Proof of Inequality Shown in 5.1

LetV be a finite dimensional vector space and letA, B₁, B₂, . . . , Bq−1, C, V₁, V₂, . . . , Vq−1, W, X, Y, Z be subspaces ofV for some positive integerq. We now list a set of functions that maps these subspaces one to another. The mapping of these functions is shown pictorially in Fig. 5.3.

for 1≤i≤q−1 :f_AVi :A→V_i f_AZ :A→Z

for 1≤i, j≤q−1, j 6=i:f_BiBjB_i →B_j for 1≤i≤q−1 :f_BiY :B_i→Y for 1≤i≤q−1 :fBiZ :Bi →Z fCA :C→A

5.3 Proof of Inequality Shown in 5.1

B_i

X

B_j

V_i

Y f_XB

i fCBⁱ

fXB f^WB j

i

f_CY f_YX

f_CB

j

fWB^j

f_YW

f_XC

f_ZC

fAZ f_ZW

f_CV

i

f_CA

f_AV

i

f_WA f_{B B}

i j

C

W

Z

A f_B

Y

j

f_{B Z}

j

fV Xi

f_{V B}

j i

fB_iY fBZ_i

f_{B B}

j i

Figure 5.3: In this figure each circle with a label inside represents a vector subspace. We assume a set of mappings between these subspaces; these mappings are shown with an arrow directed from the domain to the co-domain (the mappings are shown for one particular values of i and j where 1≤i, j≤q−1,i6=j).

for 1≤i≤q−1 :fCBi :C →Bi for 1≤i≤q−1 :fCVi :C→Vi

f_CY :C→Y for 1≤i, j≤q−1, j 6=i:f_ViBj :V_i →B_j for 1≤i≤q−1 :fViX :Vi →X for 1≤i≤q−1 :fW Bi :W →Bi

f_{W A} :W →A for 1≤i≤q−1 :f_XBi :X →Bi

f_XC :X→C f_{Y W} :Y →W

fY X :Y →X fZC :Z→C

f_ZW :Z →W.

Due to Lemma 8, we have the following results.

(i) Pq−1

i=1fAVi+fAZ =I over a subspaceA⁰ ofA where

codim_A(A⁰)≤dim(A|V₁, V₂, . . . , Vq−1, Z) (5.6)

TH-2118_136102023

(ii) for 1≤i≤q−1: Pq−1

j=1,j6=ifBiBj+fBiY +fBiZ =I over a subspaceB_i⁰ of Bi where

codim_Bi(B⁰_i)≤dim(B_i|B₁, . . . , Bi−1, B_i+1, . . . , Bq−1, Y, Z) (5.7)

(iii) f_CA+f_CY =I over a subspaceC⁰ of C where

codim_C(C⁰)≤dim(C|A, Y) (5.8)

(iv) for 1≤i≤q−1: f_CBi+f_CVi =I over a subspaceC_i of C where

codimC(Ci)≤dim(C|B_i, Vi) (5.9) (v) for 1≤i≤q−1: Pq−1

j=1,j6=ifViBj+fViX =I over a subspaceV_i⁰ of Vi where

codim_Vi(V_i⁰)≤dim(V_i|B₁, . . . , Bi−1, B_i+1, . . . , Bq−1, X) (5.10) (vi) f_{W A}+Pq−1

i=1f_{W Bi} =I over a subspaceW⁰ of W where

codimW(W⁰)≤dim(W|A, B₁, B2, . . . , Bq−1) (5.11) (vii) Pq−1

i=1fXBi+fXC =I over a subspaceX⁰ of X where

codimX(X⁰)≤dim(X|B₁, B2, . . . , Bq−1, C) (5.12)

(viii) f_{Y W} +f_{Y X} =I over a subspaceY⁰ ofY where

codim_Y(Y⁰)≤dim(Y|W, X) (5.13)

(ix) f_ZC+f_ZW =I over a subspaceZ⁰ of Z where

codimZ(Z⁰)≤dim(Z|C, W) (5.14)

Now, let’s consider the following composite functions:

f_CA+f_{W A}f_{Y W}f_CY :C →A (5.15)

for 1≤i≤q−1 : (f_{W Bi}f_{Y W} +f_XBif_{Y X})f_CY :C→B_i (5.16)

fXCfY XfCY :C→C. (5.17)

