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Abstract

We prove assorted properties of matrices over Z2{\mathbb{Z}_{2}} , and outline the complexity of the concepts required to prove these properties. The goal of this line of research is to establish the proof complexity of matrix algebra. It also presents a different approach to linear algebra: one that is formal, consisting in algebraic manipulations according to the axioms of a ring, rather than the traditional semantic approach via linear transformations.
Proving properties of matrices over Z2
Michael Soltys
March 14, 2012
Abstract
We prove assorted properties of matrices over Z2, and outline the
complexity of the concepts required to prove these properties. The goal of
this line of research is to establish the proof complexity of matrix algebra.
It also presents a different approach to linear algebra: one that is formal,
consisting in algebraic manipulations according to the axioms of a ring,
rather than the traditional semantic approach via linear transformations.
Keywords: Proof complexity, matrix identities, Frege and extended Frege.
1 Introduction
We are interested in the proof complexity of matrix algebra over the field of two
elements GF(2). In particular, we examine the identity AB =IBA =I
which has been proposed as a candidate for separating the Frege and extended
Frege propositional proof systems. We investigate the properties of matrices
that can be proven over the simplest of fields, with the hope that understanding
these properties will yield a low-complexity proof of AB =IBA =I.
All matrices are considered to be over the field of two elements {0,1}; in the
literature this field is denoted as Z2or as GF(2). Let Mn×m(F) be the set of
n×mmatrices over a field F; let M(n) := Mn×n(Z2), i.e., M(n) is the set of
all square n×nmatrices over Z2. If the size of a matrix Ais not specified, we
assume AM(n). We use Aij to denote entry (i, j) of the matrix A.
We let Atdenote the transpose of matrix A, i.e., the matrix whose (i, j )
entry is entry (j, i) of A. Let In,0nbe the identity matrix and zero matrix,
respectively, in M(n).
Given a matrix AM(n), we often find it useful to represent Ait terms of
its principal minor, denoted MA, as follows:
a RA
SAMA,
where ais the top-left entry of A, i.e., a=A11, and
RA=A12 A13 . . . A1n,
SA=A21 A31 . . . An1t.(1)
1
The results in this paper can be interpreted in various extensions of the
theory LA, for example LAP or LA, where LAP. The theory LA is capable
of rudimentary ring reasoning about matrices; the theory LAP adds matrix
powering, P(A, i) = Ai, and LA permits induction on ΣB
1-formulas, which
are formulas with existential quantification over matrices of bounded size. Over
Z2,LA corresponds to AC0(2), LAP to NC2(in fact, slightly weaker), and
LA corresponds to polynomial-time reasoning. See section 13, the appendix,
and [SC04] for more details.
Alternatively, we can employ theories corresponding to the complexity class
L=AC0(det2), as defined in [CF10] where the theory is called VL. The
advantage of these theories over the LA-family of theories is that LA is field
independent, and it is not clear that LA can put to use the fact that Z2is a
particularly simple field.
2 Matrix identities
A motivation for this paper is to understand the complexity of concepts required
to prove:
AB =IBA =I , (‘Inverse Identity’)
and related matrix identities (see [SC04]), and to understand the proof com-
plexity of combinatorial matrix algebra in general. That is, we would like to
know what is the complexity of the concepts required to reason about basic
linear algebra, and about the combinatorial applications of matrix algebra—as
presented, for example, in [BR91].
There are two main motivations for this line of research; first, in reverse
mathematics we are interested in the weakest logical theories capable of for-
malizing linear algebra. But mainly, Cook proposed the ‘Inverse Identity’ as
a candidate for separating the Frege and extended Frege propositional proof
systems—this is one of the principal open questions in theoretical computer
science.
Lemma 1 If AB =Iand we can show that Ahas some left-inverse, then
BA =I; furthermore, this can be shown in LA. In other words,
LA `(AB =I C(CA =I)) BA =I.
Proof: AB =Iimplies ABA =A, so A(BA I) = 0; so if Ahas some left
inverse, call it C, then CA(BA I) = 0, so BA =I.
