 6. A cyclic code is a code C ⊂ Fⁿq such that if c = (c1, . . . , cn) ∈ C then (cn, c1, . . . , cn−1) ∈ C.

Definition 7. The dual C^⊥ of a code C is defined as C^⊥ = {(a₁, . . . , a_n) ∈ Fⁿq :

n

X

i=1

a_ic_i = 0 ∀(c₁, . . . , c_n) ∈ C}

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Note that C^⊥ is always linear as well.

Proposition 1. Let C ⊂ Fⁿq be a code, then Dim(C) + Dim(C^⊥) = n.

Example 1. Let C ⊂ F³₂ be linear and cyclic. We can construct cyclic codes C of dimension 0, 1, 2 and 3. If ei ∈ C for any i ∈ {1, 2, 3}, then, since C is cyclic, e_i ∈ C for i = 1, 2, 3.

Hence, in this case Dim(C) = 3 and C = F³₂. If C = Span{(1, 1, 1)}, then C is a cyclic code with Dim(C) = 1. Finally, let C = {(0, 0, 0), (1, 1, 0), (0, 1, 1), (1, 0, 1)}. Then C is a cyclic code with Dim(C) = 2. These codes, along with C = {(0, 0, 0)}, are the only cyclic codes in F³₂.

Let Fq[X] denote the polynomial ring over Fq, then for any n ≥ 1 define

R := Fq[X]/(Xⁿ− 1) = {a₀+ a₁X + · · · + a_n−1Xⁿ⁻¹ mod (Xⁿ− 1) : a₀, . . . , a_n−1 ∈ Fq} Next define a mapping g : Fⁿq → R as

g((b₀, . . . , b_n−1)) = b₀+ b₁X + · · · + b_n−1Xⁿ⁻¹ mod (Xⁿ− 1)

for (b0, . . . , bn−1) ∈ Fⁿq. Then g is an isomorphism between the vector spaces Fⁿq and R, that is g is a linear bijective map. From now on, elements of R will be denoted by ¯f ∈ R where f ∈ Fq[X]. The point of the identification of Fⁿq with R will become clear when cyclic codes are considered.

Let C ⊂ Fⁿq be a cyclic code, then C gets mapped by g onto ˜C ⊂ R. Now we will determine what properties ˜C has, caused by the fact that C is a cyclic code.

That C is linear implies the following:

1. 0 ∈ ˜C, since (0, . . . , 0) ∈ C.

2. if f₁, f₂ ∈ ˜C and a ∈ Fq, then f_i= g(c_i) for a c_i∈ C. Since C is linear, also c₁−ac₂ ∈ C and hence f1− af₂ = g(c1− ac₂) ∈ ˜C.

So ˜C is linear as well.

That C is cyclic means that if (b₀, . . . , b_n−1) ∈ C, then (b_n−1, b₀, . . . , b_n−2) ∈ C. In ˜C this means that if ¯f = b0 + b1X + · · · + bn−1Xⁿ⁻¹ mod (xⁿ− 1) ∈ ˜C then also bn−1+ b0X +

· · · + b_n−2Xⁿ⁻¹ mod (Xⁿ− 1) ∈ ˜C. However in ˜C it holds that 1 mod (Xⁿ − 1) = Xⁿ mod (xⁿ−1). So b_n−1+b₀X +· · ·+b_n−2Xⁿ⁻¹ mod (Xⁿ−1) = b₀X +· · ·+b_n−2Xⁿ⁻¹+b_n−1Xⁿ mod (Xⁿ− 1) = Xf . Hence ˜C being cyclic means that if f ∈ ˜C then Xf ∈ ˜C as well.

Combining this with the linearity of ˜C the third property becomes 3. for any f ∈ ˜C, g ∈ R it holds that f g, gf ∈ ˜C.

All 3 properties together are exactly the properties that make ˜C an ideal in R. Hence, we can conclude that

Proposition 2. ˜C ⊂ R is linear and cyclic ⇔ ˜C is an ideal in R.

So to find cyclic codes in Fⁿ_q is equivalent to finding ideals in R. The following proposition states how to do that.

Proposition 3. I is an ideal in R ⇔ I = f R with f |(Xⁿ− 1).

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For an ideal I generated by a polynomial f , that is I = f R, we will use the alternative notation I = (f ).

Example 2. We will again look at F³₂ like in example 1. Now R = F2[X]/(X³ − 1) and X³ − 1 = (X + 1)(X² + X + 1). All cyclic codes can be found by looking at the ideals (1), (X +1), (X²+X +1), (X³−1) (for convenience we won’t denote the bar on each generator).

Now (1) = R and thus has dimension 3. Next, (X + 1) = {0, X + 1, X²+ X, 1 + X²} which is a 2-dimensional cyclic code and it corresponds with C = {(0, 0, 0), (1, 1, 0), (0, 1, 1, ), (1, 0, 1)}

like in example 1. Furthermore, (X²+ X + 1) = {0, X²+ X + 1} and (X³− 1) = {0}.

For a cyclic code (f ) ⊂ R with f |(Xⁿ− 1) we can also determine the dimension.

Proposition 4. Let f |(Xⁿ− 1) ∈ Fq[X], then the cyclic code (f ) has dimension n − deg(f ).

Using proposition 4 there is an easy way of constructing a basis for a cyclic code C = (f ) ⊂ R with f |(Xⁿ− 1). Let k =deg(f ) and write f = Pk

i=0aiXⁱ where each ai ∈ Fq. Since C is cyclic, each of the elements X^jf ∈ C. For 0 ≤ j ≤ n − k − 1 these elements correspond with

(a0, . . . , ak, 0, . . . , 0), (0, a0, . . . , ak, 0, . . . , 0), . . . , (0, . . . , 0, a0, . . . , ak) ∈ Fⁿq

These vectors are all linearly independent and there are n − k of these vectors which is exactly the dimension of C, so they are a basis of C.