5.3 Proof of Inequality Shown in 5.1

Using (5.11), we have:

f_{W A}f_{Y W}f_CY +

q−1

X

i=1

f_{W Bi}f_{Y W}f_CY =f_{Y W}f_CY over a subspacef_CY⁻¹f_{Y W}⁻¹ (W⁰) ofC. (5.18) Using (5.12), we have:

f_XCf_{Y X}f_CY +

q−1

X

i=1

f_XBif_{Y X}f_CY =f_{Y X}f_CY over a subspacef_CY⁻¹f_{Y X}⁻¹(X⁰) ofC. (5.19) Using (5.13), we have:

f_{Y W}f_CY +f_{Y X}f_CY =f_CY over a subspacef_CY⁻¹(Y⁰) of C. (5.20) We now add functions shown in (5.15)-(5.17) and use equations (5.8), (5.18), (5.19), and (5.20).

f_{W A}f_{Y W}f_CY +

q−1

X

i=1

f_{W Bi}f_{Y W}f_CY +f_XCf_{Y X}f_CY +

q−1

X

i=1

f_XBif_{Y X}f_CY +f_CA

=f_{Y W}f_CY +f_{Y X}f_CY +f_CA

=fCY +fCA

=I over a subspace: C⁰⁰ =f_CY⁻¹f_{Y W}⁻¹ (W⁰)∩f_CY⁻¹f_{Y X}⁻¹(X⁰)∩f_CY⁻¹(Y⁰)∩C⁰ ofC.

Using Lemma 6, we get:

codim_C(C⁰⁰)≤codim_C(f_CY⁻¹f_{Y W}⁻¹ (W⁰)) +codim_C(f_CY⁻¹f_{Y X}⁻¹(X⁰)) +codim_C(f_CY⁻¹(Y⁰)) +codim_C(C⁰) Using (7), we have:

codim_C(C⁰⁰)≤codim_W(W⁰) +codim_X(X⁰) +codim_Y(Y⁰) +codim_C(C⁰). (5.21) Now, according to Lemma 9 there exists a subspace ¯C of C⁰⁰ over which:

fCA+fW AfY WfCY = 0 (5.22)

for 1≤i≤q−1 : (f_{W Bi}f_{Y W} +f_XBif_{Y X})f_CY = 0 (5.23)

fXCfY XfCY =−I. (5.24)

where codim_C⁰⁰( ¯C)≤dim(A) +

q−1

X

i=1

dim(Bi) +dim(C)−dim(A, B1, B2, . . . , Bq−1, C). (5.25)

TH-2118_136102023

AscodimC( ¯C) =codimCC⁰⁰+codim_C⁰⁰C, we have:¯

codim_C( ¯C)≤codim_W(W⁰) +codim_X(X⁰) +codim_Y(Y⁰) +codim_C(C⁰) +dim(A) +

q−1

X

i=1

dim(Bi)

+dim(C)−dim(A, B1, B2, . . . , Bq−1, C). (5.26)

Now, let’s consider the following composite functions:

fW A(fY WfBiY +fZWfBiZ) :Bi →A (5.27)

(f_{W Bi}f_{Y W} +f_XBif_{Y X})f_BiY +f_{W Bi}f_ZWf_BiZ :Bi→Bi (5.28) for 1≤j≤q−1, j 6=i: (f_{W Bj}f_{Y W} +f_XBjf_{Y X})f_BiY +f_{W Bj}f_ZWf_BiZ+f_BiBj :B_i →B_j (5.29)

fXCfY XfBiY +fZCfBiZ :Bi →C. (5.30)

Using equation (5.11), we have:

f_{W A}f_{Y W}f_BiY +

q−1

X

i=j

f_{W Bj}f_{Y W}f_BiY =f_{Y W}f_BiY over a subspacef_B⁻¹

iYf_{Y W}⁻¹ (W⁰) ofB_i. (5.31) Using equation (5.11), we have:

fW AfZWfBiZ+

q−1

X

j=1

fW BjfZWfBiZ =fZWfBiZ over a subspacef_B⁻¹

iZf_ZW⁻¹ (W⁰) ofBi. (5.32) Using equation (5.12), we have:

f_XCf_{Y X}f_BiY +

q−1

X

j=1

f_XBjf_{Y X}f_BiY =f_{Y X}f_BiY over a subspace f_B⁻¹

iYf_{Y X}⁻¹(X⁰) of B_i. (5.33) Using equation (5.13), we have:

fY WfBiY +fY XfBiY =fBiY over a subspacef_B⁻¹

iY(Y⁰) of Bi. (5.34) Using equation (5.13), we have:

f_ZWf_BiZ+f_ZCf_BiZ=f_BiZ over a subspacef_B⁻¹

iZ(Z⁰) of B_i. (5.35) Consider the subspace B_i⁰⁰ of Bi where:

