When Does an Ensemble of Matrices with Randomly Scaled Rows Lose Rank?

0.49 MB·English

by  Navid Naderializadeh

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When Does an Ensemble of Matrices with
Randomly Scaled Rows Lose Rank?
Navid Naderializadeh, Aly El Gamal, and A. Salman Avestimehr
Department of Electrical Engineering, University of Southern California
Emails: [email protected], [email protected], [email protected]
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2 Abstract

r
p Weconsidertheproblemofdeterminingranklossconditionsforaconcatenationoffull-rankmatrices,suchthat
A
each row of the composing matrices is scaled by a random coefficient. This problem has applications in wireless

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interference management and recommendation systems. We determine necessary and sufficient conditions for the
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designofeachmatrix,suchthattherandomensemblewillalmostsurelyloserankbyacertainamount.Theresultis

] proved by converting the problem to determining rank loss conditions for the union of some specific matroids, and
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I thenusingtoolsfrommatroidandgraphtheoriestoderivethenecessaryandsufficientconditions.Asanapplication,
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s we discuss how this result can be applied to the problem of topological interference management, and characterize
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[ the linear symmetric degrees of freedom for a class of network topologies.

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v
4 I. INTRODUCTION
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We consider the problem in Figure 1, in which K users design matrices B ,...,B , where B is an n ×
5 1 K i
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m matrix, designed by the ith user, with full column rank m ≤ n. A destination obtains the matrix B =
0 i i D
1. [Λ1B1 Λ2B2 ··· ΛKBK], where Λi is an n×n diagonal matrix with random diagonal entries, and the set of
0 all random coefficients is drawn from a continuous joint distribution. It is easy to see that R=min(cid:16)(cid:80)K m ,n(cid:17)
5 i=1 i
1 is the maximum rank that the matrix B can have. However, by a careful design, one can reduce the rank of this
D
:
v
matrix. In particular, the question that we address in this work is: Under what conditions on the design of the
i
X a.s.
matrices B ,...,B will the matrix B lose rank by τ almost surely, i.e., rank(B ) ≤ R−τ, while each of the
r 1 K D D
a
matrices B ,...,B has full column rank?
1 K
B
1 ¤
B2 ¤2 1
. B = [¤B ¤B … ¤ B ]
. D 1 1 2 2 K K
.
¤
B K
K
Fig.1:Multipleusers,eachonehavingamatrixBiandanotheruserreceivingaconcatenationofthesematriceswithrandomly-scaledrows.
The aforementioned problem arises naturally when understanding the fundamentals of interference management
in wireless networks, in particular for the case where no information about the channel state is assumed to be
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available at the transmitters except for the knowledge about the network topology. In this setting, the K users in
Figure 1 represent transmitters that are interfering at a single receiver. The matrix B represents the beamforming
i
matrixattheith transmitterandthejth diagonalelementofthematrixΛ representsthechannelcoefficientinthejth
i
time slot. The column span of the matrix B represents the space of the received interference. The aforementioned
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problem is that of determining the conditions on the beamforming matrices that result in an alignment of the
interference caused by users 1 to K in a subspace whose dimension is at most R−τ.
Another motivating example arises in recommendation systems, where for a fixed set of items, each matrix B
i
represents the ratings given by the ith group of users. Each row represents one user’s ratings for the set of items,
and each diagonal element of Λ represents a random scaling factor that reflects the user’s own bias. For example,
i
one user can give a rating of 10 to his most favorite movie and 5 for his least favorite while another user can have
a highest rating of 8 and a lowest rating of 4. The problem in this setting would be to understand conditions on
the ratings of each user group that result in a rank loss of the ensemble. Understanding conditions of rank loss in
this application is useful for completion algorithms that recover missing ratings. In other words, the answer of our
question in this setting specifies the structure of matrix entries which suffice to complete the entire rating matrix.