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2 The Number of Subspaces of F

ⁿq

In this section we will count the number of d-dimensional subspaces of Fⁿq. We start with the most simple case, namely d = 1.

Example 3. To count all the 1-dimensional subspaces of Fⁿq, it is needed to count how many vectors there are that have a different span. So we have to count all the non-zero vectors up to scalar multiples. There are qⁿ− 1 non-zero vectors in Fⁿq, so the number of 1-dimensional subspaces is equal to (qⁿ− 1)/(q − 1) =Pn−1

i=1 qⁱ.

To count how many subspaces there are of arbitrary dimension, a counting argument using group theory will be used. Let V_d ⊂ Fⁿq be the linear subspace of dimension d, 0 < d ≤ n, defined as V_d= Span{e₁, . . . , e_d}. Let V be another d-dimensional linear subspace of Fⁿq with a chosen basis {f1, . . . , fd}. We extend this basis to a basis of Fⁿq: {f1, . . . , fd, fd+1, . . . , fn} is a basis of Fⁿq for certain f_d+1, . . . , fn∈ Fⁿq. Now define the matrix A : Fⁿq → Fⁿq by ei 7→ f_i for i = 1, . . . , n. Observe that A ∈ GL(n, Fq) and that AV_d = V . But V is an arbitrary d-dimensional linear subspace of Fⁿ_q and for each V we can find the matrix as defined above.

So

G(d, n)(Fq) : = {V ⊂ Fⁿq : V a linear subspace of Fⁿq, Dim(V ) = d}

= {AV_d: A ∈ GL(n, Fq)}.

Note that it is possible that A1V_d= A2V_d for A1, A2∈ GL(n, Fq), A1 6= A₂, so the same subspace could be obtained by different invertible matrices. For example, let {f₁, . . . , f_d, f_d+1, . . . , f_n} and {f1, . . . , fd, ˜fd+1, . . . , ˜fn} be two different bases of Fⁿq and let A1 : ei 7→ f_i, i = 1, . . . , n and A2 : ej 7→ f_j, for j = 1, . . . d and ej 7→ ˜fi for j = d + 1, . . . , n. Then A1Vd = A2Vd, but A16= A₂.

We now look at what happens when AV_d= V_d. We get the following lemma.

Lemma 1. The set H := {A ∈ GL(n, Fq) : AV_d= V_d} is a subgroup of GL(n, Fq).

Proof. For the identity matrix I we have IV_d = V_d, so I ∈ H. If A, B ∈ H, so AV_d = V_d and BV_d = V_d, then ABV_d = AV_d = V_d and hence AB ∈ H. Also, since AV_d = V_d we get V_d= A⁻¹V_d, so A⁻¹∈ H. Hence H is a subgroup of GL(n, Fq).

We use the following result from group theory found in [1].

Theorem 2. [Lagrange’s Theorem] If H is a subgroup of a finite group G, then #H|#G.

The proof of this theorem shows that #G = [G : H] · #H and G = Sm

i=1giH for certain g_i⁰s ∈ G where m = [G : H] is the index of H in G and g_iH ∩ g_jH = ∅ if i 6= j.

Using this result for the groups used in this section, we get

GL(n, Fq) =

[GL(n,Fq):H]

[

i=1

g_iH

for certain g_i⁰s ∈ GL(n, Fq) with g_iH ∩ g_jH = ∅ if i 6= j. We get the following lemma.

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Lemma 2. Let G1 = {g₁, . . . .g_m}, where m = [GL(n, Fq) : H], such that GL(n, Fq) = S

g∈G1gH and Vd= Span{e1, . . . , ed}. Then

G(d, n)(Fq) = {gV_d: g ∈ G1}.

Furthermore #G(d, n)(Fq) = [GL(n, Fq) : H].

Proof. We have already seen that for any g ∈ GL(n, Fq) we get gV_d ∈ G(d, n)(Fq). So {gV_d: g ∈ G1} ⊆ G(d, n)(Fq). Let V ∈ G(d, n)(Fq), then we have seen there is a g ∈ GL(n, Fq) such that V = gVd. Since GL(n, Fq) = S

gi∈G1giH, we get that g = g1h for unique g1 ∈ G1 and h ∈ H. Then V = gV_d = g1hV_d = g1V_d. So V ∈ {gV_d : g ∈ G1}. Hence G(d, n)(Fq) = {gV_d : g ∈ G1}. Now the following holds for g_i, g_j ∈ G1: g_iV_d = g_jV_d ⇔ g_j⁻¹g_iV_d = V_d ⇔ g_j⁻¹gi ∈ H ⇔ g_j⁻¹giH = H ⇔ giH = gjH ⇔ i = j. So each gi ∈ G1 gives a different giV_d∈ G(d, n)(Fq)). Hence #G(d, n)(Fq) = #{gV_d: g ∈ G1} = #G1 = [GL(n, Fq) : H].

Now we want to determine [GL(n, Fq) : H] = #GL(n, Fq)/#H.

Lemma 3. The number of elements of GL(n, Fq) is equal to Qn

i=1(qⁿ− qⁱ⁻¹).

Proof. A matrix A ∈ GL(n, Fq) ⇔ A has n linearly independent columns. Let A = [v1, . . . , vn], with vi ∈ Fⁿq, i = 1, . . . , n. Then v1 needs to have at least one non-zero element for A to be invertible. So there are qⁿ− 1 possibilities. For v₂ we require that it is not a multiple of the first column, that is v₂ 6= a₁v₁ with a₁ ∈ Fq. So for v₂ there are qⁿ− q possibilities.