B_i⁰⁰=f_B⁻¹

iYf_{Y W}⁻¹ (W⁰)∩f_B⁻¹

iZf_ZW⁻¹ (W⁰)∩f_B⁻¹

iYf_{Y X}⁻¹(X⁰)∩f_B⁻¹

iY(Y⁰)∩f_B⁻¹

iZ(Z⁰)∩B_i⁰. (5.36)

5.3 Proof of Inequality Shown in 5.1

Adding functions shown in (5.27)-(5.30) over B_i⁰⁰ and using equations (5.7) and (5.31)-(5.35), we get:

f_{W A}(f_{Y W}f_BiY +f_ZWf_BiZ) +sum^q−1_j=1f_{W Bj}f_{Y W}f_BiY +sum^q−1_j=1f_XBjf_{Y X}f_BiY +sum^q−1_j=1f_{W Bj}f_ZWf_BiZ+sum^q−1_j=1,j6=if_BiBj+f_XCf_{Y X}f_BiY +f_ZCf_BiZ

=fY WfBiY +fZWfBiZ+sum^q−1_j=1,j6=ifBiBj+fY XfBiY +fZCfBiZ

=f_BiY +f_BiZ+sum^q−1_j=1,j6=if_BiBj =I.

Then, according to Lemma 9 there exists a subspace ¯B_i of B_i⁰⁰ over which:

fW A(fY WfBiY +fZWfBiZ) = 0 (5.37)

(f_{W Bi}f_{Y W} +f_XBif_{Y X})f_BiY +f_{W Bi}f_ZWf_BiZ−I = 0 (5.38) for 1≤j ≤q−1, j 6=i: (fW BjfY W +fXBjfY X)fBiY +fW BjfZWfBiZ+fBiBj = 0 (5.39)

fXCfY XfBiY +fZCfBiZ= 0. (5.40)

such that

codim_B⁰⁰

i( ¯Bi)≤dim(A) +

q−1

X

i=1

dim(Bi) +dim(C)−dim(A, B1, B2, . . . , Bq−1, C). (5.41) Applying Lemma 6 to equation (5.36), we get:

codimBi(B_i⁰⁰)≤codimBi(f_B⁻¹

iYf_{Y W}⁻¹ (W⁰)) +codimBi(f_B⁻¹

iZf_ZW⁻¹ (W⁰)) +codimBi(f_B⁻¹

iYf_{Y X}⁻¹(X⁰)) +codim_Bi(f_B⁻¹

iY(Y⁰)) +codim_Bi(f_B⁻¹

iZ(Z⁰)) +codim_Bi(B_i⁰). (5.42) Applying Lemma 7 to equation (5.42), we get:

codimBi(B⁰⁰_i)≤codimW(W⁰) +codimW(W⁰) +codimX(X⁰) +codimY(Y⁰) +codimZ(Z⁰) +codim_Bi(B_i⁰). (5.43) From equations (5.41) and (5.43), we get:

codimBi( ¯Bi) =codimBi(B_i⁰⁰) +codim_B⁰⁰

i( ¯Bi) (5.44)

≤2codim_W(W⁰) +codim_X(X⁰) +codim_Y(Y⁰) +codim_Z(Z⁰) +codim_Bi(B_i⁰) (5.45) +dim(A) +

q−1

X

i=1

dim(Bi) +dim(C)−dim(A, B1, B2, . . . , Bq−1, C). (5.46)

TH-2118_136102023

Consider the following composite functions.

f_{W A}f_ZWf_AZ :A→A (5.47)

for 1≤i≤q−1 :f_{W Bi}f_ZWf_AZ+f_XBif_ViXf_AVi+

q−1

X

j=1,j6=i

(f_XBif_VjX +f_VjBi)f_AVj :A→B_i (5.48)

f_ZCf_AZ+

q−1

X

j=1

f_XCf_VjXf_AVj :A→C. (5.49)

Using equation (5.11), we get:

f_{W A}f_ZWf_AZ+

q−1

X

i=1

f_{W Bi}f_ZWf_AZ =f_ZWf_AZ over a subspacef_AZ⁻¹f_ZW⁻¹ (W⁰) of A. (5.50) Using equation (5.12), we get:

q−1

X

j=1

f_XCf_VjXf_AVj+

q−1

X

i=1

(f_XBi(

q−1

X

j=1

f_VjXf_AVj))