Asaspecialcaseofthisproblem,onecanconsidereachB beingonlyacolumnvector.Thisisequivalenttothe
i
casewhereinsteadofeachrow,eachindividualelementisscaledbyarandomcoefficient.Inthiscase,theproblem
can be shown to have a combinatorial structure, depending on the position of zero/non-zero elements of B ’s. In
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particular, it can be verified that in this case, the ensemble matrix B will lose rank by τ (i.e., rank(B )≤R−τ)
D D
if and only if there does not exist a matching of size greater than R−τ between rows and columns of B , where
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a row is connected to a column if and only if the element with the corresponding row and column indices is
non-zero (see e.g.,[1–3] and also the Zippel-Schwartz Lemma [4, 5]). However, once B ’s are not column vectors,
i
theproblembecomesmuchmorecomplicatedsincethereisastructureintherandomscaling;elementsofthesame
row in each matrix are scaled by the same coefficient.
In this paper, we characterize a necessary and sufficient condition for rank-loss of the matrix B . In particular,
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we determine under what condition on B ’s, the matrix B loses rank by τ almost surely. The proof consists
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of a succession of equivalence steps. In the first step, we connect the rank of B to the rank of the union of
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certain matroids. This is first done by expanding the determinants of specific submatrices of B and then using
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the Zippel-Schwartz Lemma to express the almost-sure rank loss condition of B in terms of the products of
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the determinants of certain submatrices of B ’s being equal to zero. Then we use the notion of sparse subspaces
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in order to connect the rank loss of B to the column span of B ’s and then to the rank loss of the union of
D i
carefully-defined matroids. After representing the rank condition of B in terms of the rank loss of the union of
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certain matroids, in the second step, we make use of the matroid union theorem. This theorem enables us to derive
an equivalent condition on the structure of B ,...,B for the rank loss of the union of matroids.
1 K
In the final step of the proof, we simplify the above condition by constructing appropriate bipartite graphs in
which one partite set represents the column vectors of a carefully chosen set of bases for the column span of each
ofthematricesB ,...,B andtheotherpartitesetrepresentstherowindices.Acolumnvectorvertexisconnected
1 K
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to a row vertex if and only if the vector has a non-zero entry in the corresponding row. We use a variation of the
Hall’s marriage theorem to reach a final condition expressed through matching sizes on the constructed graphs.
As an application of the derived rank-loss condition, we utilize the result in the context of the topological
interferencemanagementproblem.Thisproblemfocusesonthescenarioinwhichinterferencemanagementprimarily
relies on a coarse knowledge about channel states in the network, namely the “topology” of the network. Network
topology simply refers to the 1-bit feedback information for each link between each transmitter and each receiver,
indicatingwhetherornotthesignalofthetransmitterisreceivedabovethenoiseflooratthecorrespondingreceiver.
There have been several prior works in the literature that have considered similar scenarios. In [6], it is assumed
thattransmittersareawareofthenetworktopologyaswellastheactualchannelgainswithinalocalneighborhood;
theconceptofnormalizedcapacitywasintroducedbycharacterizingthemaximumachievablerateusingthischannel
knowledgeasafractionoftheachievablerateusingglobalchannelstateinformation.In[7],theauthorsconsidered
a more restrictive scenario, called topological interference management (TIM), in which the transmitters are only
aware of the topology with no information about the channel coefficients. It has been shown that if the channel
gains in the network remain constant for a sufficiently large time, then topological interference management is
closelyconnectedtotheclassicalindexcodingproblem(see,e.g.,[8–11]),andviathisconnection,aclassoflinear
interference management schemes has been introduced which rely on the network topology knowledge to align the
interference over time at unintended receivers. In [12, 13], the authors considered a class of retransmission-based
schemesthatonlyallowstransmitterstoresendtheirsymbolsinordertoassistwiththeneutralizationofinterference
at the receivers. Besides their simplicity, these schemes are robust to channel variations over time and were shown
to be optimal in terms of the symmetric degrees-of-freedom (DoF) in many classes of topologies via the outer
bounds developed in [12, 13].