3 Powers and products
Over the field Z2,c+c= 0, and hence, for any AM(n), we have that
A+A= 0n. Lemmas 2 and 3 come from [Cob58]. The lemmas in this section
can be shown in LAP augmented by the index function y= 2x, which we define
2
as exp(0) = 1 and exp(i+ 1) = 2 ·exp(i). This permits repeated squaring of a
matrix, polynomially many times of course.
Lemma 2 (I+A)2i=I+A2i.
Proof: (I+A)2= (I+A)(I+A) = I+A+A+A2=I+A2.
Lemma 3 (I+A)2i=I A2i= 0.
Proof: A2i= 0 I+A2i=I (I+A)2i=I, where the last equality
follows by lemma 2.
Lemma 4 If (I+A)2i1=Iand AB =I, then A2i1=I.
Proof: Suppose that (I+A)2i1=I; multiply both sides by (I+A) to obtain
(I+A)2i= (I+A) and using lemma 2 we have I+A2i=I+A, and so
A2i=AA2iB=AB A2i1=I.
Lemma 5 AB =BA (I+A)(I+B)=(I+B)(I+A).
Lemma 6 If A, B are inverses, i.e., AB =BA =I, then (A+B)2i=A2i+B2i.
Proof: First note that (I+A)(I+B) = I+A+B+AB =I+A+B+I=A+B;
similarly (I+B)(I+A) = I+B+A+BA =B+A, and in particular
(I+A)(I+B)=(I+B)(I+A). Therefore:
(A+B)2i= (I+A)2i(I+B)2i
()
= (I+A2i)(I+B2i)
=I+A2i+B2i+A2iB2i
=I+A2i+B2i+I
=A2i+B2i,
where the () equality follows from lemma 2.
4 Idempotence, nilpotence and zero-divisors
A matrix Ais idempotent if A2=A, it is nilpotent if Ai= 0 for some i > 0 and
it is a right-zero-divisor (respectively, left-zero-divisor) if there exists a matrix
C6= 0 such that AC = 0 (respectively, CA = 0). If Ais a right-zero-divisor
then it is also a left-zero-divisor, but the Cmight differ; for example, if
A=1 1
0 0 , C =1 0
1 0 , D =0 1
0 1 ,
then AC = 0, but CA 6= 0 while DA = 0.
3
If AB =Ithen LA proves that BA is idempotent, and also that neither
Acan be a left-zero-divisor nor Bcan be right-zero-divisors; for example, if
AB =Iand BC = 0, then ABC = 0, so C= 0.
By lemma 2, if Ais idempotent, so is (I+A).
If AB =I, then LAP shows that neither Anor Bcan be nilpotent. Suppose
Ai= 0. Then AiBi= 0 but AiBi=I.
5 Symmetric matrices
Let Aby a symmetric matrix, i.e., A=At, that is Aequals its transpose.
We say that x, y are A-orthogonal if xtAy = 0; we sometimes write this as
hx, yiA=xtAy.
Lemma 7 If Ais a symmetric matrix, then V={xZn
2:hx, xiA= 0}is a
vector space.
Proof: If Ais symmetric then:
hx, yiA=xtAy = (xtAy)t=ytAt(xt)t=ytAx =hy , xiA.
Then, if x, y Vthen:
h(x+y),(x+y)iA= (x+y)tA(x+y)
=xtAx +ytAx +xtAy +ytAy
=hx, xiA+hy, xiA+hx, yiA+hy , yiA,
and hx, xiA=hy, yiA= 0 since we assumed that x, y V, and
hy, xiA+hx, yiA= 0,
as well since hy, xiA=hx, yiA.
Lemma 8 LA proves that if Ais symmetric, and AB =I, then BA =I, and
furthermore Bis also symmetric.
Proof: AB =Ithen I=It= (AB)t=BtAt=BtA, as Ais symmetric.
By lemma 1 we have BA =I. On the other hand, from BtA=Iwe have
BtAB =Band so Bt=B, and hence Bis also symmetric.
Let d(A) denote the diagonal of a symmetric matrix A, i.e.,
d(A)=[A11A22 . . . Ann ].
In an unpublished note, Filmus ([Fil10]) make an interesting observation, which
we present as lemma 9. We give a simplified proof, however, in the style of
Gaussian Elimination.