=fXC q−1

X

j=1

fVjXfAVj+ (

q−1

X

i=1

fXBi)(

q−1

X

j=1

fVjXfAVj)

=

q−1

X

j=1

fVjXfAVj over a subspace(

q−1

X

j=1

fVjXfAVj)⁻¹(X⁰) ofA. (5.51) Using equation (5.10), we get:

q−1

X

j=1

f_VjXf_AVj+

q−1

X

i=1 q−1

X

j=1,j6=i

f_VjBif_AVj

=

q−1

X

j=1

f_VjXf_AVj+

q−1

X

j=1 q−1

X

i=1,i6=j

f_VjBif_AVj

=

q−1

X

j=1

(f_VjX +

q−1

X

i=1,i6=j

f_VjBi)f_AVj

=

q−1

X

j=1

f_AVj over a subspace f_AV⁻¹

1(V₁⁰)∩f_AV⁻¹

2(V₂⁰)∩ · · · ∩f_AV⁻¹

q−1(V_q−1⁰ ) of A. (5.52) Using equation (5.14), we get:

f_ZWf_AZ+f_ZCf_AZ =f_AZ over a subspacef_AZ⁻¹(Z⁰) of A. (5.53) Consider the following subspace A⁰⁰ ofA.

A⁰⁰=f_AZ⁻¹f_ZW⁻¹ (W⁰)∩(

q−1

X

j=1

fVjXfAVj)⁻¹(X⁰)∩f_AV⁻¹

1(V₁⁰)∩ · · ∩f_AV⁻¹

q−1(V_q−1⁰ )∩f_AZ⁻¹(Z⁰)∩A⁰. (5.54)

5.3 Proof of Inequality Shown in 5.1

Now adding the functions shown in equations (5.47)-(5.49) overA⁰⁰ and using equations (5.6), (5.50), (5.51), (5.52), and (5.53), we get:

fW AfZWfAZ+

q−1

X

i=1

(fW BifZWfAZ+fXBifViXfAVi +

q−1

X

j=1,j6=i

(fXBifVjX+fVjBi)fAVj) +fZCfAZ

+

q−1

X

j=1

f_XCf_VjXf_AVj

=fZWfAZ+

q−1

X

j=1

fVjXfAVj+

q−1

X

i=1 q−1

X

j=1,j6=i

fVjBifAVj+fZCfAZ

=fAZ+

q−1

X

j=1

fAVj =I. (5.55)

Then, according to Lemma 9 there exists a subspace ¯A ofA⁰⁰ over which:

fW AfZWfAZ −I = 0 (5.56)

for 1≤i≤q−1 : fW BifZWfAZ+fXBifViXfAVi+

q−1

X

j=1,j6=i

(fXBifVjX +fVjBi)fAVj = 0 (5.57)

f_ZCf_AZ +

q−1

X

j=1

f_XCf_VjXf_AVj = 0. (5.58)

such that

codim_A⁰⁰( ¯A)≤dim(A) +dim(B1) +dim(B2) +· · ·+dim(Bq−1) +dim(C)

−dim(A, B₁, B₂, . . . , Bq−1, C). (5.59) Applying Lemma 6 to equation (5.54), we get:

codimA(A⁰⁰)≤codimA(f_AZ⁻¹f_ZW⁻¹ (W⁰)) +codimA((

q−1

X

j=1

fVjXfAVj)⁻¹(X⁰)) +codimA(f_AV⁻¹

1(V₁⁰)) + · · ·+codim_A(f_AV⁻¹

q−1(V_q−1⁰ )) +codim_A(f_AZ⁻¹(Z⁰)) +codim_A(A⁰). (5.60) Applying Lemma 7 to equation (5.60), we get:

codimA(A⁰⁰)≤codimW(W⁰) +codimX(X⁰) +codimV1V₁⁰) +· · ·+codimVq−1(V_q−1⁰ ) +codim_Z(Z⁰) +codim_A(A⁰)

=codim_W(W⁰) +codim_X(X⁰) +

q−1

X

i=1

codim_ViV_i⁰) +codim_Z(Z⁰) +codim_A(A⁰). (5.61)

TH-2118_136102023

From equation (5.61) and (5.59), we have:

codim_A( ¯A) =codim_A(A⁰⁰) +codim_A⁰⁰( ¯A)