Inthispaper,weuseourderivedranklossconditioninordertosolvetwoproblemsinthetopologicalinterference
management framework. In this setting, Figure 1 can be seen as a set of K transmitters causing interference at a
destination, and the considered rank-loss problem can be seen as reducing the dimension of the subspace occupied
by interference at the receiver through careful design of the beamforming matrices at the transmitters.
Wefirstcharacterizethe“best”topologiesfortopologicalinterferencemanagement;i.e.,thetopologiesforwhich
half symmetric DoF is achievable for the case of time-varying channels. It is easy to see that in topologies with at
leastoneinterferencelink,thesymmetricDoFisupperboundedby 1.Thus,thetopologiesinwhichhalfsymmetric
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DoF is achievable represent the best topologies that one can hope for (from the degrees-of-freedom perspective).
For the case where the channel gains in the network are assumed to remain constant for a long-enough period
of time, the necessary and sufficient condition on the network topology for achieving half symmetric DoF was
characterized in [7]. For the case of case of time-varying channels, a sufficient condition for the achievability of
half symmetric DoF was derived in [14]. In this work, we close the gap in the results on half symmetric DoF by
introducing a necessary and sufficient condition under which the symmetric DoF of 1 is achievable for the case of
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time-varying channels (i.e., without requiring the channels to remain fixed for a long enough time).
Second, we use our condition to characterize the linear symmetric degrees of freedom (DoF) for a class of
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network topologies with exclusive interference sets. This helps to resolve the characterization of the symmetric
DoF for a set of previously open problems considered in [12].
II. PROBLEMFORMULATIONANDMAINRESULT
Consider K matrices B ,...,B , where each matrix B ,i ∈ [K] has size n×m , m ≤ n (we use [K] to
1 K i i i
denote the set {1,...,K} for any positive integer K). Without loss of generality, we assume that each matrix is
full-columnrank,sincethelinearly-dependentcolumnscanberemovedfromeachmatrix.Furthermore,considerK
diagonalmatricesΛ ,...,Λ ,eachofsizen×n,wheretheirdiagonalelementsaredrawnfromajointcontinuous
1 K
distribution.
Inthispaper,theproblemunderconsiderationistheranklossofthematrixB =[Λ B Λ B ··· Λ B ].
D 1 1 2 2 K K
To be precise, we aim to find an equivalent condition for when
a.s.
rank([Λ B Λ B ... Λ B ]) ≤ R−τ, (1)
1 1 2 2 K K
(cid:16) (cid:17)
where R=min (cid:80)K m ,n denotes the maximum possible rank of [Λ B Λ B ... Λ B ] and τ ∈Z+.
i=1 i 1 1 2 2 K K
Notation: For a matrix B ∈Rn×m, we use calligraphic B to denote the subspace in Rn spanned by the columns
of B. Also, for any X ⊆[n] and Y ⊆[m], B denotes the submatrix of B created by removing the rows with
X,Y
indices outside X and removing the columns with indices outside Y, and B denotes the submatrix of B created
∗,Y
by removing the columns with indices outside Y. Besides, for any X ⊆ [n], PK denotes the set of all partitions
X
of X to K disjoint subsets (each one possibly empty). Finally, for any J ⊆[n], Jc denotes the complement of J
in [n] (i.e., [n]\J) and S denotes the subspace of Rn spanned by the columns of the n×n identity matrix with
J
indices in J. In other words, S is the subspace of Rn which includes all the vectors that have zero entries in Jc.
J
We call S the sparse subspace of the set J.
J
We now state the main result in the following theorem.