Lemma 9 LA proves that for all symmetric A,vsuch that Av =d(A).
4
Proof: The proof is by induction on n,AM(n). For n= 1 let v= 1. For
n > 1 let
A=a X
XtM,
and we consider the following cases: if X= 0 then by IH vMsuch that MvM=
d(M), where Mis the principal submatrix of A(and Mis also symmetric if A
is symmetric). So let v= [1 vM], and then Av =d(A). If X6= 0 and a= 1
then let
C=1 0
XtIn1,
and observe that
A0=CAC t
=1 0
XtIn1 a X
XtM 1X
0In1
=1 0
XtIn1 a aX +X
XtXtX+M
=a aX +X
aXt+XtaXtX+XtX+XtX+M
=a aX +X
aXt+XtaXtX+M
=1 0
0XtX+M
where we remind the reader that over Z2,x+x= 0, and so XtX+XtX= 0
and since a= 1, aX +X=X+X= 0. Using the IH we know that wsuch
that A0w=d(A0), i.e., CAC tw=d(CACt), and since CC =In(over Z2), we
have
A(Ctw) = Cd(C ACt)
=C1
d(XtX+M)
=C1
Xt+d(M)
were d(XtX+M) = Xt+d(M) follows (over Z2) from the fact that the diagonal
of XtXequals [ x1x1x2x2. . . xn1xn1] which in turn is just Xsince
xixi=xi,
=1
Xt+Xt+d(M)
=1
d(M).
5
Since a= 1, A(Ctw) = d(A) and so letting v=Ctwwe are done showing that
in the case X6= 0 and a= 1, vsuch that Av =d(A).
If, on the other hand, X6= 0 and a= 0, there are two possibilities. First,
d(M) = 0, in which case v= 0 and Av = 0 = d(A), or there is some diagonal
entry of M, say Mii, which is not equal to zero; let Pibe the identity matrix
with row 1 and row ipermuted. Then, we can repeat the argument for the
second case with A0= (CPi)A(C Pi)t, since A0=C(PiAP t
i)Ctwhich has the
effect of bringing Mii 6= 0 (and hence Mii = 1) to the position (1,1) of A.
Symmetric matrices over finite fields have been considered in [Mac69], where,
in section I, the author shows the following interesting results—originally due
to A. A. Albert, and can be found in [Alb38]:
Theorem 1 If AM(n)is an invertible symmetric matrix of GF(2m)then A
can be factored in the form A=MtMif and only if d(A)6= 0, i.e., some entry
on the diagonal of Ais not zero.
The proof of theorem 1, as presented in [Mac69], can be formalized in LA.
6 Trace
The proof of lemma 10 can be formalized in LAP with exp (as defined in
section 3), while lemma 11 can be (obviously) formalized in LA.
Lemma 10 tr(A) = tr(A2i)for all i.
Proof: Note that
a RA
SAMA2
=a2+RASAaRA+MARA
aSA+MASASARA+M2
A,
so
tr(A2) = a2+RASA+ tr(SARA+M2
A)
=a+RASA+ tr(SARA) + tr(M2
A)
=a+RASA+RASA+ tr(M2
A),
and again RASA+RASA= 0 and tr(M2
A) = tr(MA) by induction; an induction
that can be carried out in LA. On the other hand, tr(A) = tr(A2i) can be
carried out in LAP.
Lemma 11 tr(AtA) = tr(AAt) = Pi,j Aij .
Proof: tr(AtA) = Pi(AtA)ii =Pi,j At
ij Aji =Pi,j Aj iAj i =Pij Aij .
In fact, from the proof of lemma 11 we see that (AtA)ii is the sum of the
elements in row iof A.
6
7 Annihilating polynomials
We say that p(x)6= 0 is an annihilating polynomial of Aif p(A) = 0. Of course,
the annihilating polynomial par excellence of any matrix is its characteristic
polynomial; this is the famous Cayley-Hamilton theorem that can be shown in
LA (see [SC04]).
Lemma 12 LA proves that p(A)2=p(A2).
Proof: p(A)2=PiaiAi2=Pi,j aiajAi+j=Pia2
iA2i, and we lost terms
where i6=jsince aiajAi+j+ajaiAj+i= 0, so p(A)2=Piai(Ai)2=p(A2).