≤codim_W(W⁰) +codim_X(X⁰) +

q−1

X

i=1

codim_ViV_i⁰) +codim_Z(Z⁰) +codim_A(A⁰) +dim(A)

+dim(B₁) +dim(B₂) +· · ·+dim(Bq−1) +dim(C)−dim(A, B₁, B₂, . . . , Bq−1, C). (5.62) For 1≤i≤q−1 consider the following composite functions. (note that these equations hold for each different values of iin the given range)

fXBifViXfCVi +fCBi :C→Bi (5.63)

for 1≤j ≤(q−1), j6=i: (f_XBjf_ViX +f_ViBj)f_CVi :C→B_j (5.64)

fXCfViXfCVi :C →C. (5.65)

Using equations (5.12), we have:

f_XCf_ViXf_CVi+

q−1

X

j=1

f_XBjf_ViXf_CVi =f_ViXf_CVi over a subspacef_CV⁻¹

if_V⁻¹

iX(X⁰) of C. (5.66) Using equations (5.10), we have:

fViXfCVi+

q−1

X

j=1,j6=i

fViBjfCVi =fCVi over a subspacef_CV⁻¹

i(V_i⁰) of C. (5.67) Consider a subspace C_i⁰ where

C_i⁰ =f_CV⁻¹

if_V⁻¹

iX(X⁰)∩f_CV⁻¹

i(V_i⁰)∩Ci. (5.68)

Over C_i⁰ adding the functions shown in equations (5.63)-(5.65) and using equations (5.9), (5.66) and (5.67), we get:

f_XBif_ViXf_CVi+f_CBi+

q−1

X

j=1,j6=i

(f_XBjf_ViX +f_ViBj)f_CVi +f_XCf_ViXf_CVi

=f_ViXf_CVi+f_CBi+

q−1

X

j=1,j6=i

f_ViBjf_CVi

=f_CVi+f_CBi =I. (5.69)

Applying Lemma 9 to equation (5.69), we get that over a subspace ¯Ci we have:

fXBifViXfCVi+fCBi = 0 (5.70)

5.3 Proof of Inequality Shown in 5.1

for 1≤j≤(q−1), j 6=i: (fXBjfViX+fViBj)fCVi = 0 (5.71)

f_XCf_ViXf_CVi−I = 0. (5.72)

such that

codim_C⁰

i( ¯Ci)≤dim(B1) +dim(B2) +· · ·+dim(Bq−1) +dim(C)

−dim(B₁, B₂, . . . , Bq−1, C). (5.73) Applying Lemma 6 to equation (5.68), we get:

codimC(C_i⁰)≤codimC(f_CV⁻¹

if_V⁻¹

iX(X⁰)) +codimC(f_CV⁻¹

i(V_i⁰)) +codimC(Ci). (5.74) Applying Lemma 7 to equation (5.74), we get:

codim_C(C_i⁰)≤codim_X(X⁰) +codim_Vi(V_i⁰) +codim_C(C_i). (5.75) From equations (5.73) and (5.75), we get:

codim_C( ¯C_i) =codim_C(C_i⁰) +codim_C⁰

i( ¯C_i)≤codim_X(X⁰) +codim_Vi(V_i⁰) +codim_C(C_i)

+dim(B₁) +dim(B₂) +· · ·+dim(Bq−1) +dim(C)−dim(B₁, B₂, . . . , Bq−1, C). (5.76) Consider the following vector subspaces.

for 1≤i≤q−1 : S_Bi ={u∈B_i|f_BiY(u)∈f_CY( ¯C)}. (5.77) Hence, equation (5.23) holds overS_Bi whenf_CY is replaced byf_BiY. So overS_Bi we have:

for 1≤i≤(q−1) : (fW BifY W +fXBifY X)fBiY = 0 (5.78) Since equation (5.38) holds over ¯B_i, from equations (5.38) and (5.78), over a subspace ¯B_i ∩S_Bi we have:

fW BifZWfBiZ=I (5.79)

Now consider the following subspaces.

for 1≤i≤q−1 : RBi ={u∈Bi|f_BiZ(u)∈fAZ( ¯A)}

for 1≤i≤q−1 : L_Bi ={u∈B_i|f_{Y W}f_BiY(u)∈f_ZWf_AZ( ¯A)}.

