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A ROW-REDUCED FORM FOR COLUMN-PARTITIONED
MATRICES
STEPHAN FOLDES AND ERKKO LEHTONEN
Abstract. Let the columns of a p × q matrix M over any field be partitioned
into n blocks, M = [M1 , . . . , Mn ]. All p × p submatrices of M with columns
from distinct blocks Mi are singular if and only if there is a nonsingular p × p
matrix Q and a positive integer m ≤ p such that in QM = [QM1 , . . . , QMn ]
the last m rows are null in all but at most m − 1 blocks QMi .
1. Singularity of partition-constrained submatrices and the
row-reduced form
Theorem. Let the columns of a p × q matrix M with entries in any field be partitioned into n blocks, M = [M1 , . . . , Mn ]. The following are equivalent.
(1) All p × p submatrices extracted from M with columns from distinct blocks
Mi are singular.
(2) There is a nonsingular p × p matrix Q and a positive integer m ≤ p such
that in QM = [QM1 , . . . , QMn ] the last m rows are null in all but at most
m − 1 blocks QMi .
Remark 1. If n = q, then this is just the characterization of singular matrices M
by the property of having a null row in their reduced row-echelon form.
Remark 2. If n < p, then (1) is vacuously true and (2) holds with the identity
matrix as Q and m = p.
In order to prove the Theorem, it will be convenient to recast it in a somewhat
more general form, using the following definitions and notation for purposes of
precision and simplicity in the proof.
An A × B matrix with entries in a field F is any map M : A × B → F , where A,
B are finite sets of positive integers. The matrix product M N of M : A × B → F
and N : B × C → F is a map A × C → F whose value on (a, c) ∈ A × C is defined
by the usual convolution formula. For A0 ⊆ A, B 0 ⊆ B, we denote by M [A0 , B 0 ]
the restriction of M (as a map) to A0 × B 0 ; thus M = M [A, B]. If any of A0 or B 0
is a singleton {a}, then we may omit the set braces and write a for {a}.
For any set B, a partition is a set Π of nonempty pairwise disjoint subsets of B
the union of which is B. A partial transversal of Π is a subset J of B intersecting
every partition class K ∈ Π in at most one element.
Reformulation. Let M be an A × B matrix. Consider a partition Π of B into n
classes, Π = {B1 , . . . , Bn }. The following are equivalent.
(1) For all partial transversals J of the partition Π, the rank of M [A, J] is less
than |A|.
Date: March 5, 2006.
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STEPHAN FOLDES AND ERKKO LEHTONEN
(2) There is a nonsingular A × A matrix Q and a nonempty subset A0 ⊆ A
such that at most |A0 | − 1 of the matrices (QM )[A0 , Bi ] are non-null.
2. Proof of the reformulation
(2) ⇒ (1). The implication is straightforward, especially if considered in the first
formulation of the theorem.
(1) ⇒ (2). If the rank of M is less than |A|, then clearly there is a nonsingular
Q and a singleton A0 ⊆ A such that (QM )[A0 , B] is null, which clearly implies (2),
even without calling on the hypothesis. Therefore, suppose that the rank of M is
|A|.
Let P ⊆ B. For any nonsingular A × A matrix Q, the matrices M [A, P ] and
(QM )[A, P ] have the same rank. Call a subset P ⊆ B independent if this rank is
|P |. Let P ⊆ B be an independent set of size |A| meeting as many blocks of Π as
possible, say r blocks; clearly r < |A|. Let KP denote the union of those blocks of
Π that meet P .
Let Q be any nonsingular A × A matrix such that for every i ∈ P the matrix (QM )[A, i] has exactly one entry equal to 1 and the other entries equal to 0.
Since P is independent, there is a bijection k : P → A such that for all i ∈ P ,
(QM )(k(i), i) = 1. We will denote QM by M 0 .
Let K0 be any block of Π intersecting P in at least two elements. Let us choose
two distinct elements l0 , l00 in P ∩ K0 . Let A0 = {k(l0 ), k(l00 )}. For each block
K 6= K0 meeting P , choose a fixed lK ∈ K ∩ P . If M 0 [A0 , B \ KP ] 6= 0, we could
choose a j ∈ B \ KP such that M 0 [A0 , j] 6= 0, and then either (P ∪ {j}) \ {l0 } or
(P ∪ {j}) \ {l00 } would be an independent set meeting more blocks of Π than P , a
contradiction to the optimal choice of P . Therefore M 0 [A0 , B \ KP ] = 0. If there
were no blocks K 6= K0 meeting P for which M 0 [A0 , K] 6= 0, we would be done;
thus we can assume that there is such a K.