Theorem 1. The following two statements are equivalent.
a.s.
rank([Λ B Λ B ... Λ B ]) ≤ R−τ. (C1)
1 1 2 2 K K
K K
(cid:88) (cid:88)
∀Y ⊆[m ],i∈[K] s.t. |Y |=R,∃J ⊆[n]: dim(S ∩B )≥|J|+τ. (C2)
i i i J i,∗,Yi
i=1 i=1
Example 1. As a simple example, consider the case where K = 2 and m = m = n. In this case, B and
1 2 2 1
B will each be of size n× n and the only possible choice for Y and Y will be [n]. Theorem 1 implies that
2 2 1 2 2
a.s.
rank([Λ B Λ B ]) ≤ n−τ if and only if there exists a set J ⊆[n] such that,
1 1 2 2
dim(S ∩B )+dim(S ∩B )≥|J|+τ. (2)
J 1 J 2
 T
1 1 1 0
Example 2. Continuing Example 1, assume n=2m1 =2m2 =4 and suppose we fix B1 =  . In
1 2 3 0
this case, Theorem 1 implies that the only way for the matrix [Λ B Λ B ] to lose rank by τ =1 almost-surely
1 1 2 2
5
is that both columns of B have zeros in their 4th entries. The set J ⊆[4] satisfying (2) would be {1,2,3} in this
2
case.
Remark1. ThesignificanceoftheaboveresultliesinfindingaconditiononthestructureofthematricesB ,i∈[K],
i
such that the statistical rank loss condition is met. Checking the structural condition of (C2) does not involve
statistical analysis and relies only on the combinatorial structure of B ’s.
i
Remark 2. When each element of the matrices B ,...,B is scaled by a random coefficient, (C1) corresponds to
i K
the size of a bipartite graph matching representing rows and columns of B as the two partite sets. This condition
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will be similar to (C2), but instead of dim(S ∩B ), the number of columns in B inside the subspace S is
J i i J
considered. This is intuitive as in the considered setting, each row is scaled by the same coefficient, and hence, the
angles between column vectors are preserved. Our result can be seen as a generalization of the classical rank-loss
results for random matrices.
Remark 3. We show in Section IV how the result of Theorem 1 can be used to characterize the linear degrees of
freedom for the topological interference management problem of a class of networks that had been considered as
an open problem before.
III. PROOFOFTHEMAINRESULT
The proof of Theorem 1 is composed of four steps, in each of which we present a condition equivalent to
condition (C1).
A. Step 1: Expansion of Determinant
We begin by the following equivalence lemma.
Lemma 1. Condition (C1) is equivalent to the following.
K
(cid:88)
∀X ⊆[n]:|X|>R−τ,∀Y ⊆[m ],i∈[K] s.t. |Y |=|X|,
i i i
i=1
K
(cid:89)
∀(I ,I ,...,I )∈PK s.t. |I |=|Y |: det(B )=0. (C3)
1 2 K X i i i,Ii,Yi
i=1
Proof:Bythedefinitionofmatrixrank,theconditionin(C1)isequivalenttothefactthatanysquaresubmatrix
of B with size greater than (R−τ)×(R−τ) should have a zero determinant almost-surely. This means that for
D
any subset of rows X ⊆[n]:|X|>R−τ and any subsets of columns Y ⊆[m ], i∈[K] s.t. (cid:80)K |Y |=|X|,
i i i=1 i
a.s.
det([Λ B ... Λ B ]) = 0.
1,X,X 1,X,Y1 K,X,X K,X,YK
It can be shown that this determinant is composed of monomials in the channel gains whose coefficients are
in the form of (cid:81)K det(B ) for some (I ,I ,...,I ) ∈ PK s.t. |I | = |Y |. By the Zippel-Schwartz Lemma
i=1 i,Ii,Yi 1 2 K X i i
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[4, 5], for the whole multivariate polynomial to be equal to zero for almost all values of the channel gains, each
of these coefficients should be equal to zero, which gives (C3).
(cid:104) (cid:105)T
Example3. ConsiderthecasewhereK =2,n=3,m1 =1,m2 =2,andwehaveB1 = a11 a21 a31 ,B2 =
 T
a(cid:48) a(cid:48) a(cid:48)
 11 21 31 . Also, let λi, λ(cid:48)i, i∈{1,2,3}, be the ith diagonal element of Λ1 and Λ2, respectively. We then
a(cid:48) a(cid:48) a(cid:48)
12 22 32
have,
 
λ a λ(cid:48)a(cid:48) λ(cid:48)a(cid:48)
1 11 1 11 1 12
 
BD =λ2a21 λ(cid:48)2a(cid:48)21 λ(cid:48)2a(cid:48)22.