TH-2118_136102023

So fBiZ(RBi) is a subspace of fAZ( ¯A). Then, since from equation (5.56) fW A is invertible over fZWfAZ( ¯A); fW A is also invertible over fZWfBiZ(RBi). Similarly, fY WfBiY(LBi) is a subspace of f_ZWf_AZ( ¯A). Hencef_{W A}is also invertible over f_{Y W}f_BiY(L_Bi). Hence over a subspaceR_Bi∩L_Bi from equation (5.37) we have:

f_{Y W}f_BiY +f_ZWf_BiZ= 0 (5.80)

Applying this equation in equation (5.38), over a subspace ¯Bi∩RBi∩LBi we have:

f_XBif_{Y X}f_BiY =I (5.81)

Now consider the following subspace:

for 1≤i≤q−1 : SAi ={u∈A|f_AVi(u)∈fCVi( ¯Ci)}.

Hence for 1 ≤ i, j ≤ (q −1), j 6= i, (f_XBjf_ViX +f_ViBj)f_AVi(S_Ai) is a subspace of (f_XBjf_ViX + f_ViBj)f_CVi( ¯C_i). Hence from equation (5.71), overS_Ai we have:

for 1≤j≤(q−1), j 6=i: (f_XBjf_ViX +f_ViBj)f_AVi = 0 (5.82) Applying equation (5.82) on equation (5.57), over a subspace ∩^q−1_i=1SAi∩A¯ we have:

for 1≤j≤q−1 : f_{W Bj}f_ZWf_AZ+f_XBjf_VjXf_AVj = 0 (5.83) Let us now consider the following subspaces:

for 1≤i≤q−1 : LAi ={u∈A|f_AZ(u)∈( ¯Bi∩SBi)} (5.84) for 1≤i≤q−1 : R_Ai ={u∈A|f_ViXf_AVi(u)∈f_{Y X}f_BiY( ¯B_i∩R_Bi∩L_Bi)} (5.85) S = ¯A∩(∩^q−1_i=1L_Ai)∩(∩^q−1_i=1R_Ai)∩(∩^q−1_i=1S_Ai). (5.86) For anya∈S, from equation (5.58), we have:

fZCfAZ(a) +

q−1

X

i=1

fXCfViXfAVi(a) = 0

From (5.85) we know there exists a b_i ∈( ¯B_i∩R_Bi∩L_Bi) such thatf_ViXf_Ri(a) =f_{Y X}f_BiY(b_i) So,fZCfAZ(a) +

q−1

X

i=1

fXCfY XfBiY(bi) = 0

From equation (5.81) we know that bi=fXBifY XfBiY(bi) for any bi∈( ¯Bi∩RBi∩LBi). So,

5.3 Proof of Inequality Shown in 5.1

f_ZCf_AZ(a) +

q−1

X

i=1

f_XCf_{Y X}f_BiYf_XBif_{Y X}f_BiY(bi) = 0 or, f_ZCf_AZ(a) +

q−1

X

i=1

f_XCf_{Y X}f_BiYf_XBif_ViXf_AVi(a) = 0 Using equation (5.83), we have:

fZCfAZ(a)−

q−1

X

i=1

fXCfY XfBiYfW BifZWfAZ(a) = 0

From (5.84) we know there exists a b⁰_i∈( ¯Bi∩SBi) such thatfAZ(a) =fBiZ(b⁰_i). So, f_ZCf_AZ(a)−

q−1

X

i=1

f_XCf_{Y X}f_BiYf_{W Bi}f_ZWf_BiZ(b⁰_i) = 0

From equation (5.79) we know thatb⁰_i =f_{W Bi}f_ZWf_BiZ(b⁰_i) for any b⁰_i∈( ¯B_i∩S_Bi). So, f_ZCf_AZ(a)−

q−1

X

i=1

f_XCf_{Y X}f_BiY(b⁰_i) = 0. (5.87)

Since b⁰_i ∈B¯_i, using equation (5.40) in equation (5.87), we have:

fZCfAZ(a) +

q−1

X

i=1

fZCfBiZ(b⁰_i) = 0 or,f_ZCf_AZ(a) +

q−1

X

i=1

f_ZCf_AZ(a) = 0

qf_ZCf_AZ(a) = 0. (5.88)

We now argue that for equation (5.88) to hold for any a ∈ S, S must be a zero subspace. From equation (5.56) we know thatf_AZ is one-to-one over ¯A. From equation (5.77) we know thatf_BiY(S_Bi) for 1 ≤ i ≤ (q −1) is a subspace of f_CY( ¯C). Because of equation (5.24), f_XCf_{Y X} is one-to-one over f_CY( ¯C). So f_XCf_{Y X} is also one-to-one over f_BiY(S_Bi). Then, from equation (5.40) it can be concluded that f_ZCf_BiZ is one-to-one over S_Bi. Now, from (5.84) we know f_AZ(S) is a subspace of f_BiZ( ¯B_i∩S_Bi) for any 1≤i≤q−1. Sof_ZC is one-to-one overf_AZ(S). Moreover, as a pre-condition, since the characteristic of the finite field does not belong to{p₁, p2, . . . , p_l},q 6= 0 over the finite field.

