From moments to sparse representations, a geometric, algebraic and algorithmic viewpoint. B. Mourrain.part 1

From moments to sparse representations,
a geometric, algebraic and algorithmic viewpoint
Bernard Mourrain
Inria M´editerran´ee, Sophia Antipolis
Bernard.Mourrain@inria.fr
Part I

Sparse representation problems
1 Sparse representation problems
2 Duality
3 Artinian algebra
B. Mourrain From moments to sparse representations 2 / 27

Sparse representation of signals
Given a function or signal f (t):
decompose it as
f (t) =
r
i=1
(ai cos(µi t) + bi sin(µi t))eνi t
=
r
i=1
ωieζit

Prony’s method (1795)
For the signal f (t) = r
i=1 ωieζit, (ωi , ζi ∈ C),
Evaluate f at 2 r regularly spaced points: σ0 := f (0), σ1 := f (1), . . .
Compute a non-zero element p = [p0, . . . , pr] in the kernel:





σ0 σ1 . . . σr
σ1 σr+1
...
...
σr−1 . . . σ2r−1 σ2r−1










p0
p1
...
pr





= 0
Compute the roots ξ1 = eζ1 , . . . , ξr = eζr of p(x) := r
i=0 pi xi .
Solve the system





1 . . . . . . 1
ξ1 ξr
...
...
ξr−1
1 . . . . . . ξr−1
r










ω1
ω2
...
ωr





=





σ0
σ1
...
σr−1





.

Symmetric tensor decomposition
and Waring problem (1770)
Symmetric tensor decomposition problem:
Given a homogeneous polynomial ψ of degree d in the variables
x = (x0, x1, . . . , xn) with coeﬃcients ∈ K:
ψ(x) =
|α|=d
σα
d
α
xα
,
ﬁnd a minimal decomposition of ψ of the form
ψ(x) =
r
i=1
ωi (ξi,0x0 + ξi,1x1 + · · · + ξi,nxn)d
with ξi = (ξi,0, ξi,1, . . . , ξi,n) ∈ K
n+1
spanning disctint lines, ωi ∈ K.
The minimal r in such a decomposition is called the rank of ψ.

Sylvester approach (1851)
Theorem
The binary form ψ(x0, x1) = d
i=0 σi
d
i xd−i
0 xi
1 can be decomposed as a
sum of r distinct powers of linear forms
ψ =
r
k=1
ωk (αkx0 + βkx1)d
iﬀ there exists a polynomial p(x0, x1) := p0xr
0 + p1xr−1
0 x1 + · · · + pr xr
1 s.t.





σ0 σ1 . . . σr
σ1 σr+1
...
...
σd−r . . . σd−1 σd










p0
p1
...
pr





= 0
and of the form p = c r
k=1(βkx0 − αkx1) with (αk : βk) distinct.

Sparse interpolation
Given a black-box polynomial function f (x)
ﬁnd what are the terms inside from output values.

Find r ∈ N, ωi ∈ C, αi ∈ N such that f (x) = r
i=1 ωi xαi .

Choose ϕ ∈ C
Compute the sequence of terms σ0 = f (1), . . . , σ2r−1 = f (ϕ2r−1);
Construct the matrix H = [σi+j ] and its kernel p = [p0, . . . , pr ] s.t.





σ0 σ1 . . . σr
σ1 σr+1
...
...
σr−1 . . . σ2r−1 σ2r−1










p0
p1
...
pr





= 0
Compute the roots ξ1 = ϕα1 , . . . , ξr = ϕαr of p(x) := r
i=0 pi xi and
deduce the exponents αi = logϕ(ξi).
Deduce the weights W = [ωi ] by solving VΞ W = [σ0, . . . , σr−1]
where VΞ is the Vandermonde system of the roots ξ1, . . . , ξr .

Decoding
− − − − − − − − − − →
An algebraic code:
E = {c(f ) = [f (ξ1), . . . , f (ξm)] | f ∈ K[x]; deg(f ) ≤ d}.
Encoding messages using the dual code:
C = E⊥
= {c | c · [f (ξ1), . . . , f (ξl )] = 0 ∀f ∈ V = xa
⊂ F[x]}
Message received: r = m + e for m ∈ C where e = [ω1, . . . , ωm] is an
error with ωj = 0 for j = i1, . . . , ir and ωj = 0 otherwise.

Find the error e.

Berlekamp-Massey method (1969)
Compute the syndrome σk = c(xk) · r = c(xk) · e = r
j=1 ωij
ξk
ij
.
Compute the matrix





σ0 σ1 . . . σr
σ1 σr+1
...
...
σr−1 . . . σ2r−1 σ2r−1










p0
p1
...
pr





= 0
and its kernel p = [p0, . . . , pr ].
Compute the roots of the error locator polynomial
p(x) = r
i=0 pi xi = pr
r
j=1(x − ξij
).
Deduce the errors ωij
.