It follows from lemma 12 that p(A)2i=p(A2i), and that this can be shown
in LAP with exp. The following lemma is an immediate consequence of the
previous one.
Lemma 13 If p(x)is an annihilating polynomial for A, then p(x)is also an
annihilating polynomial for A2ifor all i.
Proof: Suppose p(A) = 0. Then 0 = p(A)2i=p(A2i).
Lemma 14 If AB =Iand p(x)is an annihilating polynomial for A, then we
can prove, in LAP, that BA =I. That is,
LAP `(AB =Ip6= 0 p(A) = 0) BA =I.
Proof: Suppose that p(A) = a0I+a1A+· · · +akAk= 0. As p6= 0, let aibe
the smallest non-zero coefficient of this polynomial; so, in effect,
p(A) = aiAi+ai+1Ai+1 +· · · +akAk.
Consider
0 = p(A)Bi=aiI+ai+1A+· · · +akAki
=I+A(ai+1I+· · · +akAki1)
=I+ (ai+1I+· · · +akAki1)A,
i.e., (ai+1I+· · · +akAki1) is the (two-sided) inverse of A, and so AB =I
implies ABA =Awhich implies A(BA I) = 0, and now using the inverse of
Awe obtain BA I= 0 and so BA =I.
8 Pigeonhole principle and counting
Let PHP denote the Pigeonhole principle. In this section we present three
proofs of the ‘Inverse Identity’ that can be formalized in LAP augmented with
PHP over sets of exponential size. What is interesting about these “counting
arguments” is that they dispense with linear algebra in proving the ‘Inverse
Identity’; they rely on the finiteness of the underlying field, and basic ring
properties of matrix addition and multiplication. As such, they offer a limited
proof-complexity insight into the ‘Inverse Identity’.
7
Proof I of the ‘Inverse Identity’
This is a simple proof of the ‘Inverse Identity’, which extends easily to any finite
field. It uses PHP over sets of size 2n2for matrices in M(n).
Consider the sequence I, A, A2, A3, . . . , A2n2
. Since |M(n)|= 2n2it follows
by PHP that there exist 0 i < j 2n2such that Ai=Aj; but then, using
AB =I, we have AiBi=AjBiI=Aji, where ji > 0, and so Aji1is
a (left and right) inverse of A, and using lemma 1 we are done.
Proof II of the ‘Inverse Identity’
Let Φ : M(n) M(m) be a mapping.
Lemma 15 If n>m, then YM(m)such that
|{XM(n) : Φ(X) = Y}| 2(nm)2.
Proof: Let SΦ(Y) := {XM(n) : Φ(X) = Y}. Since
M(n) = [
YM(m)
SΦ(Y)
it follows that |M(n)|≤|M(m)| · max{|SΦ(Y)|:YM(m)}, and so we have
that 2(nm)22n2m2max{|SΦ(Y)|:YM(m)}.
Suppose that AB =I, and let ΦA:M(n+ 1) M(n) be a mapping
defined as follows:
ΦA(C) := c0A+c1A2+· · · +c(n+1)21A(n+1)21,(2)
where Ais assumed to be n×n, and entry cjis C1+div(j,n),rem(j,n), i.e., given
q, r such that j=qn +rwhere 0 r < n, then cjis entry cq+1,r. By lemma 15
we have that there is a YM(n) such that there exist C6=C0mapping to it,
i.e., ΦA(C) = ΦA(C0), and so ΦA(C)+ΦA(C0) = 0. Therefore, we obtain an
annihilating polynomial for Aof degree (n+ 1)21; by lemma 14 we are done.
Proof III of the ‘Inverse Identity’
We define ΦA(C) as in (2), and we present a variation of the argument in
proof II. Since |M(n+ 1)|= 2(n+1)2, one of the following two must hold:
|{CM(n+ 1) : A(C))11 = 0}| 2(n+1)2/2,
|{CM(n+ 1) : A(C))11 = 1}| 2(n+1)2/2.
We pick the set for which it is true; we now repeat the argument with the C’s
in that set and A(C))12. We end up with C, C 0such that ΦA(C) = ΦA(C0),
and once again obtain an annihilating polynomial for A.