We now define inductively sets K0 ⊂ K1 ⊂ · · · ⊂ Ki and A0 ⊂ A1 ⊂ · · · ⊂ Ai
such that each set K0 , . . . , Ki is a union of certain blocks of Π, all of which meet
P , Ki is the union of i + 1 blocks, and Ai is a subset of A having i + 2 elements.
If Ki and Ai have been defined and there is no block K ∈ Π meeting P such
that K 6⊆ Ki and M 0 [Ai , K] 6= 0, then the construction stops and Ki+1 and Ai+1
are not defined. If there is such a block K, then let Ki+1 = Ki ∪ K and let
Ai+1 = Ai ∪ {k(lK )} = Ai ∪ {k(lKi+1 \Ki )}. Clearly there is an index m ≥ 0 such
that K0 , . . . , Km and A0 , . . . , Am are defined but Km+1 and Am+1 are not. In fact,
due to the aforementioned assumption, m ≥ 1. For 1 ≤ i ≤ m, Ki \ Ki−1 is a block
of Π and Ai \ Ai−1 is a singleton.
For each element j of a block K = Ki \Ki−1 , 1 ≤ i ≤ m, such that M 0 [Ai−1 , j] 6=
0 (there are always such elements j in K) we define κ(j) to be the smallest t ≥ 0
such that M 0 [At , j] 6= 0. We clearly have 0 ≤ t ≤ i − 1. Among all elements j of K
with M 0 [At , j] 6= 0, we choose one with minimal κ(j), and we denote this element
of K by hK .
We claim that for all 1 ≤ i ≤ m, M 0 [Ai , B \ KP ] = 0. To show that this is true,
let i be a minimal index for which this would fail. We shall derive a contradiction.
Let β ∈ B \ KP with M 0 [Ai , β] 6= 0. Recall that Ai \ Ai−1 is a singleton. Let
a1 = k −1 [Ai \Ai−1 ] and b1 = hK where a1 is in block K. We now define inductively
sequences of elements a1 , . . . , at of P and b1 , . . . , bt of B. If at and bt have been
defined and bt ∈
/ KP then the construction stops and bt+1 and at+1 are not defined.
A ROW-REDUCED FORM FOR COLUMN-PARTITIONED MATRICES
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If bt ∈ KP , then define at+1 and bt+1 as follows. Let at+1 be an element of P such
that k(at+1 ) ∈ Aκ(bt ) and M 0 (k(at+1 ), bt ) 6= 0 (there are at least one and at most
two such elements, depending on whether κ(bt ) is 0 or not). If this at+1 ∈
/ K0 ,
then let bt+1 = hK where at+1 is in block K, and if at+1 ∈ K0 , then let bt+1 = β.
Clearly there is an index s ≥ 2 such that a1 , . . . , as and b1 , . . . , bs are defined but
as+1 and bs+1 are not. Then
(P \ {a1 , a2 , . . . , as }) ∪ {b1 , b2 , . . . , bs }
is independent and meets r + 1 blocks. We have reached a contradiction and the
claim is proved.
It follows from the claim established above and the definition of m that M 0 [Am ,
B \ Km ] = 0. |Am | = m + 2 and the only blocks K of Π for which M 0 [Am , K] 6= 0
are among the m + 1 blocks K0 , K1 \ K0 , . . . , Km \ Km−1 . To conclude that (2)
holds, we can take A0 = Am .
3. Concluding remarks
From the proof given in the preceding section, we can draw the following additional conclusion strengthening the Theorem:
Corollary. If the columns of a p × q matrix over a field are partitioned into blocks
so that each nonsingular square submatrix is contained in at most r blocks, for some
r < p, then there is a non-negative m ≤ r such that in some row-equivalent matrix
the last m + 1 rows are null in all but at most m blocks of the given partition.
The Theorem as applied to matrices over the two-element field is used in [2]
to establish the descending chain condition in a particular quasi-order of k-valued
logic functions (k ≥ 3). The quasi-ordering is based on the composition of functions
from inside with the quasi-linear functions of Burle [1].
Acknowledgements
The authors thank Miguel Couceiro for useful discussions of the subject.
References
[1] G. A. Burle, The classes of k-valued logics containing all one-variable functions, Diskretnyi
Analiz 10 (1967) 3–7 (in Russian).
[2] E. Lehtonen, Subfunction relations defined by Burle’s clones, manuscript, Mar. 2006, http:
//math.tut.fi/algebra/.
Institute of Mathematics, Tampere University of Technology, P.O. Box 553, FI-33101
Tampere, Finland
E-mail address: [email protected]
Institute of Mathematics, Tampere University of Technology, P.O. Box 553, FI-33101
Tampere, Finland
E-mail address: [email protected]