PreliminariesProperties of tensor rank
Open problems
The role of tensor rank in the complexity analysisof bilinear forms
Dario A. Bini
Dipartimento di Matematica, Universita di Pisawww.dm.unipi.it/˜bini
ICIAM07, Zurich, 16-20 July 2007
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
1 PreliminariesTensorsTensor rankBilinear forms
2 Properties of tensor rankBorder rankLower bounds
3 Open problemsThe direct sum conjectureSome tensors of unknown rank
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
TensorsTensor rankBilinear forms
Tensors
Let F be a number field, say, R, C; tensors of the kind
A = (ai1,i2,...,ih) ∈ Fn1×n2×···×nh ,
that is, h-way arrays, are encountered in many problems of verydifferent nature [Comon 2001], [Comon, Golub, Lim, Mourrain,2006], [De Silva, Lim 2006]
Blind source separation
High order factor analysis
Independent component analysis
Candecomp/Parafac model
Complexity analysis
Psycometric, Chemometric, Economy,...
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
TensorsTensor rankBilinear forms
Tensors
It is surprising that little interplay occurred among thesedifferent research areas
Some properties have been rediscovered in different contexts
Apparently, results obtained in one field have not migrated tothe other fields
In this talk I wish to provide an overview of the main resultsconcerning tensors obtained in the research field of computationalcomplexity (starting from 1969) with the aim of
creating a synergic exchange of information between theseresearch areas
presenting problems which might be solved with the morerecent tools
presenting old results that might be adapted and extended tothe new needs
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
TensorsTensor rankBilinear forms
Tensors
It is surprising that little interplay occurred among thesedifferent research areas
Some properties have been rediscovered in different contexts
Apparently, results obtained in one field have not migrated tothe other fields
In this talk I wish to provide an overview of the main resultsconcerning tensors obtained in the research field of computationalcomplexity (starting from 1969) with the aim of
creating a synergic exchange of information between theseresearch areas
presenting problems which might be solved with the morerecent tools
presenting old results that might be adapted and extended tothe new needs
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
TensorsTensor rankBilinear forms
Tensor rank
Definition (Hitchcock 1927)
A tensor T = (ti1,...,ih) has rank 1 if there exist vectors
u(k) = (u(k)i ) ∈ Fnk , k = 1 : h such that ti1,...,ih = u
(1)i1
u(2)i2· · · u(h)
ih,
T = u(1) ◦ u(2) ◦ · · · ◦ u(h)
Definition (Hitchcock 1927)
The tensor rank rk(A) of A = (ai1,...,ih) is the minimum number rof rank-1 tensors Ti ∈ Fn1×···×nh such that
A = T1 + T2 + · · ·+ Tr canonical decomposition
For matrices, the tensor rank coincides with the customary rank
For simplicity of notation we restrict ourselves to the case h = 3,i.e., to three-way arrays.
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
TensorsTensor rankBilinear forms
Some remarks
For A ∈ Fm×n×p a canonical decomposition
A = u(1) ◦ v(1) ◦w(1) + u(2) ◦ v(2) ◦w(2) + · · ·+ u(r) ◦ v(r) ◦w(r)
is defined by three matrices
U = (ui ,j) ∈ Fm×r , V = (vi ,j) ∈ Fn×r , W = (wi ,j) ∈ Fp×r ,
whose columns are the vectors u(i), v(i), w(i), i = 1 : r ,respectively.
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
TensorsTensor rankBilinear forms
Some remarks
A tensor A = (ai ,j ,k) ∈ Fm×n×p, can be represented by means ofthe set of the 3-slabs Ak = (ai ,j ,k)i ,j ∈ Fm×n of A or by a singlematrix of variables
A =
p∑k=1
skAk
A =
[[1 00 −1
];
[0 11 0
]]↔
[s1 s2s2 −s1
]
This suggests a different point of view for tensor rank:
rk(A) is the minimum set of rank one matrices which span the linearspace generated by the 3-slabs A1, . . . ,Ap [Gastinel 71, Fiduccia 72].