 
λ a λ(cid:48)a(cid:48) λ(cid:48)a(cid:48)
3 31 3 31 3 32
Now, rank(B )≤2 is equivalent to det(B )=0. Note that,
D D
  
a(cid:48) a(cid:48)
det(BD)=λ1λ(cid:48)2λ(cid:48)3a11det 21 22
a(cid:48) a(cid:48)
31 32
  
a(cid:48) a(cid:48)
−λ2λ(cid:48)1λ(cid:48)3a21det 11 12
a(cid:48) a(cid:48)
31 32
  
a(cid:48) a(cid:48)
+λ3λ(cid:48)1λ(cid:48)2a31det 11 12.
a(cid:48) a(cid:48)
21 22
Therefore, the Zippel-Schwartz Lemma implies that this determinant is almost-surely equal to zero if and only if
each of the products inside the brackets is equal to zero.
B. Step 2: Sparse Subspaces
We now express the condition (C3) in terms of sparse subspaces. Recall that a sparse subspace is defined by
column vectors that have zero entries for a specific set of rows.
Lemma 2. Condition (C3) is equivalent to the following.
∀X ⊆[n]:|X|>R−τ,
K
(cid:88)
∀Y ⊆[m ],i∈[K] s.t. |Y |=|X|,
i i i
i=1
∀(I ,I ,...,I )∈PK s.t. |I |=|Y |,∀i∈[K]:
1 2 K X i i
K
(cid:88)
dim(SIic ∩Bi,∗,Yi)>0. (C4)
i=1
Proof: First assume that the condition in (C3) holds. Then for some i∈[K] there exists a linear combination
of the columns in B which is equal to the zero vector. Therefore, if we apply this linear combination to the
i,Ii,Yi
entire matrix Bi,∗,Yi, we end up with a vector in the sparse subspace SIic since Bi,∗,Yi is full-column rank. This
implies (C4).
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If the condition in (C4) holds, then for some i∈[K], there exists a linear combination of the columns in B
i,∗,Yi
whichiszerointhecoordinatesinI .ThismeansthatforanyI :|I |=|Y |,det(B )=0,implyingthat(C3)
i i i i i,Ii,Yi
holds.
C. Step 3: Connection to Rank of Union of Matroids
In this step, we show how to represent condition (C4) in terms of the rank loss of the union of certain matroids.
To this end, for any choice of X ⊆[n] and Y ⊂[m ], ∀i∈[K] which satisfy (cid:80)K |Y |=|X|>R−τ and
i i i=1 i
dim(S ∩B )=0,∀i∈[K], (3)
Xc i,∗,Yi
define I as
i,X,Yi
I ={I ⊆X :dim(S ∩B )=0},i∈[K]. (4)
i,X,Yi I∪Xc i,∗,Yi
We have the following claim that is proved in Appendix A.
Claim1. M =(X,I )isamatroidwithrankfunctionr (J)=|J|−dim(S ∩B ),i∈[K].
i,X,Yi i,X,Yi i,X,Yi J∪Xc i,∗,Yi
For any matroid M , the dual matroid M∗ is defined as M∗ = (X,I∗ ) whose basis sets are
i,X,Yi i,X,Yi i,X,Yi i,X,Yi
the complements of the basis sets of M [15].1 Using Theorem 39.2 in [15] and Claim 1, the rank of this dual
i,X,Yi
matroid is
r∗ (J)=|J|−r (X)+r (X\J)
i,X,Yi i,X,Yi i,X,Yi
=|J|−(|X|−|Y |)+(|X\J|−dim(S ∩B ))
i Jc i,∗,Yi
=|Y |−dim(S ∩B ). (5)
i Jc i,∗,Yi
(cid:16) (cid:17)
Now, consider the union of the dual matroids M∗ , i∈[K], denoted by (cid:87)K M∗ = X,(cid:87)K I∗ ,
i,X,Yi i=1 i,X,Yi i=1 i,X,Yi
where
K (cid:40)K (cid:41)
(cid:95) (cid:91)
I∗ = I :I ∈I∗ .
i,X,Yi i i i,X,Yi
i=1 i=1
Let r∗ (.) denote the rank of (cid:87)K M∗ . Then we have the following lemma.