Continuing in this way, we find that vi cannot be of the form vi = a1v1 + · · · + ai−1vi−1

with a_j ∈ Fq, j = 1, . . . , i − 1. So, for v_i there are qⁿ− qⁱ⁻¹ possibilities. Then we get

#GL(n, Fq) =Qn

i=1(qⁿ− qⁱ⁻¹).

Now that #GL(n, Fq) is known, we need to know how many elements H has.

Lemma 4. The number of elements of H = {A ∈ GL(n, Fq) : AVd= Vd} is equal to

#H = #GL(d, Fq) · #GL(n − d, Fq) · q^d(n−d)=

d

Y

i=1

(q^d− qⁱ⁻¹) ·

n−d

Y

i=1

(q^n−d− qⁱ⁻¹) · q^d(n−d) Proof. To compute the number of elements of H, we first determine what the matrices of H look like as a block matrix. For A ∈ H we have that AVd= Vd, where Vd= Span{e1, . . . , ed}.

So A maps each e_i, i = 1, . . . d, to a linear combination of e₁, . . . , e_d. Taking into account that A has to be invertible, we get that A is of the form

A =A₁₁ A₁₂ 0 A22

where A11∈ GL(d, Fq), A22∈ GL(n − d, Fq) and A12a d × (n − d) matrix. Since each aij ∈ Fq, there are q^d(n−d)possibilities for A12. So we get

#H = #GL(d, Fq) · #GL(n − d, Fq) · q^d(n−d)=

d

Y

i=1

(q^d− qⁱ⁻¹) ·

n−d

Y

i=1

(q^n−d− qⁱ⁻¹) · q^d(n−d)

Now we can count #G(d, n)(Fq).

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Corollary 1. The number of elements of G(d, n)(Fq) is equal to

#G(d, n)(Fq) =

Q_n

i=1(qⁿ− qⁱ⁻¹) q^d(n−d)·Qd

i=1(q^d− qⁱ⁻¹) ·Qn−d

i=1(q^n−d− qⁱ⁻¹) (2) Proof. From lemma 2 we have #G(d, n)(Fq) = [G : H] = #GL(n, Fq)/#H. The result then follows from lemma 3 and 4.

Observe that #G(d, n)(Fq) is symmetric in the variable d around n/2, that is #G(d, n)(Fq) =

#G(n − d, n)(Fq). Another way to see this, is by using proposition 1. That is, if V ∈ G(d, n)(Fq), then V^⊥∈ G(n − d, n)(Fq). So #G(d, n)(Fq) ≤ #G(n − d, n)(Fq). Interchanging d and n − d, it shows that #G(n − d, n)(Fq) ≤ #G(d, n)(Fq) and hence #G(d, n)(Fq) =

#G(n − d, n)(Fq).

2.1 Growth Pattern of #G(d, n)(Fq)

Now that we know the number of d-dimensional subspaces of Fⁿ_q, we wish to determine how that number grows as q increases. If q is large enough, then qⁿ qⁿ⁻¹ for any n ∈ N, so approximately #GL(n, Fq) =Qn

i=1(qⁿ− qⁱ⁻¹) ≈ qⁿ². To be exact, it holds that

#GL(n, Fq) qⁿ² =

Qn

i=1(qⁿ− qⁱ⁻¹)

qⁿ² =

n

Y

i=1

qⁿ− qⁱ⁻¹ qⁿ

=

n

Y

i=1

(1 − qⁱ⁻¹⁻ⁿ) =

n

Y

i=1

(1 − q⁻ⁱ) (3)

Hence

q→∞lim

#GL(n, Fq)

qⁿ² = lim

q→∞

n

Y

i=1

(1 − q⁻ⁱ) = 1

Using the same argument for #G(d, n)(Fq), if q is large enough, we get approximately, using (2),

#G(d, n)(Fq) ≈ qⁿ²

q^d(n−d)· q^d² · q^(n−d)² = qⁿ²^−nd−n²^+2nd−d² = q^d(n−d) So for fixed d and n this gives

q→∞lim

#G(d, n)(Fq)

q^d(n−d) = 1 (4)

Note that #G(d, n)(Fq) has a maximum in the variable d when d = ⁿ₂ if n is even and when d = ^n±1₂ if n is odd. This allows us to determine the growth rate of the total number of subspaces. If n is even we write n = 2m for m ∈ N and use (4) to obtain

q→∞lim Pn

d=0#G(d, n)(Fq)

#G(m, n, q) = lim

q→∞

P2m

d=0q^d(2m−d)

q^m² = lim

q→∞

2m

X

d=0

q^−(m−d)² = 1

Hence, as q → ∞ and n is even, it is enough to look at the number of subspaces of dimension

n

2, instead of the total number of subspaces. This will be useful later on. If n is odd instead,

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we write n = 2m + 1. When d = ^n±1₂ we know #G(d, n)(Fq) has a maximum. Here that is when d = m or d = m + 1. Using (4) again we obtain

q→∞lim Pn

d=0#G(d, n)(Fq) 2#G(m, n, q) = lim

q→∞

P2m+1

d=0 q^d(2m+1−d) 2q^m(m+1) = 1

2 lim

q→∞

2m+1

X

d=0

q−(d−m)(d−m−1)

= 1

Thus for n odd and q → ∞ only the subspaces of dimension ^n±1₂ are important for counting the total number of subspaces.