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
TensorsTensor rankBilinear forms
Some remarks
A tensor A = (ai ,j ,k) ∈ Fm×n×p, can be represented by means ofthe set of the 3-slabs Ak = (ai ,j ,k)i ,j ∈ Fm×n of A or by a singlematrix of variables
A =
p∑k=1
skAk
A =
[[1 00 −1
];
[0 11 0
]]↔
[s1 s2s2 −s1
]This suggests a different point of view for tensor rank:
rk(A) is the minimum set of rank one matrices which span the linearspace generated by the 3-slabs A1, . . . ,Ap [Gastinel 71, Fiduccia 72].
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
TensorsTensor rankBilinear forms
Bilinear forms
Problem: Given matrices Ak = (ai ,j ,k)i=1:m,j=1:n, k = 1 : pcompute the set of bilinear forms
fk(x, y) = xTAky, k = 1 : p
with the minimum number of nonscalar multiplications with no useof commutativity (noncommutative bilinear complexity)
A nonscalar multiplication is a multiplication of the kind
s = (m∑
i=1
αixi )(n∑
j=1
βjyj), αi , βj ∈ F
Remark: The set of bilinear forms is uniquely determined by thetensor A = (ai ,j ,k).
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
TensorsTensor rankBilinear forms
Bilinear forms
A canonical decomposition of the tensor A associated with the setof bilinear forms provides an algorithm of complexity r . In fact
A =r∑
`=1
u(`) ◦ v(`) ◦w(`) → Ak =r∑
`=1
wk,`u(`) ◦ v(`)
→ xTAky =r∑
`=1
wk,`(xTu(`))(v(`)Ty)
Theorem (Strassen 1975)
The noncommutative bilinear complexity of a set of bilinear formsfk(x, y) = xTAky, k = 1 : p, Ak = (ai ,j ,k), is given by the tensorrank of the associated tensor A = (ai ,j ,k).
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
TensorsTensor rankBilinear forms
An example
Multiplication of complex numbers
(x1 + ix2)(y1 + iy2) = (x1y1 − x2y2) + i(x1y2 + x2y1) = f1 + if2
Apparently, 4 multiplications are needed
Tensor: A =
[[1 00 −1
];
[0 11 0
]]The tensor rank is at most 3:[
1 00 −1
]=
[1 00 0
]−
[0 00 1
][
0 11 0
]=
[1 11 1
]−
[1 00 0
]−
[0 00 1
]Algorithm:
s1 = (x1 + x2)(y1 + y2),s2 = x1y1,s3 = x2y2
f1 = s2 − s3,f2 = s1 − s2 − s3
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
TensorsTensor rankBilinear forms
An example
Multiplication of complex numbers
(x1 + ix2)(y1 + iy2) = (x1y1 − x2y2) + i(x1y2 + x2y1) = f1 + if2
Apparently, 4 multiplications are needed
Tensor: A =
[[1 00 −1
];
[0 11 0
]]
The tensor rank is at most 3:[1 00 −1
]=
[1 00 0
]−
[0 00 1
][
0 11 0
]=
[1 11 1
]−
[1 00 0
]−
[0 00 1
]Algorithm:
s1 = (x1 + x2)(y1 + y2),s2 = x1y1,s3 = x2y2
f1 = s2 − s3,f2 = s1 − s2 − s3
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
TensorsTensor rankBilinear forms
An example
Multiplication of complex numbers
(x1 + ix2)(y1 + iy2) = (x1y1 − x2y2) + i(x1y2 + x2y1) = f1 + if2
Apparently, 4 multiplications are needed
Tensor: A =
[[1 00 −1
];
[0 11 0
]]The tensor rank is at most 3:[
1 00 −1
]=
[1 00 0
]−
[0 00 1
][
0 11 0
]=
[1 11 1
]−
[1 00 0
]−
[0 00 1
]Algorithm:
s1 = (x1 + x2)(y1 + y2),s2 = x1y1,s3 = x2y2
f1 = s2 − s3,f2 = s1 − s2 − s3
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
Tensor rank
Main problem: For a given tensor, compute its rank and a canonicaldecomposition.
Two ways of attacking the problem
looking for lower bounds to the tensor ranklooking for upper bounds to the tensor rank
Things get more complicate: unlike the case of m × n matrices
The rank depends on the ground field FHigh rank tensors can be approximated by low rank tensors
Rational algorithms for tensor rank, like Gaussian elimination, can-not exist
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
Topological properties of tensors: Border Rank
For m × n matrices, any sequence {Ak} of m × n matrices of rankr with limit A = limk Ak is such that rank(A) ≤ r .