X,Y1,...,YK i=1 i,X,Yi
Lemma 3. For any X ⊆ [n],Y ⊆ [m ],i ∈ [K] s.t. (cid:80)K |Y | = |X| > R−τ and (3) is satisfied, the following
i i i=1 i
are equivalent
K
(cid:88)
∀(I1,I2,...,IK)∈PXK s.t. |Ii|=|Yi|: dim(SIic ∩Bi,∗,Yi)>0. (6)
i=1
r∗ (X)<|X|. (7)
X,Y1,...,YK
1ForamatroidM =(X,I),J ⊆X iscalledabasissetifJ ∈I andthereisnoJ(cid:48)∈I suchthatJ ⊂J(cid:48)⊆X [15,Ch.39].
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Proof: Condition (6) is equivalent to the following for any valid choice of X and Y ’s:
i
(cid:64)(I ,I ,...,I )∈PK :|I |=|Y | and X\I ∈I ,∀i∈[K]. (8)
1 2 K X i i i i,X,Yi
Now, let us focus on a specific i∈[K] and the corresponding set I whose size satisfies |I |=|Y |,
i i i
|X\I |=|X|−|Y |
i i
=|X|−dim(S ∩B )
X∪Xc i,∗,Yi
=r (X), (9)
i,X,Yi
where (9) follows from Claim 1. By the definition of the rank of a matroid (see, e.g. [15, Ch. 39]), (9) implies that
all the members of I (i.e., the independent sets of M ) are of size at most |X \I |. This means that if
i,X,Yi i,X,Yi i
X\I ∈I , then it is a basis for M . This, in turn, is equivalent to I being a basis for the dual matroid
i i,X,Yi i,X,Yi i
M∗ .
i,X,Yi
Ontheotherhand,from(5)wehavethattherankofthedualmatroidM∗ isupperboundedby|Y |,implying
i,X,Yi i
that all the members of I∗ have size at most |Y |. Thus, under the constraint |I | = |Y |, I being a basis for
i,X,Yi i i i i
the dual matroid M∗ is equivalent to I ∈I∗ . Consequently, (8) (hence (6)) is equivalent to the following
i,X,Yi i i,X,Yi
for any valid choice of X and Y ’s:
i
(cid:64)(I ,I ,...,I )∈PK :|I |=|Y | and I ∈I∗ ,∀i∈[K]. (10)
1 2 K X i i i i,X,Yi
Finally, it is easy to verify that (10) is equivalent to r∗ (X) < |X|, since there cannot exist any
X,Y1,...,YK
(I ,I ,...,I )∈PK such that I ∈I∗ at the same time for all i∈[K]. This completes the proof.
1 2 K X i i,X,Yi
Notethatusingthislemma,wecannowreplacethelasttwolinesofcondition(C4)bysimplyr∗ (X)<
X,Y1,...,YK
|X|.
D. Step 4: Matroid Union Theorem
Inthisstepoftheproof,wemakeuseoftheMatroidUnionTheorem[15]tocharacterizeanequivalentcondition
to (7). We start by stating the Matroid Union Theorem.
Theorem 2. (Matroid Union Theorem [15, Chapter 42]) Let M = (E ,I ), i ∈ [K], be K matroids with rank
i i i
functions r (.). Then (cid:87)K M is also a matroid with rank function
i i=1 i
(cid:32) K (cid:33)
(cid:88)
r(U)= min |U \T|+ r (T ∩E ) . (11)
i i
T⊆U
i=1
Equipped with the above theorem, we can now state the following equivalence lemma.