8
9 Gaussian Elimination
We give a proof of the ‘Inverse Identity’ based on the Gaussian elimination
algorithm; this proof can be formalized in LA, and hence it is a polynomial
time proof. In fact, in [Sol02, TS05] it has been shown that polysize extended
Frege can prove the correctness of the Gaussian elimination procedure (over Z2,
and over bigger fields as well). LA allows induction over formulas asserting
the existence of matrices; such formulas can express the existence of row and
column operations necessary to carry out the Gaussian elimination algorithm.
Hence LA proves the correctness of Gaussian elimination (every matrix can
be put in upper triangular form), and this in turn can be employed to prove the
‘Inverse Identity’.
Suppose that AB =I; we show that Ahas some left-inverse (lemma 1).
Recall the definition of SAin (1). If SAis zero then repeat the argument
inductively on MAMB=In1.
Otherwise, if SA6= 0, let Pbe a permutation matrix defined as follows: if
a= 1, P=In; if a= 0, then Pswaps the first row of Awith a row whose
first entry is non-zero; Pis just the identity matrix with the corresponding rows
swapped. Also let
C=1 0
SP A In1,
that is, Cis the identity matrix where the first column, except for the top entry,
is replaced by the corresponding entries in P A. Observe that CC =P P =In.
Then
AB =In(CP )AB(P C )=(C P )In(P C)
(CP A)(B P C) = C(P P )C=C C =In,
and
CP A =1 0
SP A In1 1RP A
SP A MP A =1RP A
0SP ARP A +MP A .
We now repeat the argument inductively on MCP AMB P C =In1. Notice that
this proof is intrinsically sequential, as at each step we construct a new matrix,
and we need the previous steps to do that.
10 Quasi-triangular matrices
In this section we explore the following question: what is the weakest condition
on matrices A, B that allows a proof of the ‘Inverse Identity’ in LA? We already
know from lemma 8 that if A, B are symmetric matrices then LA proves the
‘Inverse Identity’ for such matrices. Here, inspired by the Gaussian elimination
proof in the previous section, we define the notion of a “quasi-triangular pair”
of matrices.
9
Given AM(n), let MA,i be its i-th minor. That is MA,0=A,MA,1=MA,
and for i2, MA,i =MMA,i1, i.e., MA,i is Awith the first irows and columns
removed. Let ai, RA,i , SA,i be defined as follows: ai= (MA,i1)ii =Aii, and
RA,i =RMA,i1and SA,i =SMA,i1. We say that a matrix Ais quasi-triangular
if for all iwe have that RA,i is zero or SA,i is zero; it is clear how this is a
generalization of the notion of a triangular matrix.
We define recursively what it means for a matrix Ato be a quasi-transpose
of a matrix B; if A, B M(1), then Ais a quasi-transpose of Bif A=B. For
i > 1, Ais a quasi-transpose of Bif A11 =B11 and either RA=RBSA=SB,
or RA=St
BSA=Rt
B, and MAis a quasi-transpose of MB.
Lemma 16 The matrix Ais quasi-triangular if Ais the quasi-transpose of a
triangular matrix.
Finally, we say that matrices (A, B) are a quasi-triangular pair if for all i
at least one of {RA,i, SA,i , RB,i , SB,i }is zero. Observe that if (A, B) is a quasi-
triangular pair then so is (MA, MB); indeed, the point of this definition is to
find as weak a condition on A, B as possible to ensure that if AB =In, then
MAMB=In1, which allows induction on LA formulas and consequently an
LA proof of the ‘Inverse Identity’.
Lemma 17 LA proves that if (A, B)is a quasi-triangular pair, and AB =I,
then BA =I.
Proof: Suppose that A, B M(n), and they are a quasi-triangular pair. We
prove that LA `AB =IBA =I, by induction on n, by cases on the
definition of a “quasi-triangular pair.”