Simultaneous decomposition
Simultaneous decomposition problem
Given symmetric tensors ψ1, . . . , ψm of order d1, . . . , dm, ﬁnd a
simultaneous decomposition of the form
ψl =
r
i=1
ωl,i (ξi,0x0 + ξi,1x1 + · · · + ξi,nxn)dl
where ξi = (ξi,0, . . . , ξi,n) span distinct lines in K
n+1
and ωl,i ∈ K for
l = 1, . . . , m.

Proposition (One dimensional decomposition)
Let ψl = dl
i=0 σ1,i
dl
i xdl −i
0 xi
1 ∈ K[x0, x1]dl
for l = 1, . . . , m.
If there exists a polynomial p(x0, x1) := p0xr
0 + p1xr−1
0 x1 + · · · + pr xr
1 s.t.

















σ1,0 σ1,1 . . . σ1,r
σ1,1 σ1,r+1
...
...
σ1,d1−r . . . σ1,d1−1 σ1,d1
...
...
σm,0 σm,1 . . . σm,r
σm,1 σr+1
...
...
σm,dm−r . . . σm,dm−1 σm,dm






















p0
p1
...
pr





= 0
of the form p = c r
k=1(βkx0 − αkx1) with [αk : βk] distinct, then
ψl =
r
i=1
ωi,l (αl x0 + βl x1)dl
for ωi,l ∈ K and l = 1, . . . , m.

Duality
2 Duality
3 Artinian algebra

Duality
Sequences, series, duality (1D)
Sequences: σ = (σk)k∈N ∈ KN indexed by k ∈ N.
Formal power series:
σ(y) = ∞
k=0 σk
yk
k! ∈ K[[y]] σ(z) = ∞
k=0 σk zk ∈ K[[z]]
Linear functionals: K[x]∗ = {Λ : K[x] → K linear}.
Example:
p → coeﬃcient of xi in p = 1
i! ∂i (p)(0)
eζ : p → p(ζ).

Duality
Series as linear functionals: For σ(y) = ∞
k=0 σk
yk
k! ∈ K[[y]] or
σ(z) = ∞
k=0 σk zk ∈ K[[z]],
σ : p =
k
pkxk
→ σ|p =
k
σkpk
(yk
k! ) (resp. (zk)) is the dual basis of the monomial basis (xk)k∈N.
Example:
eζ(y) = ∞
k=0 ζk yk
k! = eζy ∈ K[[y]] eζ(z) = ∞
k=0 ζk zk = 1
1−ζz ∈ K[[z]]
Structure of K[x]-module: p Λ : q → Λ(p q).
x σ(y) =
∞
k=1 σk
yk−1
(k−1)! = ∂(σ(y))
p(x) σ(y) = p(∂)(σ(y)) p(x) σ(z) = π+(p(z−1
)(σ(z)))

Duality
Sequences, series, duality (nD)
Multi-index sequences: σ = (σα)α∈Nn ∈ KNn
indexed by
α = (α1, . . . , αn) ∈ Nn.
Taylor series:
σ(y) = α∈Nn σα
yα
α! ∈ K[[y1, . . . , yn]] σ(z) = α∈Nn σαzα
∈ K[[z1, . . . , zn]]
where α! = αi ! for α = (α1, . . . , αn) ∈ N.
Linear functionals: σ ∈ R∗ = {σ : R → K, linear}
σ : p =
α
pαxα
→ σ|p =
α
σαpα
The coeﬃcients σ|xα = σα ∈ K, α ∈ Nn are called the moments of σ.
Structure of R-module: ∀p ∈ R, σ ∈ R∗, p σ : q → σ|p q :
p σ = p(∂1, . . . , ∂n)(σ)(y) p σ = π+(p(z−1
1 , . . . , z−1
n )σ(z))

Duality
Symmetric tensor and apolarity
Apolar product: For f = |α|=d fα
d
α xα
, g = |α|=d gα
d
α xα
∈ K[x]d ,
f , g d =
|α|=d
fα gα
d
α
.
Property: f , (ξ0x0 + · · · + ξnxn)d = f (ξ0, . . . , ξn)
Duality: For ψ ∈ Sd , we deﬁne ψ∗ ∈ S∗
d = HomK(Sd , K) as
ψ∗ : Sd → K
p → ψ, p d
Example: ((ξ0x0 + · · · + ξnxn)d )∗ = eξ : p ∈ Sd → p(ξ) (evaluation at ξ)
Dual symmetric tensor decomposition problem:
Given ψ∗
∈ S∗
d , ﬁnd a decomposition of the form ψ∗
=
r
i=1 ωi eξi
where
ξi = (ξi,0, ξi,1, . . . , ξi,n) span distinct lines in K
n+1
, ωi ∈ K (ωi = 0).