For higher order tensor this property does not hold anymore
Example (Bini, Capovani, Lotti, Romani 1980)
The tensor
A =
[[1 00 1
];
[0 10 0
]]has rank 3. The tensor
Aε =
[[1 0ε 1
];
[0 10 0
]]has rank 2 for any ε 6= 0
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
Topological properties of tensors: Border Rank
For m × n matrices, any sequence {Ak} of m × n matrices of rankr with limit A = limk Ak is such that rank(A) ≤ r .
For higher order tensor this property does not hold anymore
Example (Bini, Capovani, Lotti, Romani 1980)
The tensor
A =
[[1 00 1
];
[0 10 0
]]has rank 3. The tensor
Aε =
[[1 0ε 1
];
[0 10 0
]]has rank 2 for any ε 6= 0
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
Canonical decomposition
[1 0ε 1
]=
[1 ε−1
ε 1
]− ε−1
[0 10 0
][
0 10 0
]=
[0 10 0
]
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
Border rank
Definition (Bini, Capovani, Lotti, Romani 1980)
The border rank of A ∈ Fm×n×p (F = R, C) is
brk(A) = min{r : ∀ε > 0 ∃E ∈ Fm×n×p : ||E|| < ε, rk(A+E) = r}
where || · || is any norm
Some properties:
brk(A) ≤ rk(A)
brk(A) is the minimum number of nonscalar multiplicationssufficient to approximate the set of bilinear forms associatedwith A with arbitrarily small nonzero error
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
More on rank and border rank
Let Aε = A+ Eε be such that
rk(Aε) = brk(A)the entries of Eε are polynomials of degree d
Then
rk(A) ≤ (d + 1)brk(A)
Proof:
Write d + 1 copies of the canonical decomposition of lengthrk(Aε) obtained with (d + 1) pairwise different values of εTake linear combinations of these decompositions withcoefficients γj , j = 1 : d + 1 in order to kill the terms in εi ,i = 1 : d and to have
∑γj = 1
Obtain a decomposition of length (d + 1)brk(A) with no error
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
More on rank and border rank
Let Aε = A+ Eε be such that
rk(Aε) = brk(A)the entries of Eε are polynomials of degree d
Then
rk(A) ≤ (d + 1)brk(A)
Proof:
Write d + 1 copies of the canonical decomposition of lengthrk(Aε) obtained with (d + 1) pairwise different values of ε
Take linear combinations of these decompositions withcoefficients γj , j = 1 : d + 1 in order to kill the terms in εi ,i = 1 : d and to have
∑γj = 1
Obtain a decomposition of length (d + 1)brk(A) with no error
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
More on rank and border rank
Let Aε = A+ Eε be such that
rk(Aε) = brk(A)the entries of Eε are polynomials of degree d
Then
rk(A) ≤ (d + 1)brk(A)
Proof:
Write d + 1 copies of the canonical decomposition of lengthrk(Aε) obtained with (d + 1) pairwise different values of εTake linear combinations of these decompositions withcoefficients γj , j = 1 : d + 1 in order to kill the terms in εi ,i = 1 : d and to have
∑γj = 1
Obtain a decomposition of length (d + 1)brk(A) with no error
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
More on rank and border rank
Let Aε = A+ Eε be such that
rk(Aε) = brk(A)the entries of Eε are polynomials of degree d
Then
rk(A) ≤ (d + 1)brk(A)
Proof:
Write d + 1 copies of the canonical decomposition of lengthrk(Aε) obtained with (d + 1) pairwise different values of εTake linear combinations of these decompositions withcoefficients γj , j = 1 : d + 1 in order to kill the terms in εi ,i = 1 : d and to have
∑γj = 1
Obtain a decomposition of length (d + 1)brk(A) with no error
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