Now that the growth patterns of #G(d, n)(Fq) are known as q → ∞, we are interested in the same growth pattern as n tends to infinity, both for fixed dimensions and the total number of subspaces. We start again with (3), that is what can be said about the convergence of

n→∞lim

n

Y

i=1

(1 − q⁻ⁱ)

First, note that for any i ∈ N it holds that 1 − q⁻ⁱ < 1. Using this bound for each i > 1 it gives the upper bound

n→∞lim

n

Y

i=1

(1 − q⁻ⁱ) < 1 −1 q. Second, for the convergence it is required that P∞

i=1log(1 − q⁻ⁱ) converges where log is the natural logarithm. Now, the logarithm satisfies log(1 − x) ≥ −2x for any x ∈ [0,¹₂]. To show this, define g : [0,¹₂] → R by g(x) = log(1 − x) + 2x. Then g(0) = 0 and

g⁰(x) = −1

1 − x+ 2 = 2x − 1

x − 1 ≥ 0 for x ∈ [0,1 2].

The claim now follows. Since 0 < q⁻ⁱ≤ ¹₂ we find log(1 − q⁻ⁱ) ≥ −2q⁻ⁱ and

−2

∞

X

i=1

q⁻ⁱ= −2( 1

1 −¹_q − 1) = −2( q

q − 1− 1) = −2 q − 1 so P∞

i=1log(1 − q⁻ⁱ) converges by the comparison test and P∞

i=1log(1 − q⁻ⁱ) ≥_q−1⁻². HenceQ∞

i=1(1 − q⁻ⁱ) converges, so define φ as φ(q) :=Q∞

i=1(1 − q⁻ⁱ) and then φ(q) = lim

n→∞

n

Y

i=1

(1 − q⁻ⁱ) = exp( lim

n→∞log(1 − q⁻ⁱ)) ≥ exp( −2 q − 1) See figure 1 for a plot of φ.

We can conclude that

n→∞lim

#GL(n, Fq) φ(q)qⁿ² = 1

φ(q) lim

n→∞

#GL(n, Fq) qⁿ² = 1

φ(q) lim

n→∞

n

Y

i=1

(1 − q⁻ⁱ) = 1

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Figure 1: Plot of φ

For the growth rate #G(d, n)(Fq) for d and q fixed while n tends to infinity we will find the growth rate of each term separately and then combine them. We have the following

n→∞lim Q_n

i=1(1 − q⁻ⁱ) φ(q)qⁿ² = 1

n→∞lim

q^d(n−d)Qd

i=1(q^d− qⁱ⁻¹) q^d(n−d)Qd

i=1(q^d− qⁱ⁻¹) = 1

n→∞lim

φ(q)q^(n−d)² Qn−d

i=1(q^n−d− qⁱ⁻¹) = 1 Now

n→∞lim #G(d, n)(Fq)/ φ(q)qⁿ² q^d(n−d)Qd

i=1(q^d− qⁱ⁻¹)φ(q)q^(n−d)²

!

= lim

n→∞

Qn

i=1(1 − q⁻ⁱ)

φ(q)qⁿ² ·q^d(n−d)Qd

i=1(q^d− qⁱ⁻¹) q^d(n−d)Qd

i=1(q^d− qⁱ⁻¹) · φ(q)q^(n−d)² Qn−d

i=1(q^n−d− qⁱ⁻¹) = 1 Combining the powers of q in the first denominator gives

qⁿ²

qd(n−d)+(n−d)² = qⁿ²

q^nd−d²⁺ⁿ²^−2nd+d² = q^nd Hence the growth rate of #G(d, n)(Fq) for d and q fixed is

n→∞lim

#G(d, n)(Fq) q^nd/Qd

i=1(q^d− qⁱ⁻¹) = 1 (5)

If now d → ∞ as well, this gives us

n→∞lim

d→∞

#G(d, n)(Fq) q^d(n−d)

φ(q)

= 1 (6)

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For the growth rate of the total number of subspaces as n tends to infinity, we again have to distinguish the cases n even and n odd. Since for n even only subspaces of dimension ⁿ₂ need to be considered, but for n odd the subspaces of dimensions ¹₂(n ± 1) are the ones that are important. We first take n = 2m for m ∈ N and use both (5) and (6) and the symmetry of G(d, n)(Fq) to obtain

m→∞lim P2m

d=0#G(d, 2m)(Fq)

#G(m, 2m)(Fq) = lim

m→∞

φ(q)P2m

d=0q^2md/Qd

i=1(q^d− qⁱ⁻¹) q^m²

= lim

m→∞φ(q)

2m

X

d=0

q^m(2d−m) Qd

i=1(q^d− qⁱ⁻¹)

= lim

m→∞φ(q) q^m² Qm

i=1(q^m− qⁱ⁻¹)+ lim

m→∞2φ(q)

2m

X

d=m+1

q^m(2d−m) Qd

i=1(q^d− qⁱ⁻¹)

= 1 + 2 lim

m→∞

2m

X

d=m+1

q^−(m−d)² = 1 + 2 lim

m→∞

m

X

d=1

q^−d²

The series limm→∞Pm

d=1q^−d² = P∞

d=1q^−d² converges since q^−d² ≤ q^−d for any d ≥ 1 and P∞

d=1q^−d = _q−1¹ , so converges. Now define the function φeven by φeven(q) = 1 + 2P∞ d=1q^−d², for a plot of φ_even see figure 2, then we find

m→∞lim P2m

d=0#G(d, 2m)(Fq) φeven(q)q^m²

φ(q)