Duality
Symmetric tensors and secants
The evaluation eξ ∈ S∗
d at ξ ∈ K
n+1
represented by the vector (ξα)|α|=d
deﬁnes a point of the Veronese variety Vn
d ⊂ P(S∗
d ).
ψ∗ = r
i=1 ωi eξi
iﬀ the corresponding point [ψ∗] in P(S∗
d ) is in the linear
span of the evaluations [eξi
] ∈ Vn
d .
Let So
r (Vn
d ) = {[ψ∗] ∈ P(S∗
d ) | ψ∗ = r
i=1 ωi ei with [ei ] ∈ Vn
d , ωi ∈ K}.
The closure Sr (Vn
d ) = So
r (Vn
d ) is the rth-secant of Vn
d .

Duality
Dehomogeneization (apart´e)
S = K[x0, . . . , xn] R = K[x1, . . . , xn]
ι0 : p(x0, . . . , xn) → p(1, x1, . . . , xn)
x
deg(p)
0 p(x1
x0
, . . . , xn
x0
) → p(x1, . . . , xn) : h0
Dual action
h∗
0 : σ ∈ S∗
d → σ ◦ h0 ∈ R∗
≤d
ι∗
0 : σ ∈ R∗
≤d → σ ◦ ι ∈ S∗
d
ι∗
0(σ(y) + O(y)d
) = [e0 σ(y)]d
where e0 = i
yk
0
k! .
For I ⊂ R, let [e0 I⊥]∗ be the vector space of homogeneous components of
e0 σ(y) for σ ∈ I⊥ ⊂ R∗, then
[e0 I⊥
]∗ = (J : x∗
0 )⊥
for any J such that ι0(J) = I (e.g. J = (Ih0 )).

Duality
Inverse systems
For I an ideal in R = K[x],
I⊥
= {σ ∈ R∗
| ∀p ∈ I, σ|p = 0}.
In K[[y]], I⊥ is stable by derivations with respect to yi .
In K[[z]], I⊥ is stable by “division” by variables zi .
Inverse system generated by ω1, . . . , ωr ∈ K[y]
ω1, . . . , ωr = ∂α
y (ωi ), α ∈ Nn
resp. π+(z−α
ωi(z)), α ∈ Nn
Example: I = (x2
1 , x2
2 ) ⊂ K[x1, x2]
I⊥
= 1, y1, y2, y1y2 = y1y2 resp. 1, z1, z2, z1z2 = z1z2
Dual of quotient algebra: for A = R/I, A∗ = I⊥.

Duality
Hankel operators
Hankel operator: For σ = (σ1, . . . , σm) ∈ (R∗)m,
Hσ : R → (R∗
)m
p → (p σ1, . . . , p σm)
σ is the symbol of Hσ.
Truncated Hankel operator: V , W1, . . . , Wm ⊂ R,
HW ,V
σ : p ∈ V → ((p σi )|Wi
)
Example: V = xα, α ∈ A = xA , W = xβ, β ∈ B = xB ⊂ R,
σ ∈ R∗,
HA,B
σ = [ σ|xα
xβ
]α∈A,β∈B = [σα+β]α∈A,β∈B.
Ideal:
Iσ = ker Hσ = {p ∈ K[x] | p σ = 0},
= {p =
α
pαxα
| ∀β ∈ Nn
α
pασα+β = 0}
Linear recurrence relations on the sequence σ = (σα)α∈Nn .
Quotient algebra: Aσ = R/Iσ

Studied case: dim Aσ < ∞

Artinian algebra
2 Duality
3 Artinian algebra

Artinian algebra
Structure of an Artinian algebra A
Deﬁnition: A = K[x]/I is Artinian if dimK A < ∞.
Hilbert nullstellensatz: A = K[x]/I Artinian ⇔ VK(I) = {ξ1, . . . , ξr } is
ﬁnite.
Assuming K = K is algebraically closed, we have
I = Q1 ∩· · ·∩Qr where Qi is mξi
-primary where VK(I) = {ξ1, . . . , ξr }.
A = K[x]/I = A1 ⊕ · · · ⊕ Ar , with
Ai = ui A ∼ K[x1, . . . , xn]/Qi ,
u2
i = ui , ui uj = 0 if i = j, u1 + · · · + ur = 1.
dim R/Qi = µi is the multiplicity of ξi .

Artinian algebra
Structure of the dual A∗
Sparse series:
PolExp = σ(y) =
r
i=1
ωi (y) eξi
(y) | ωi (y) ∈ K[y],
where eξi
(y) = ey·ξi = ey1ξ1,i +···+ynξn,i with ξi,j ∈ K.
Inverse system generated by ω1, . . . , ωr ∈ K[y]
ω1, . . . , ωr = ∂α
y (ωi ), α ∈ Nn
Theorem
For K = K algebraically closed,
A∗
= ⊕r
i=1 Di eξi
(y) ⊂ PolExp
VK(I) = {ξ1, . . . , ξr }
Di = ωi,1, . . . , ωi,li
with ωi,j ∈ K[y], Q⊥
i = Di eξ where I = Q1 ∩ · · · ∩ Qr
µ(ωi,1, . . . , ωi,li
) := dimK(Di ) = µi multiplicity of ξi .

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