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Lemma 4. Condition (C4) is equivalent to the following.
∀X ⊆[n]:|X|>R−τ,
K
(cid:88)
∀Y ⊆[m ],i∈[K] s.t. |Y |=|X|,
i i i
i=1
K
(cid:88)
∃J ⊆X : dim(S ∩B )>|J|. (C5)
J∪Xc i,∗,Yi
i=1
Proof:IfforaspecificchoiceofX andY ’s,(3)isnotsatisfied(i.e.,dim(S ∩B )>0forsomei∈[K]),
i Xc i,∗,Yi
then it is clear that both statements in (C4) and (C5) hold. Hence, w.l.o.g. we assume that (3) is satisfied. For any
X ⊆[n],Y ⊆[m ],i∈[K] s.t. (cid:80)K |Y |=|X|>R−τ and (3) is satisfied, we will get from Lemma 3 that the
i i i=1 i
condition in (C4) is equivalent to,
r∗ (X)<|X|
X,Y1,...,YK
(cid:32) K (cid:33)
(a) (cid:88)
⇔ min (|X|−|J|)+ (|Y |−dim(S ∩B )) <|X|
J⊆X i Jc i,∗,Yi
i=1
K (cid:32) K (cid:33)
(cid:88) (cid:88)
⇔|X|+ |Y |−max |J|+ dim(S ∩B ) <|X|
i J⊆X Jc i,∗,Yi
i=1 i=1
(cid:18) K (cid:19)
(b) (cid:88)
⇔ max |J|+ dim(S ∩B ) >|X|
J⊆X Jc i,∗,Yi
i=1
K
(cid:88)
⇔ ∃J ⊆X : dim(S ∩B )>|X\J|
Jc i,∗,Yi
i=1
K
(c) (cid:88)
⇔ ∃J ⊆X : dim(S ∩B )>|J|,
J∪Xc i,∗,Yi
i=1
where (a) follows from (5) and Theorem 2, (b) follows from the assumption that (cid:80)K |Y |=|X| and (c) follows
i=1 i
by changing J to X\J. This completes the proof.
E. Step 5: Hall’s Marriage Theorem
In the final step of the proof, we prove the equivalence between (C5) and (C2) by constructing an appropriate
bipartite graph. One partite set represents the column vectors of a carefully chosen set of bases for the subspaces
B and the other partite set has n elements, each corresponding to a row. A column vector vertex is connected
i,∗,Yi
to a row vertex if and only if the vector has a non-zero entry in the corresponding row. We then use the following
variation of Hall’s marriage theorem to complete the proof by representing the final condition (C2) in terms of a
matching size on the constructed bipartite graph.
Theorem 3. Let G = (A∪B,E) be a bipartite graph. G has a matching of size k if and only if the size of the
neighboring set |N(I)|≥|I|−|B|+k for any I ⊆B.