AB =a RA
SAMA b RB
SBMB=ab +RASBaRB+RAMB
bSA+MASBSARB+MAMB(3)
Case 1: SA= 0 or RB= 0, then we can see from equation (3) that
AB =ab +RASBaRB+RAMB
bSA+MASBMAMB,
since SARB= 0 SA= 0 RB= 0, and since AB =In, it follows that
MAMB=In1, and given that (A, B) is a quasi-triangular pair, so is (MA, MB),
and hence by induction MBMA=In1.
Suppose now that SA= 0. Then,
BA =ba bRA+RBMA
aSBSBRA+In1,
and from (3) we see that 0 = bSA+MASB=MASB, and since MBMA=In1
it follows that SB= 0, so ab = 1 and so ba = 1. Finally,
bRA+RBMA=RA+RBMA=RA+RAMBMA=RA+RA= 0.
10
Therefore, BA =In. The case where RB= 0 is symmetric.
Case 2: SB= 0 or RA= 0, then since ab+RASB= 1 it follows that ab = 1,
and so a=b= 1. Therefore,
In=AB =1RB+RAMB
SA+MASBSARB+MAMB,
thus MASB=SAand RAMB=RB, and hence
MASBRA
| {z }
=0
MB+MAMB=In1,
and so MAMB=In1. By induction we conclude that MBMA=In1, and we
are done.
11 Lanczos algorithm
The “Block Lanczos Algorithm for Finding Dependencies over a Finite Field”
was invented by Peter L. Montgomery ([Mon95]). Recall that if AB =I, then
from basic algebraic manipulations we obtain that A(BA I) = 0; thus, if we
could show that Ahas any left-inverse, we would be able to show the ‘Inverse
Identity’. As symmetric matrices have a lot of nice properties, perhaps one could
work with the symmetric matrix ˆ
A=AtAinstead of working with A. Note that
showing that ˆ
Ahas a left-inverse would still prove, from A(BA I) = 0, that
BA =I. This is because A(BA I) = 0 implies AtA(BA I) = 0, and so
ˆ
A(BA I) = 0. This section suggests a connection between the “Block Lanczos
Algorithm” and our ‘Inverse Identity’.
Let Abe a symmetric matrix and suppose that we have a set of mvectors
{w1, w2, . . . , wm}, and that they satisfy the following three conditions:
wt
iAwi6= 0 1 im
wt
iAwj= 0 i6=j
X(AW =W X )
(4)
where the second condition says that the wi’s are A-orthogonal, as defined in
section 5, and where in the last condition W=w1w2. . . wm, i.e.,
Wis a matrix whose rows are the wi’s. Note that saying X(AW =W X)
is equivalent to stating that for all wi,Awispan{w1, w2, . . . , wm}. Suppose
further that y(b=W y) and define the vector vas follows:
v:=
m
X
i=1
wt
ib
wt
iAwi
wi.(5)
Theorem 2 LA `[(4) y(b=W y)(5)] Av =b.
11
Proof: We prove the theorem in the case F=Z2. Over the field Z2we have the
implication wt
iAwi6= 0 wt
iAwi= 1, and hence definition (5) can be restated
as v:= Pm
i=1(wt
ib)wi. Then:
Wt(Av b) = Wt(A(
m
X
i=1
(wt
ib)wi)b)()
= (
m
X
i=1
(wt
ib)WtAwi)Wtb
= (
m
X
i=1
(wt
ib)(wt
iAwi)ei)Wtb(∗∗)
= (
m
X
i=1
(wt
ib)ei)Wtb=WtbWtb= 0,
where () and (∗∗) follow from (4) since wt
jAwi= 0 for j6=iand wt
iAwi= 1,
respectively.
Thus we obtain Wt(Av b) = 0 and hence, for any n×mmatrix X,
(W X )t(Av b) = 0. By (4) X(AW =W X ), and so (AW )t(Av b) = 0, and
since Ais symmetric it follows that
WtA(Av b)=0.(6)
By assumptions y(Av b=W y), i.e., Av b=Pm
i=1 yiwi, and so multiplying
on the left by wt
iAwe conclude, for every i, that yi= 0. Hence Av b= 0 and
so Av =b.
Note that given (4) and (5) and y(b=W y), then Av =bfor any field F,
finite or infinite.