= 1

φeven(q) lim

m→∞

φ(q)P2m

d=0#G(d, 2m)(Fq)

q^m² = 1

Figure 2: Plot of φeven

Now we take n = 2m + 1, then the maximum number of subspaces occur at d = m and d = m + 1. Since G(d, n)(Fq) is symmetric, we take 2#G(m, 2m + 1)(Fq) in the limit. Using (5) and (6) again we obtain

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m→∞lim

P2m+1

d=0 #G(d, 2m + 1)(Fq)

2#G(m, 2m + 1)(Fq) = lim

m→∞

φ(q)P2m+1

d=0 q^(2m+1)d/Qd

i=1(q^d− qⁱ⁻¹) 2q^m(m+1)

= lim

m→∞

1 2φ(q)

2m+1

X

d=0

q(d−m)(m+1)+md

Qd

i=1(q^d− qⁱ⁻¹)

= lim

m→∞φ(q) q^(m+1)² Q_m+1

i=1 (q^m− qⁱ⁻¹) + lim

m→∞φ(q)

2m+1

X

d=m+2

q(d−m)(m+1)+md

Qd

i=1(q^d− qⁱ⁻¹)

= 1 + lim

m→∞

2m+1

X

d=m+2

q−(d−m)(d−m−1) = 1 + lim

m→∞

m

X

d=1

q^−d(d+1)

The series lim_m→∞P_m

d=1q^−d(d+1)=P∞

d=1q^−d(d+1) converges as well since q^−d(d+1)≤ q^−d for any d ≥ 1. Now define the function φ_odd by φ_odd(q) = 1 +P∞

d=1q^−d(d+1), for a plot of φ_odd see figure 3, then we find

m→∞lim

P2m+1

d=0 #G(d, 2m + 1)(Fq) 2φ_odd(q)q^m(m+1)

φ(q)

= 1

φodd(q) lim

m→∞

φ(q)P2m+1

d=0 #G(d, 2m + 1)(Fq)

2q^m(m+1) = 1

Figure 3: Plot of φ_odd

From figure 2 and figure 3 we can see that φodd(q) → 1 much faster than φeven(q) → 1. This is not surprising, since for n even we took the subspaces of dimension ⁿ₂, but for n odd we took all the subspaces of dimensions ¹₂(n ± 1). We present the conclusions in the following proposition.

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Proposition 5. Define the functions φ and φ₂ as φ(q) = lim_n→∞Qn

i=1(1 − q⁻ⁱ), φ_even(q) = 1 + 2P∞

d=1q^−d² and φodd(q) = 1 +P∞

d=1q^−d(d+1), then the following limits for the number of subspaces hold.

q→∞lim

#G(d, n)(Fq)

q^d(n−d) = 1

q→∞lim P2m

d=0#G(d, 2m)(Fq)

q^m² = 1

q→∞lim

P2m+1

d=0 #G(d, 2m + 1)(Fq)

2q^m(m+1) = 1

n→∞lim

#G(d, n)(Fq) q^nd/Qd

i=1(q^d− qⁱ⁻¹) = 1

n→∞lim

d→∞

#G(d, n)(Fq) q^d(n−d)

φ(q)

= 1

m→∞lim P2m

d=0#G(d, 2m)(Fq) φeven(q)q^m²

φ(q)

= 1

m→∞lim

P2m+1

d=0 #G(d, 2m + 1)(Fq) 2φodd(q)q^m(m+1)

φ(q)

= 1

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3 Schur Product

In this section the Schur product of vectors and subspaces of Fⁿq will be defined and some basic properties will be shown. We start with the definition.

Definition 8. Let v = (v₁, . . . , v_n), w = (w₁, . . . , w_n) ∈ Fⁿq. The Schur product between v and w, denoted by v ∗ w, is defined as

v ∗ w := (v₁w₁, . . . , v_nw_n) For V, W linear subspaces of Fⁿq the Schur product is defined as

V ∗ W := Span{v ∗ w : v ∈ V, w ∈ W }

In this thesis we only consider the Schur product V ∗ V for a V ∈ G(d, n)(Fq).

Lemma 5. For any u, v, w ∈ Fⁿq and α ∈ Fq the following properties of the Schur product hold.

1. v ∗ w = w ∗ v 2. (αv) ∗ w = α(v ∗ w)

3. (u + v) ∗ w = u ∗ w + v ∗ w

4. v ∗ w = 0 if and only if for any i ∈ {1, . . . , n} it holds that vi= 0 or wi = 0.

5. For V, W subspaces of Fⁿq with W ⊂ V , then W ∗ W ⊂ V ∗ V . Proof. 1. Since v_i, w_i ∈ Fq, we have that v_iw_i = w_iv_i for i = 1, . . . n. So

v ∗ w = (v1w1, . . . , vnwn) = (w1v1, . . . , wnvn) = w ∗ v 2. (αv) ∗ w = (αv1w1, . . . , αvnwn) = α(v1w1, . . . , vnwn) = α(v ∗ w)

3. (u+v)∗w = ((u1+v1)w1, . . . , (un+vn)wn) = (u1w1+v1w1, . . . , unwn+vnwn) = u∗w +v ∗w 4. v ∗ w = 0 if and only if v_iw_i = 0 for i = 1, . . . , n. Since Fq is a domain, v_iw_i = 0 if and only if vi = 0 or wi= 0.

5. Let v, w ∈ W , then v, w ∈ V . So v ∗ w ∈ V ∗ V . Hence W ∗ W ⊂ V ∗ V .

If V ⊂ Fⁿq is given as V = Span{v1, . . . , vm} with v_i ∈ Fⁿq, i = 1, . . . , m, then any u, w ∈ V can be written as u = P_m

j=1a_jv_j and w = P_m

j=1b_jv_j for certain a_j, b_j ∈ Fq, j = 1, . . . , m.