Hence for equation (5.88) to hold, S must be a zero subspace. Now,

dim(A) =dim(A)−dim(S) =codim_A(S) =codim_A( ¯A∩(∩^q−1_i=1L_Ai)∩(∩^q−1_i=1R_Ai)∩(∩^q−1_i=1S_Ai)) Applying Lemma 6, we have:

dim(A)≤codimA( ¯A) +

q−1

X

i=1

codimA(LAi) +

q−1

X

i=1

codimA(RAi) +

q−1

X

i=1

codimA(SAi). (5.89)

TH-2118_136102023

We now calculate some values that would help us in computing a bound overdim(A).

codim_Bi(S_Bi) =codim_Bi(f_B⁻¹

iY(f_CY( ¯C)))

Applying Lemma 7; and noting that from equation (5.24)f_CY is one-to-one over ¯C, we have:

codimBi(SBi)≤codimY(fCY( ¯C)) =dim(Y)−dim(fCY( ¯C)) =dim(Y)−dim( ¯C)

or,codim_Bi(S_Bi)≤dim(Y) +codim_C( ¯C)−dim(C). (5.90)

codim_Bi(R_Bi) =codim_Bi(f_B⁻¹

iZ(f_AZ( ¯A)))

Applying Lemma 7; and noting that from equation (5.56) f_AZ is one-to-one over ¯A, we have:

codimBi(RBi)≤codimZ(fAZ( ¯A)) =dim(Z)−dim(fAZ( ¯A)) =dim(Z)−dim( ¯A)

or, codim_Bi(R_Bi)≤dim(Z) +codim_A( ¯A)−dim(A). (5.91)

codimBi(LBi) =codimBi(f_B⁻¹

iYf_{Y W}⁻¹ (fZWfAZ( ¯A)))

Applying Lemma 7, and since from equation (5.56)f_ZWf_AZ is one-to-one over ¯A, we have:

codimBi(LBi)≤codimW(fZWfAZ( ¯A)) =dim(W)−dim(fZWfAZ( ¯A)) =dim(W)−dim( ¯A) or,codimBi(LBi)≤dim(W) +codimA( ¯A)−dim(A). (5.92)

codimA(SAi) =codimA(f_AV⁻¹

i(fCVi( ¯Ci)))

Applying Lemma 7; and noting from equation (5.72) thatf_CVi is one-to-one over ¯C, we have:

codim_A(S_Ai)≤codim_Vi(f_CVi( ¯C_i)) =dim(V_i)−dim(f_CVi( ¯C_i)) =dim(V_i)−dim( ¯C_i)

or,codimA(SAi)≤dim(Vi) +codimC( ¯Ci)−dim(C). (5.93)

codim_A(R_Ai) =codim_A(f_AV⁻¹

if_V⁻¹

iX(f_{Y X}f_BiY( ¯B_i∩R_Bi∩L_Bi))) Applying Lemma 7, we have:

codim_A(R_Ai)≤codim_X(f_{Y X}f_BiY( ¯B_i∩R_Bi∩L_Bi))

= dim(X)−dim(fY XfBiY( ¯Bi∩RBi∩LBi))

From equation (5.81) we know that f_{Y X}f_BiY is one-to-one over ¯Bi∩R_Bi∩L_Bi. So, codim_A(R_Ai)≤dim(X)−dim( ¯B_i∩R_Bi∩L_Bi)

5.3 Proof of Inequality Shown in 5.1

= dim(X) +codimBi( ¯Bi∩RBi∩LBi)−dim(Bi)

Applying Lemma 6 and then substitutingcodim_Bi(R_Bi) andcodim_Bi(L_Bi) from equations (5.91) and (5.92), we have:

codimA(RAi)≤dim(X) +codimBi( ¯Bi) +codimBi(RBi) +codimBi(LBi)−dim(Bi) or, codim_A(R_Ai)≤dim(X) +codim_Bi( ¯B_i) +dim(Z) +codim_A( ¯A)−dim(A)