Suppose that AB =I, and consider the matrix ˆ
A=AtA. Represent Bas
b1b2. . . bnwhere biis the i-th column of B. Then observe:
bt
iˆ
Abj=bt
iAtAbj= (Abi)t(Abj) = et
iej=(1 if i=j
0 if i6=j
and thus Bsatisfies the first two Lanczos conditions for ˆ
A=AtAgiven in (4).
The third Lanczos condition is equivalent to BA =I.
12 Open questions
If AB =I, can we show that AtA(which is symmetric) is also invertible? Since
LA `AB =IA(BA I) = 0, it follows that (AtA)(BA I) = 0 is
provable from AB =Iin LA. Once we have a left-inverse for AtAwe would
have BA =I. It would be interesting to see if Lanczos’ algorithm can be of
help here.
Matrices over Z2are often employed as adjacency matrices; that is, given a
graph G= ([n], E), where E[n]×[n], it can be represented by AGM(n)
as follows: (i, j)E Aij = 1. But matrices over Z2can be also seen
as incidence matrices where given a collection of nelements, and a collection
X1, X2, . . . , Xmof subsets of [n], iXk Aij = 1, where AMn×m(Z2).
An interesting paper examining incidence matrices is [Rys60].
12
13 Appendix
The logical theory LA is strong enough to prove the ring properties of matrices
such as A(BC)=(AB)C, A +B=B+A, but weak enough so that the theorems
of LA translate into propositional tautologies with short Frege proofs. LA
has three sorts of object: indices (i.e., natural numbers), ring elements, and
matrices, where the corresponding variables are denoted i, j, k, ...;a, b, c, ...; and
A, B, C, ..., respectively. The semantic assumes that objects of type ring are from
a fixed but arbitrary ring (for the purpose of this paper we are only interested
in Z2, which is a field), and objects of type matrix have entries from that ring.
Terms and formulas are built from the following function and predicate sym-
bols, which together comprise the language LLA:
0index,1index ,+index,index ,index,div,rem,
0ring,1ring ,+ring,ring ,ring,1,r,c,e,Σ,
index,=index ,=ring,=matrix ,condindex,condring
(7)
The intended meaning should be clear, except in the case of index, cut-off
subtraction, defined as ij= 0 if i<j. For a matrix A:r(A),c(A) are the
numbers of rows and columns in A,e(A, i, j) is the ring element Aij (where
Aij = 0 if i= 0 or j= 0 or i > r(A) or j > c(A)), Σ(A) is the sum of the
elements in A. Also cond(α, t1, t2) is interpreted if αthen t1else t2, where α
is a formula all of whose atomic sub-formulas have the form mnor m=n,
where m, n are terms of type index, and t1, t2are terms either both of type
index or both of type ring. The subscripts index ,ring, and matrix are usually
omitted, since they ought to be clear from the context.
We use n, m for terms of type index, t, u for terms of type ring, and T , U for
terms of type matrix. Terms of all three types are constructed from variables
and the symbols above in the usual way, except that terms of type matrix are
either variables A, B, C, ... or λ-terms λijhm, n, ti. Here iand jare variables
of type index bound by the λoperator, intended to range over the rows and
columns of the matrix. Also m, n are terms of type index not containing i, j
(representing the numbers of rows and columns of the matrix) and tis a term
of type ring (representing the matrix element in position (i, j)).
Atomic formulas have the forms mn, m =n, t =u, T =U, where the
three occurrences of = formally have subscripts index,ring ,matrix, respectively.
General formulas are built from atomic formulas using the propositional con-
nectives ¬,,and quantifiers ,.
13.1 Axioms and rules of LA
For each axiom listed below, every legal substitution of terms for free variables
is an axiom of LA. Note that in a λterm λijhm, n, tithe variables i, j are
bound. Substitution instances must respect the usual rules which prevent free
variables from being caught by the binding operator λij. The bound variables
i, j may be renamed to any new distinct pair of variables.
13
13.1.1 Equality Axioms
These are the usual equality axioms, generalized to apply to the three-sorted
theory LA. Here = can be any of the three equality symbols, x, y, z are variables
of any of the three sorts (as long as the formulas are syntactically correct). In
A4, the symbol fcan be any of the non-constant function symbols of LA.
However A5 applies only to , since this in the only predicate symbol of LA
other than =.