Then

v ∗ w = (

m

X

j=1

a_jv_j) ∗ (

m

X

k=1

b_kv_k) =

m

X

j=1 m

X

k=1

a_jb_kv_j ∗ v_k

So V ∗ V is spanned by {v_j∗ v_k : j = 1, . . . , m, k = 1, . . . , m}. Note, that if all vectors in {v₁, . . . , v_m} are linearly independent, it is in most cases not true that all vectors in {v_j∗v_k : j = 1, . . . , m, k = 1, . . . , m} are linearly independent. As a consequence of this, we have the following result.

Corollary 2. If V ⊂ Fⁿq has dimension d, then Dim(V ∗ V ) ≤ ^d₂ + d.

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Proof. Let {v₁. . . . , v_d} be a basis of V . Then V ∗ V = Span{v_j ∗ v_k : j = 1, . . . , m, k = 1, . . . , m} and since the Schur product is symmetric the set {vj ∗ v_k |j = 1, . . . , m, k = 1, . . . , m} consists of at most ^d₂ + d elements. Hence Dim(V ∗ V ) ≤ ^d₂ + d.

Some examples to show what the Schur product of subspaces can be will now be given.

Example 4. 1. Let V ⊂ F³q be given as V = Span{v₁ = (1, 1, 0), v₂ = (0, 1, 1)}. Then v1∗ v₁ = (1, 1, 0) = v1

v₁∗ v₂ = (0, 1, 0) v₂∗ v₂ = (0, 1, 1) = v₂ These are all linearly independent, and so V ∗ V = F³q.

2. Let W ⊂ F³q be given as W = Span{w1= (1, 1, 0), w2 = (0, 0, 1)}. Then w₁∗ w₁ = (1, 1, 0) = w₁

w₁∗ w₂ = (0, 0, 0) w2∗ w₂ = (0, 0, 1) = w2

We see that here W ∗ W = W .

3. Let U ⊂ F³₃ be given as U = Span{u₁= (1, 0, 0), u₂ = (0, 2, 1)}. Then u₁∗ u₁= (1, 0, 0) = u₁

u₁∗ u₂= (0, 0, 0) u2∗ u₂= (0, 1, 1)

From this we get U ∗ U = Span{u1, (0, 1, 1)} and we notice that now U ∗ U 6= U . From these examples it is clear that the Schur product of subspaces can vary greatly. The Schur product of a subspace can be the whole space Fⁿq, the subspace itself, and it can be different from the subspace but with the same dimension. In particular, if V ⊂ Fⁿq, then it does not hold in general that V ⊂ V ∗ V . However, over F2 we have that v ∗ v = v for any v ∈ Fⁿ2. So then it is always true that V ⊂ V ∗ V .

The Schur product of vectors was defined with respect to the standard basis of Fⁿq. The next example shows what can happen if a different basis is chosen.

Example 5. Let (e₁, e₂, e₃) be the standard basis of F³₃ and take v₁ = (2, 1, 0), v₂ = (0, 1, 2) with respect to this basis. If the subspace V is defined as V = Span{v₁, v₂}, we have

v₁∗ v₁ = (1, 1, 0) v1∗ v₂ = (0, 1, 0) v2∗ v₂ = (0, 1, 1) and thus V ∗ V = F³3.

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However, if we take B = {v₁, v₂, e₂} as a basis of F³₃, then, with respect to the new basis, v1 = (1, 0, 0)B, v2 = (0, 1, 0)B. For the Schur product of V we now get

v₁∗ v₁= (1, 0, 0)_B v₁∗ v₂= (0, 0, 0)_B v₂∗ v₂= (0, 1, 0)_B and so, with respect to the basis B, V ∗ V = V .

From example 5 it is clear that the Schur product depends on the basis chosen for Fⁿq. This won’t pose much of a problem though, since if B1, B2 are two different basis of Fⁿq and V, W ⊂ Fⁿq such that if v = (v₁, . . . , v_n)_B₁ ∈ V then there is a w ∈ W with w = (v₁, . . . , v_n)_B₂ and vice versa. Let (V ∗ V )Bi denote the Schur product of V with respect to the basis Bi. Then for any u = (u1, . . . , un)B1 ∈ (V ∗ V )_B₁) there is a x ∈ (W ∗ W )B2 such that x = (u₁, . . . , u_n)_B₂. So if for a V ⊂ Fⁿq the Schur product V ∗ V has a certain property with respect to a basis B of Fⁿq, then there is a W ⊂ Fⁿq such that W ∗ W has the same property with respect to the standard basis of Fⁿq.

In example 4 there were examples of subspaces for which the Schur product either increased in dimension or for which the dimension stayed the same. This leads us to wondering whether the Schur product can also decrease in dimension. The following lemma deals with this.

Lemma 6. Let V be a d-dimensional subspace of Fⁿq, then Dim(V ∗ V ) ≥ d.

Proof. Let V ∈ G(d, n)(Fq), then from (1) there is a basis of V consisting of vectors of the form vi = e_σ(i)+Pn

j=d+1ai,je_σ(j) where σ ∈ Sn, i = 1, . . . , d and ai,j ∈ Fq ∀i, j. Then v_i ∗ v_i = e_σ(i)+P_n

j=d+1a²_i,je_σ(j). Since e_σ(i) ∈ Span{e/ _σ(1), . . . , e_σ(i−1), e_σ(i+1), . . . , e_σ(d)} for any i ∈ {1, . . . , d}, it follows that v_i∗ v_i ∈ Span{v/ ₁∗ v₁, . . . , v_i−1∗ v_i−1, v_i+1∗ v_i+1, . . . , v_d∗ v_d} for any i ∈ {1, . . . , d}. Hence the vectors v1∗ v₁, . . . , vd∗ v_d are all linearly independent and thus Dim(V ∗ V ) ≥ d.

