About higher-order L ¨ owner tensors
Otto Debals , KU Leuven April 2017
In this report, we give a definition of a L¨ owner tensor and prove that a L¨ owner tensor satisfies similar low-rank properties as L¨ owner matrices.
Matrix case
We have the following definition of a L¨ owner matrix:
Definition 1. Given a function f (t) sampled on points T = {t
1, t
2, . . . , t
N}. We partition the point set T into two point sets X = {x
1, x
2, . . . , x
I} and Y = {y
1, y
2, . . . , y
J} with I + J = N , and define the entries of the L¨ owner matrix L ∈ K
I×Jas
∀i, j : l
ij= f (x
i) − f (y
j)
x
i− y
j. (1)
The following has been proven in [1, 2, 5]:
Theorem 1. Given a L¨ owner matrix L of size I
1×I
2constructed from a function f (t) sampled in a point set T = {t
1, . . . , t
N} with N = I
1+ I
2. If f (t) is a rational function of degree δ and if I
1, I
2≥ δ, then L has rank δ:
rank(L) = δ = deg(f ).
Tensor case
We now define the L¨ owner tensor of arbitrary order K as a generalization of the L¨ owner matrix:
Definition 2. Given a function f (t) sampled on points T = {t
1, t
2, . . . , t
N}. We partition the point set T into K point sets T
(k)= {t
(k)1, t
(k)2, . . . , t
(k)Ik
} with P
Kk=1
I
k= N . We define the entries of the L¨ owner tensor L ∈ K
I1×···×IKas
∀i
1, . . . , i
K: l
i1i2···iK=
K
X
k=1
f (t
(k)ik
)
K
Q
m=1 m6=k
t
(k)ik
− t
(m)im
.
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Let us consider two special cases:
• For K = 2, we obtain the L¨ owner matrix L from (1):
∀i, j : l
ij= f (x
i)
x
i− y
j+ f (y
j)
y
j− x
i= f (x
i) − f (y
j) x
i− y
j.
• For K = 3, we partition a point set T into three point sets X, Y and Z with I, J and K points, respectively. We then obtain a L¨ owner tensor L ∈ K
I×J ×Kwith the following entries:
∀i, j, k : l
ijk= f (x
i)
(x
i− y
j) (x
i− z
k) + f (y
j)
(y
j− x
i) (y
j− z
k) + f (z
k)
(z
k− x
i) (z
k− y
j) . Interestingly enough, as we will prove, the low-rank property from the matrix case partly holds in the tensor case.
Theorem 2. If f (t) is a degree-1 rational function, then the L¨ owner tensor L ∈ K
I1×···×IKconstructed from evaluations of f (t) in the point set T = {t
1, . . . , t
N} with P
Kk=1
= N has rank 1.
Proof. A degree-1 rational function can be parametrized as f (t) = a
t−p1+ b. It can easily be seen that the entries of the L¨ owner tensor are invariant to the constant term b. Furthermore, if we show that L has rank 1 for a = 1, L also has rank 1 for nonzero a 6= 1. Hence, it remains to show that L has rank 1 if f (t) =
t−p1.
Let us consider Lagrange interpolation with corresponding Lagrange basis polynomials:
l
k(u) =
K
Y
m6=km=1
(u − u
m)
(u
k− u
m) for k = 1, . . . , K. (2) Given a set of K points (u
k, v
k) for k = 1, . . . , K, let us define the polynomial
g(u) =
K
X
k=1
v
kl
k(u). (3)
The function g(u) is the interpolating polynomial of degree at most K − 1 interpolating the points (u
k, v
k). It is well-known that g(u) is the unique interpolating polynomial of degree at most K − 1.
Let us consider the special case with K points (u
k, 1) for k = 1, . . . , K. We find that the con- stant function g(u) = 1 interpolates these points. Hence, this function is the only interpolating polynomial of degree at most K − 1 and we obtain the following equality by substituting (2) in (3):
g(u) =
K
X
k=1 K
Y
m=1 m6=k
(u − u
m) (u
k− u
m) = 1.
Dividing both sides with
K
Q
m=1
(u − u
m) gives
K
X
k=1
1 (u − u
k)
K
Y
m=1 m6=k
1
(u
k− u
m) =
K
Y
m=1
1 (u − u
m) .
About higher-order L¨ owner tensors 2
We can safely replace the index m in the right-hand side with k, and replace each (u − u
k) with −(u
k− u):
K
X
k=1
1 (u
k− u)
K
Y
m=1m6=k
1
(u
k− u
m) = (−1)
(K+1)K
Y
k=1
1 (u
k− u) .
Instead of the set {u
1, . . . , u
K}, let us consider the set n t
(1)i1
, . . . , t
(K)iK
o
. Let us also replace u with p. Substitution gives:
K
X
k=1
1 (t
(k)ik
− p)
K
Y
m6=km=1
1 (t
(k)ik
− t
(m)im
)
= (−1)
K+1K
Y
k=1
1 (t
(k)ik
− p) . The term 1/(p − t
(k)ik
) is equal to the function f (t) =
t−p1evaluated in t
(k)ik
. Hence, we obtain
K
X
k=1
f (t
(k)ik
)
K
Q
m6=km=1
t
(k)ik
− t
(m)im
= (−1)
K+1K
Y
k=1
1 (t
(k)ik
− p) .
The left-hand side is equal to the entry l
i1···iK. Hence, the tensor L can be written as the outer product
L = a
(1)⊗· · ·
⊗a
(K)with nonzero vectors a
(k)∈ K
Ikwith entries a
(k)ik
=
1(t(k)ik −p)
. Hence, we have shown that L has rank 1.
Theorem 3. If f (t) is a degree-R rational function with distinct poles, then the L¨ owner tensor L ∈ K
I1×···×IKconstructed from evaluations of f (t) in the point set T = {t
1, . . . , t
N} with P
Kk=1
= N and with I
k≥ δ for 1 ≤ k ≤ K has rank δ.
Proof. Based on Theorem 2, it is easy to show that L can be written as a sum of δ rank-1 terms. To show that L has rank δ, we have to show that it is not possible to write L as a sum of less rank-1 terms. This can be done by showing that the decomposition is unique. A sufficient condition for uniqueness is to show that the factor matrices of the CPD have full column rank [4]
1.
The columns of each factor matrix of L are evaluated degree-1 rational functions of the form
1
t−pr
as shown in Theorem 2. As the poles p
rare assumed to be distinct, it follows that the factor matrices have full column rank.
Note that the condition in Theorem 3 can be significantly relaxed. Unlike the matrix case, the rank of a L¨ owner tensor can be equal to the degree of the rational function even if the latter is larger than one or more dimensions of the tensor [3].
The rank properties of L¨ owner tensors in the situation of coinciding poles remain to be shown.
It is expected that the L¨ owner tensor has a low multilinear rank, similar to the Hankel case.
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