+dim(W) +codimA( ¯A)−dim(A)−dim(Bi) or, codim_A(R_Ai)≤dim(W) +dim(X) +dim(Z) +codim_Bi( ¯Bi) + 2codim_A( ¯A)−2dim(A)

−dim(B_i). (5.94)

codim_A(L_Ai) =codim_A(f_AZ⁻¹(f_BiZ( ¯Bi∩S_Bi))) Applying Lemma 7, we have:

codimA(LAi)≤codimZ(fBiZ( ¯Bi∩SBi)) =dim(Z)−dim(fBiZ( ¯Bi∩SBi)) From equation (5.79) we know thatf_BiZ is one-to-one over ¯B_i∩S_Bi. So,

codim_A(L_Ai)≤dim(Z)−dim( ¯B_i∩S_Bi) =dim(Z) +codim_Bi( ¯B_i∩S_Bi)−dim(B_i) Applying Lemma 6 and then substitutingcodimBi(SBi) from equation (5.90), we have:

codim_A(L_Ai)≤dim(Z) +codim_Bi( ¯B_i) +codim_Bi(S_Bi)−dim(B_i)

or, codimA(LAi)≤dim(Z) +codimBi( ¯Bi) +dim(Y) +codimC( ¯C)−dim(C)−dim(Bi). (5.95) Substituting equation (5.95), (5.94), and (5.93) in equation (5.89), we have:

dim(A)≤(q−1)(dim(W) +dim(X) +dim(Y) + 2dim(Z)) +

q−1

X

i=1

dim(V_i)−2(q−1)dim(A)

−2(q−1)dim(C)−

q−1

X

i=1

2dim(B_i) + (2q−1)codim_A( ¯A) + (q−1)codim_C( ¯C)

+

q−1

X

i=1

2codimBi( ¯Bi) +

q−1

X

i=1

codimC( ¯Ci). (5.96)

TH-2118_136102023

Now substituting equations (5.26), (5.46), (5.62) and (5.76) in equation (5.96), we have:

dim(A)≤(q−1)(dim(W) +dim(X) +dim(Y) + 2dim(Z)) +

q−1

X

i=1

dim(Vi)−2(q−1)dim(A)

−2(q−1)dim(C)−

q−1

X

i=1

2dim(B_i) + (7q−6)codim_W(W⁰) + (6q−5)codim_X(X⁰)

+

q−1

X

i=1

(2q)codimVi(V_i⁰) + (3q−3)codimY(Y⁰) + (4q−3)codimZ(Z⁰) + (2q−1)codimA(A⁰) + (q−1)codim_C(C⁰) +

q−1

X

i=1

2codim_B(B_i⁰) +

q−1

X

i=1

codim_C(C_i⁰) + (5q−4)(dim(A)−dim(A, B₁, . . . , Bq−1, C))

+ (6q−5)(

q−1

X

i=1

dim(B_i) +dim(C))−(q−1)dim(B₁, . . . , Bq−1, C).

Substituting values from equations (5.11), (5.12), (5.10), (5.13), (5.14), (5.6), (5.7), (5.8), and (5.9), we get:

dim(A)≤(q−1)(dim(W) +dim(X) +dim(Y) + 2dim(Z)) +

q−1

X

i=1

dim(V_i)−2(q−1)dim(A)

−2(q−1)dim(C)−

q−1

X

i=1

2dim(Bi) + (7q−6)dim(W|A, B₁, . . . , Bq−1) + (6q−5)dim(X|B₁, . . . , Bq−1, C) +

q−1

X

i=1

(2q)dim(V_i|X,∪^q−1_j=1,j6=iB_j) + (3q−3)dim(Y|W, X) + (4q−3)dim(Z|W, C) + (2q−1)dim(A|Z, V₁, . . . , Vq−1) + (q−1)dim(C|A, Y)

+

q−1

X

i=1

2dim(B_i|B₁, . . . , Bi−1, B_i+1, . . . , Bq−1, Y, Z) +

q−1

X

i=1

dim(C|V_i, B_i) + (5q−4)(dim(A)−dim(A, B₁, . . . , Bq−1, C))

+ (6q−5)(

q−1

X

i=1

dim(B_i) +dim(C))−(q−1)dim(B₁, . . . , Bq−1, C).