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4 Schur Product of Subspaces

Now that some of the basic properties of Schur products are known, we wish to answer the main question: How many linear subspaces of Fⁿq are there for which the Schur product is not equal to Fⁿq. An exact answer might not be possible, but upper or lower bounds might be able to be found. In this section various methods will be used to try and find these upper or lower bounds.

Definition 9. The set of all d-dimensional linear subspaces of Fⁿq for which the Schur product is, respectively is not, equal to Fⁿq is denoted by W(d, n, q), respectively V(d, n, q), i.e.

W(d, n, q) = {W ⊂ Fⁿq : Dim(W ) = d, W ∗ W = Fⁿq} and V(d, n, q) = G(d, n)(Fq)\W(d, n, q).

A simple case is given by Corollary 2, since if ^d₂ + d < n it holds automatically that for a d-dimensional subspace the Schur product is not the whole space. Determining when

d 2

+ d − n = d(d + 1)

2 − n = 0 we find that d = ¹₂ ± ¹₂√

1 + 8n. Since d ≥ 0, the condition ^d₂ + d < n holds when 0 ≤ d <

−¹₂+¹₂√

1 + 8n. Since√

1 + 8n >√

8n = 2√

2n, a more intuitive bound is 0 ≤ d < −¹₂+√ 2n.

Lemma 7. Let V ∈ G(d, n)(Fq) with 0 ≤ d < −¹₂ +¹₂√

1 + 8n then V ∈ V(d, n, q).

Next, from lemma 5 (5), if W ⊂ V ⊂ Fⁿq, then W ∗ W ⊂ V ∗ V . Hence, if V ∗ V 6= Fⁿq, then W ∗ W 6= Fⁿq as well. Thus, if we can count #V(n − 1, n, q), then we can find a lower bound for all the lower dimensional subspaces for which the property holds. This can be done by counting how many different subspaces there are in all the (n − 1)-dimensional subspaces for which the Schur product is not the whole space. However, counting this amount may be quite difficult as one needs to take into account all the subspaces which are counted multiple times.

Using Magma Calculator to count the number of (n − 1)-dimensional subspaces for which the Schur product is not the whole space for various n and q, the following results were obtained.

n \ q 2 3 4 5

1 1 1 1 1

2 3 4 5 6

3 6 9 12 15

4 10 16 22 28

5 15 25 35 45

6 21 49 51 -

From the table, there seems to be a pattern in each column. For q = 2 it seems to be

#V(n − 1, n, 2) =Pn

k=1k. We will prove this is true to gain insight in the general case.

Example 6. Let V ∈ G(n − 1, n)(F2), then V ⊂ V ∗ V . Thus, V ∈ V(n − 1, n, 2) is equivalent to V ∗ V = V in this case. There is a basis of V of the form

{v_i = e_σ(i)+ a_ie_σ(n): a_i ∈ F2, i = 1, . . . , n − 1}

for a certain permutation σ ∈ Sn. Then vi∗ v_j = aiaje_σ(n) if i 6= j and vi∗ v_i = vi. So, if V ∗ V = V , this implies that v_i∗ v_j = 0 for all i 6= j. Hence, there is at most one a_i 6= 0.

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There are two possibilities now: either a_i = 0 for all i, or there is exactly one a_i 6= 0. If all ai = 0, then V is just the span of n−1 standard basis vectors. So there are _n−1ⁿ = n different subspaces for which this is true. Now assume there is exactly one a_i 6= 0. Without loss of generality we can take a₁ 6= 0, so a₁ = 1. Then V = Span{e_σ(1)+ e_σ(n), e_σ(2), . . . , e_σ(n−1)}.

If σ(1) and σ(n) are chosen, V is fixed. For σ(1) and σ(n) there are ⁿ₂ different choices.

Hence

#V(n − 1, n, 2) = n +n 2

= n + n(n − 1)

2 = n(n + 1)

2 =

n

X

k=1

k.

From the Magma Calculator data, the number of elements of V(n − 1, n, q) seems to increase as follows: for q = 3 the data suggest #V(n − 1, n, 3) =Pn

k=1(2k − 1), for q = 4 it is #V(n − 1, n, 4) =Pn

k=1(3k − 2). In general this seems to be #V(n − 1, n, q) =Pn

k=1((q − 1)k − (q − 2).

This can be rewritten as

n

X

k=1

((q − 1)k − (q − 2) = (q − 1)n(n + 1)

2 − n(q − 2)

= q(n(n + 1)

2 − n) −n(n + 1) 2 + 2n

= qn 2

−n 2

+ n

= (q − 1)n 2

+ n

We will prove this is true, using mostly the same arguments used in example 6 but adjusted where needed.

Lemma 8. The number of elements of V(n − 1, n, q) is equal to

#V(n − 1, n, q) = qn(n − 1)

2 − n(n − 3)

2 .

Proof. Let V ∈ V(n − 1, n, q). We choose a basis of V of the form {v_i = e_σ(i)+ a_ie_σ(n): a_i ∈ Fq, i = 1, . . . , n − 1}

for a certain σ ∈ S_n. Then v_i∗ v_i = e_σ(i)+ a²_ie_σ(n) and v_i∗ v_j = a_ia_je_σ(n) when i 6= j. The vectors v1∗ v₁, . . . , vn−1∗ v_n−1 are linearly independent and vi∗ v_j ∈ Span{v/ ₁∗ v₁, . . . , vn−1∗ vn−1}. Since V ∈ V(n − 1, n, q), this implies v_i ∗ v_j = 0 for all i 6= j. Hence there is at most one a_i 6= 0. If all a_i = 0, then V is just the span of n − 1 standard basis vectors.

