6.3.1 Necessary and Sufficient Conditions
The decomposition ofX = A S, withA> 0,S > 0,states that each row ofX satisfies xj ∈ CS = cone(S)Denoting Kthe convex hull of the data matrixX, the necessary and sufficient condition for the uniqueness of the non-negative factorization is given by the following theorem stated by Chen [84]. Theorem 1 The decomposition ofX according to
X =A S with A>0,S >0, (6.9) is unique if an only if the simplicial coneCSsuch asK ⊂ CS is unique.
The key point is that this theorem does not provide any numerical tool to check whether a particular matrix admits a unique non-negative decomposition or not.
Chapter 4. Uniqueness of Non-negative matrix Factorization: An introduction 81
6.3.2 Sufficient Uniqueness Conditions
Proposition 1:The decomposition ofX intoAandS according to
X =A SwithA>0,S >0, (6.10) is unique if the following conditions are satisfied:
(B1) There exists a submatrix ofS of dimension(p, p)which is monomial.
(B2) There exists a submatrix ofAof dimension(p, p)which is monomial.
Proof 1: Suppose that conditions (B1) and (B2) are satisfied. After permutation of the its columns, the matrixS may be rewritten as
S =
s11 0 0 s1(p+1) . . . s1n
0 . .. 0 ... ... 0 0 spp sp(p+1) . . . spn
.
Similarly, after permutation of its rows, the matrixAmay be written as
A=
a11 0
. ..
0 app
a(p+1)1 . . . a(p+1)p ... ... am1 . . . amp
.
Let us consider the regular (p × p) matrix T = [tij] whose inverse is noted T−1 =h
t]iji
. We have
T S =
t11s11 . . . t1pspp . . . ... ...
tp1s11 . . . tppspp . . .
.
AT−1 =
a11t]11 . . . a11t]1p ... ... appt]1p . . . appt]pp
... ...
.
Chapter 4. Uniqueness of Non-negative matrix Factorization: An introduction 82 from which it turns out thatT S > 0and AT−1 > 0 if and only ifT > 0 and T−1 > 0. From property (B1), this is equivalent to say that T is a monomial matrix and according to property (B2), it shows that the solution is unique up to scaling and ordering indeterminacies.
From this result, we may deduce immediately the following corollary which gives a sufficient condition onX to admit a unique non-negative factorization.
Corollary 1:The decomposition ofX intoAandS according to
X =A SwithA>0,S >0, (6.11) is unique if the following condition is satisfied
(C1) There exists a submatrix ofX of dimension(p, p)which is monomial.
One can also give a geometrical interpretation of the sufficient uniqueness con- ditions. DefinePS =ImS∩Rn+, which is a simplicial cone with ap-dimensional frame. Denoting CS = cone(S), the condition (B1) states that CS = PS while (B2) imposes that at leastpvectors (rows) ofXare collinear with thepdifferent vectors of the frame ofPS. We can also link these sufficient conditions to those of [88]. Among these conditions: the first one is the generative model condition which imposes the existence of the positive factorization. The separability con- dition of that paper is equivalent to the condition (B1) of proposition 3, that is CS = PS. The last condition (Complete factorial sampling) imposes that there are at least (p−1)different vectors (rows) of X in the pfacets (of dimension (p−1)) of the cone PS. In this case, it is clear that the only cone of dimension pwhich includes all the vectors of X is the conePS itself. The sufficient con- ditions of proposition 3 can be recovered by migrating thepvectors generating this cone. Each facet of the cone being of dimension (p−1), each generating vector of the cone belongs to (p −1) different facets. Thus, we still have the condition of having (p−1) different vectors of X in the (p−1) facets of the cone. But only p different vectors are needed instead of p(p−1). According to this geometrical interpretation, we can deduce the different situations where the uniqueness of the solution of non-negative source separation is ensured.
These situations are represented in the figure6.1.
Chapter 4. Uniqueness of Non-negative matrix Factorization: An introduction 83
6
+
- ...
...
...
...
...
...
...
... ...6...
+
- PS PS
Case 2 : Donoho-Stodden Case 1 : Theorem 3
FIGURE6.1: Different cases for the uniqueness of the non-negative factoriza- tion. The hollow dots represent the vectors ofX which are needed to ensure uniqueness.
6.3.3 Necessary Uniqueness Conditions
Consider the case ofpsource signals and let
A= [a1 · · · ap] and S = [s1 · · · sp]T , (6.12) where{ai}pi=1are vectors of dimension(m,1)and{sj}pj=1are vectors of dimen- sion(n,1). For convenience, in this section, we will noteai(`)the`-th element of vectorai and similarly,sj(k)thek-th element of vectorsj.
Proposition 2:If the decomposition ofXintoAandS according to
X =A SwithA>0,S >0, (6.13) is unique, then the following conditions are satisfied
(A1) ∃k1, . . . , kpdistincts, such that∀i,si(ki) = 0,∀j 6=isj(ki)6= 0.
(A2) ∃l1, . . . , lpdistincts, such that∀i,ai(li) = 0,∀j 6=iaj(li)6= 0.
Proof 2:The proof of this proposition is achieved by contradiction: suppose that conditions (A1)-(A2) are not satisfied and the decompositionX =A S withA, S >0, is unique. Suppose that condition (A1) is not satisfied and let the(p×p)
Chapter 4. Uniqueness of Non-negative matrix Factorization: An introduction 84 elementary transformationTij(λ)defined by
tkk = 1,∀k = 1, . . . , p tk` =λ, if(k, `) = (i, j) tk` = 0 elsewhere.
(6.14)
Note thatT−1ij (λ) =Tij(−λ).
Define K = {k,si(k)6= 0,sj(k)6= 0}. The samples where the sources si and sj are simultaneously zero are not taken into account since they are not affected by the possible transformations.
The following decomposition
X =A Tij(λ)Tij(−λ)S (6.15) withλ > 0ensures the non-negativity ofA˜ =ATij(λ)and the matrixTij(−λ) leads to a transformation of thei-th source signal according to
˜
si =si−λsj, (6.16)
while the other source signals are unaltered.
Defining(si, sj)as
si = min
k∈K
si(k), sj = max
k∈K
sj(k),
(6.17)
there exists0< λ < si/sj such that
∀k,si(k)−λsj(k)>si−λsj >0
=⇒ Tij(λ)S >0.
Therefore, the decomposition is not unique. This is in contradiction with the as- sumptions. Concerning condition (A2), the same reasoning is employed using λ <0.
Remark 1: This necessary conditions shows that the separation using only non negativity assumptions may provide an incorrect solution. Particularly, since condition A2 imposes the existence of observed mixtures where the amount of
Chapter 4. Uniqueness of Non-negative matrix Factorization: An introduction 85 one source is exactly zero, which is unrealistic for example in the case of kinetic reactions.
Remark 2: This is not a sufficient condition. Indeed, consider the following example
S =
0.53 0 0.24 0.23 0.13 0.52 0 0.35 0 0.32 0.34 0.34
,
and letT−1 be the matrix defined by
T−1 =
0.66 0 0.34 0.16 0.84 0
0 0.52 0.49
.
Thus, we have
T =
1.37 0.60 −0.97
−0.26 1.08 0.18 0.28 −1.15 1.87
,
from which we get
A˜=AT−1 >0, ∀A>0;
S˜=T S=
0.80 0 0 0.20 0 0.62 0 0.38 0 0 0.70 0.30
>0.