10
Wefirstshowthat(C2)⇒(C5).Assumethat(C2)holdsandfixY ⊆[m ],i∈[K]:(cid:80)K |Y |=R.LetJ∗ ⊆[n]
i i i=1 i
besuchthat(cid:80)K dim(S ∩B )≥|J∗|+τ.Foreachi∈[K],letc(i),...,c(i) be|Y |columnvectorsthatform
i=1 J∗ i,∗,Yi 1 |Yi| i (cid:110) (cid:111)
a basis for B and a subset of these vectors form a basis for S ∩B . Now, let C = c(i),...,c(i) and
i,∗,Yi J∗ i,∗,Yi i 1 |Yi|
constructabipartitegraphG=(A∪C,E)whereA={1,2,...,n}andC ={C ,...,C }isamultisetconsisting
1 K
of the elements of the sets C ,i∈[K]. A and C are the two partite vertex sets of the graph G. For any i∈[n] and
i
c ∈ C, (i,c) ∈ E if and only if the vector c has a non-zero entry in the ith position, i.e., if c = (c ,...,c ) then
1 n
(a ,c)∈E ⇔c (cid:54)=0.Now,notethatalltheverticesinC thatcorrespondtovectorsinS canonlybeconnectedto
i i J∗
verticesinAthatcorrespondtothesetJ∗.Since(cid:80)K dim(S ∩B )≥|J∗|+τ,itfollowsthatthereisasubset
i=1 J∗ i,∗,Yi
of the partite set C of size at least |J∗|+τ whose neigboring set has size at most |J∗|. It follows from Theorem 3
that there is no matching of size R−τ+1 in the bipartite graph G. It follows that for any X ⊆[n]:|X|>R−τ
andanyK setsY(cid:48) ⊆C,i∈[K]:(cid:80)K |Y(cid:48)|=|X|,thereisnomatchingbetweentheverticescorrespondingtothe
i i=1 i
set X in the partite set A and the vertices in {Y(cid:48),...,Y(cid:48) }, and hence, there is a set J ⊆X such that the vertices
1 K
corresponding to the set X\J are connected to less than |X\J| vertices in (cid:83)K Y(cid:48). It follows that for this choice
i=1 i
of the set J, there are more than |J| vertices in (cid:83)K Y(cid:48) that are not connected to any vertex in the set X\J in the
i=1 i
partite set A. Now, if ∀i∈[K],Yi(cid:48) ⊆Yi, then (cid:80)Ki=1dim(SJ∪X¯ ∩Bi,∗,Yi(cid:48))>|J|. We now use the above argument
to prove that (C5) holds as follows. For all Y(cid:48) ⊆[m ],i∈[K] such that (cid:80)K |Y(cid:48)|>R−τ, we find Y ,i∈[K]
i i i=1 i i
such that Y(cid:48) ⊆Y ,∀i∈[K] and (cid:80)K |Y |=R, and then use the above argument to show that,
i i i=1 i
K
(cid:88)
∃J ⊆[n]: dim(S ∩B )≥|J|+τ ⇒
J i,∗,Yi
i=1
K
(cid:88)
∀X ⊆[n]:|X|= |Y(cid:48)|,
i
i=1
K
(cid:88)
∃J ⊆X : dim(SJ∪X¯ ∩Bi,∗,Y(cid:48))>|J|,
i
i=1
and hence, (C5) follows.
We now show that (C5) ⇒ (C2) by contradiction. Suppose that (C2) does not hold and fix Y ⊆[m ],i∈[K]:
i i
(cid:80)K |Y | = R such that for any J ⊆ [n], (cid:80)K dim(S ∩B ) < |J|+τ. For each i ∈ [K], let c(i),...,c(i)
i=1 i i=1 J i,∗,Yi (cid:110) (cid:111) 1 |Yi|
be |Y | column vectors that form a basis for B and define C = c(i),...,c(i) . Now, we construct a bipartite
i i,∗,Yi i 1 |Yi|
graph G = (A∪C,E) where A = {1,2,...,n} and C = {C ,...,C } is a multiset consisting of the elements
1 K
of the sets C ,i∈[K]. Also, for any i∈[n],c∈C, (i,c)∈E ⇔c (cid:54)=0.
i i
From the selection of the sets Y ,i∈[K], we have that for any I ⊆C, the neighboring set N(I)⊆A has size
i
|N(I)|>|I|−τ. It follows from Theorem 3 that there is a matching in G of size |C|−τ +1. Such a matching
will include R−τ +1 edges incident on R−τ +1 nodes in A (which we denote by X) and R−τ +1 nodes in
C (which we denote by Y(cid:48)). For each i ∈ [K], let Y(cid:48) = {j : c(i) ∈ Y(cid:48)}. Now, in order to show that (C5) is not
i j
true, it suffices to show that for the specific choice of X, Y(cid:48),i∈[K] mentioned above, the following holds:
i
K
(cid:88)
∀J ⊆X : dim(S ∩B )≤|J| (12)
J∪Xc i,∗,Y(cid:48)
i
i=1