A1 x=x
A2 x=yy=x
A3 (x=yy=z)x=z
A4 x1=y1, ..., xn=ynfx1...xn=f y1...yn
A5 i1=j1, i2=j2, i1i2j1j2
13.1.2 Axioms for indices
These are the axioms that govern the behavior of index elements. The index
elements are used to access the entries of matrices, and so we need to define
some basic number theoretic operations.
A6 i+ 1 6= 0
A7 i(j+ 1) = (ij) + i
A8 i+ 1 = j+ 1 i=j
A9 ii+j
A10 i+ 0 = i
A11 ijji
A12 i+ (j+ 1) = (i+j)+1
A13 [ijji]i=j
A14 i0=0
A15 [iji+k=j]ji=k
A16 ¬(ij)ji= 0
A17 [αcond(α, i, j) = i][¬αcond(α, i, j ) = j]
13.1.3 Axioms for a ring
These are the axioms that govern the behavior for ring elements; addition and
multiplication, as well as additive inverses. We do not need multiplicative in-
verses.
A18 06= 1 a+ 0 = a
A19 a+ (a)=0
A20 1a=a
A21 a+b=b+a
A22 ab=ba
A23 a+ (b+c)=(a+b) + c
A24 a(bc)=(ab)c
14
A25 a(b+c) = ab+ac
A26 [αcond(α, a, b) = a][¬αcond(α, a, b) = b]
13.1.4 Axioms for matrices
Axiom A27 states that e(A, i, j) is zero when i, j are outside the size of A. Axiom
A28 defines the behavior of constructed matrices. Axioms A29-A32 define the
function Σ recursively by first defining it for row vectors, then column vectors
(At:= λijhc(A),r(A), Aj ii), and then in general using the decomposition (8).
Finally, axiom A33 takes care of empty matrices.
A27 (i= 0 r(A)< i j= 0 c(A)< j)e(A, i, j )=0
A28 r(λijhm, n, ti) = mc(λijhm, n, ti) = n[1 iim1jjn]
e(λijhm, n, ti, i, j ) = t
A29 r(A)=1,c(A) = 1 Σ(A) = e(A, 1,1)
A30 r(A)=11<c(A)Σ(A) = Σ(λijh1,c(A)1, Aij i) + A1c(A)
A31 c(A)=1Σ(A) = Σ(At)
A32 1<r(A)1<c(A)Σ(A) =e(A, 1,1) + Σ(R(A)) + Σ(S(A)) + Σ(M(A))
A33 r(A)=0c(A)=0ΣA= 0
Where
R(A) := λijh1,c(A)1,e(A, 1, i + 1)i,
S(A) := λijhr(A)1,1,e(A, i + 1,1)i,
M(A) := λijhr(A)1,c(A)1,e(A, i + 1, j + 1)i.
(8)
13.1.5 Rules for LA
In addition to all the axioms just presented, LA has two rules: matrix equality
and induction.
Matrix equality rule
From the three premises:
1. e(T, i, j) = e(U, i, j )
2. r(T) = r(U)
3. c(T) = c(U)
we conclude T=U.
The only restriction is that the variables i, j may not occur free in T=U;
other than that, Tand Ucan be arbitrary matrix terms. Our semantics implies
that iand jare implicitly universally quantified in the top formula. The rule
allows us to conclude T=U, provided that Tand Uhave the same numbers of
rows and columns, and corresponding entries are equal.
15
Induction rule α(i)α(i+ 1)
α(0) α(n)
Here α(i) is any formula, nis any term of type index, and α(n) indicates nis
substituted for free occurrences of iin α(i). (Similarly for α(0).)
This completes the description of LA. We finish this section by observing
the substitution property in the lemma below. We say that a formula S0of
LA is a substitution instance of a formula Sof LA provided that S0results by
substituting terms for free variables of S. Of course each term must have the
same sort as the variable it replaces, and bound variables must be renamed as
appropriate.
Lemma 18 Every substitution instance of a theorem of LA is a theorem of
LA.
This follows by straightforward induction on LA proofs. The base case
follows from the fact that every substitution instance of an LA axiom is an LA
axiom.
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17
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