So there are _n−1ⁿ

= n different subspaces for which this is true. Now suppose there is exactly one ai 6= 0. Without loss of generality we can take a₁ 6= 0. Then V = Span{e_σ(1)+ a₁e_σ(n), e_σ(2), . . . , e_σ(n−1)}. For a₁ there are q − 1 possibilities. If σ(1) and σ(n) are chosen, V is fixed. For σ(1) and σ(n) there are ⁿ₂

different choices. Hence #V(n − 1, n, q) = (q − 1) ⁿ₂ + n.

From the proof it is seen that

V(n − 1, n, q) = {V = Span{e_σ(1)+ a₁e_σ(n), e_σ(2), . . . , e_σ(n−1)} : σ ∈ S_n, a₁∈ Fq}

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For (n − 1)-dimensional subspaces we could determine #V(n − 1, n, q) exact. One could wonder if, using similar arguments, #V(d, n, q) can be determined for arbitrary d. This might be possible for some specific d’s, however the combinatorics involved quickly become complicated as d approaches ⁿ₂. For d = n − 2 we will compute #V(n − 2, n, q) analytically, which will also make clear how the computations are more difficult from the n − 1-dimensional case.

Any element V of G(n − 2, n)(Fq) has a basis of the form

{v_i= e_σ(i)+ aie_σ(n−1)+ bie_σ(n) : ai, bi∈ Fq, i = 1, . . . , n − 2, σ ∈ Sn}

If V ∈ V(n − 2, n, q), then either Dim(V ∗ V ) = n − 2 or Dim(V ∗ V ) = n − 1. When n = 2 or n = 3 it is trivial to compute #V(n − 2, n, q), since then #V(n − 2, n, q) = #G(n − 2, n)(Fq).

So let n ≥ 4, we will look at both cases separately. Let Dim(V ∗ V ) = n − 2. Since the vectors v1∗ v₁, . . . , vn−2∗ v_n−2 are linearly independent, this means that vi∗ v_j = 0 whenever i 6= j.

Hence there are at most one a_i 6= 0 and one b_i 6= 0. If there is one a_i6= 0, then we can choose i arbitrarily since permutations are used to count all the different possibilities. There are the following cases:

1. ai = bi = 0 for i = 1, . . . , n − 2, so V = Span{e_σ(1), . . . , e_σ(n−2)} for which there are ⁿ₂ possibilities.

2. an−2 6= 0 and b_j = 0 for j = 1, . . . , n − 2. So V = Span{e_σ(1), . . . , e_σ(n−3), e_σ(n−2)+ a_n−2e_σ(n−1)}. There are q − 1 choices for a_n−2, ⁿ₂ different choices for σ(n − 2) and σ(n − 1). There are then n − 2 choices left for σ(n), which each give rise to a different subspace. Hence there are ⁿ₂(q − 1)(n − 2) different subspaces of this form.

3. There is exactly one a_i 6= 0 and b_j 6= 0 with i 6= j. Take i = n − 3 and j = n − 2, so V = Span{e_σ(1), . . . , e_σ(n−4), e_σ(n−3)+ an−3e_σ(n−1), e_σ(n−2)+ bn−2e_σ(n)}

There are (q − 1)² different choices for an−3 and bn−2 together and ⁿ₄ different possibilities for σ(n − 3), . . . , σ(n). To determine how many different subspaces there are for σ(n − 3), . . . , σ(n) fixed, it is equal to counting how many different two 2cycles there are on S4, which is 3. So in this case there are 3 ⁿ₄(q − 1)² different possibilities.

4. a_n−26= 0 and b_n−2 6= 0 and a_i = b_i = 0 when i 6= n−2. Here V = Span{e_σ(1), . . . , e_σ(n−2)+ an−2e_σ(n−1)+ bn−2e_σ(n)}. For the a_n−2 and bn−2 there are (q − 1)² choices. Switching any of the σ(n − 2), σ(n − 1) and σ(n) doesn’t give any new subspaces and when they are chosen the rest is fixed. Hence there are ⁿ₃(q − 1)² different subspaces of this form.

These were all the cases of Dim(V ∗ V ) = n − 2. So now we let Dim(V ∗ V ) = n − 1. Since the vectors v₁ ∗ v₁, . . . , v_n−2∗ v_n−2 are linearly independent, this means that there are k, l, k 6= l such that v_k∗ v_l∈ Span{v/ ₁∗ v₁, . . . , v_n−1∗ v_n−1} and v_i∗ v_j = αv_k∗ v_l for all i 6= j and α ∈ Fq. This happens in the following cases.

5. There are at least two ai6= 0 and b_j = 0 for all j. So V is of the form V = Span{e_σ(1)+ a₁e_σ(n−1), . . . , e_σ(n−2)+ a_n−2e_σ(n−1)}. First note that each choice of σ(n) gives rise to the same number of subspaces with all the choices for the ai’s and σ(1), . . . , σ(n − 1) and all these subspaces are different. Consider now the case with σ the identity, then the matrix of V with respect to the basis used here is:

2 The Number of Subspaces of F

On the Schur Product of Vector Spaces over Finite Fields

Bachelor Thesis Mathematics

Contents

1 Introduction

2 The Number of Subspaces of F

3 Schur Product

4 Schur Product of Subspaces