Lower bounds and linear algebra
Simple criteria for providing lower bounds on tensor rank andborder rank can be givenTrivial bounds
brk(A) ≥ dim(span(A1, . . . ,Ap))
Similar inequalities are valid w.r.t. the other coordinates
Assume for simplicity p = dim(span(A1, . . . ,Ap))
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
Lower bounds and linear algebra
Let U,V ,W be the matrices defining a canonical factorization ofA of length rk(A)
A =
rk(A)∑`=1
u(`) ◦ v(`) ◦w(`)
Assume w.l.o.g. that the first p columns of W are linearlyindependent, that is
W =[
W1 W2
], W1 ∈ Fp×p, det W1 6= 0
Let A′k =
∑pj=1 w
(−1)k,j Aj , define
A′ = [A′1,A
′2, . . . ,A
′p] = A •3 W−1
1 ,
i.e., choose a different basis to represent the space spanned by theslabs A1, . . . ,Ap
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
Lower bounds and linear algebra
Let U,V ,W be the matrices defining a canonical factorization ofA of length rk(A)
A =
rk(A)∑`=1
u(`) ◦ v(`) ◦w(`)
Assume w.l.o.g. that the first p columns of W are linearlyindependent, that is
W =[
W1 W2
], W1 ∈ Fp×p, det W1 6= 0
Let A′k =
∑pj=1 w
(−1)k,j Aj , define
A′ = [A′1,A
′2, . . . ,A
′p] = A •3 W−1
1 ,
i.e., choose a different basis to represent the space spanned by theslabs A1, . . . ,Ap
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
Lower bounds and linear algebra
Let U,V ,W be the matrices defining a canonical factorization ofA of length rk(A)
A =
rk(A)∑`=1
u(`) ◦ v(`) ◦w(`)
Assume w.l.o.g. that the first p columns of W are linearlyindependent, that is
W =[
W1 W2
], W1 ∈ Fp×p, det W1 6= 0
Let A′k =
∑pj=1 w
(−1)k,j Aj , define
A′ = [A′1,A
′2, . . . ,A
′p] = A •3 W−1
1 ,
i.e., choose a different basis to represent the space spanned by theslabs A1, . . . ,Ap
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
Evidently, A′ has the same rank of A and a canonicaldecomposition is given by U ′ = U,V ′ = V ,W ′ = W−1
1 W , where
W ′ =[
I W−11 W2
]=
1 0 0 0 ∗ ∗ ∗0 1 0 0 ∗ ∗ ∗0 0 1 0 ∗ ∗ ∗0 0 0 1 ∗ ∗ ∗
Remark
Any subtensor A formed by k 3-slabs [A′σ1
, . . . ,A′σk
] is such that
rk(A) ≤ rk(A)− (p − k)
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
More generally,
Theorem
rk(A) ≥ maxT
minS
(p − k + rk(A •3 TS))
T: k × p submatrix of Ip, S: p × p nonsingular
Corollary
If for any basis of span(A1, . . . ,Ap) there exists a matrix of rank qthen
rk(A) ≥ p + q − 1
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
More generally,
Theorem
rk(A) ≥ maxT
minS
(p − k + rk(A •3 TS))
T: k × p submatrix of Ip, S: p × p nonsingular
Corollary
If for any basis of span(A1, . . . ,Ap) there exists a matrix of rank qthen
rk(A) ≥ p + q − 1
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
Higher order generalization:
Corollary
Let A = [A1, . . . ,Anh] ∈ Fn1×···×nh , Ak ∈ Fn1×···×nh−1 be such
that nh = dim(span(A1, . . . ,Anh)). If for any basis of
span(A1, . . . ,Anh) there exists a tensor in the basis of rank q then
rk(A) ≥ nh + q − 1
Example: For
A =
[[1 00 1
];
[0 10 0
]]any basis must contain a nonsingular matrix, therefore
rk(A) ≥ 2 + 2− 1 = 3
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
Higher order generalization:
Corollary
Let A = [A1, . . . ,Anh] ∈ Fn1×···×nh , Ak ∈ Fn1×···×nh−1 be such
that nh = dim(span(A1, . . . ,Anh)). If for any basis of
span(A1, . . . ,Anh) there exists a tensor in the basis of rank q then
rk(A) ≥ nh + q − 1
Example: For
A =
[[1 00 1
];
[0 10 0
]]any basis must contain a nonsingular matrix, therefore
rk(A) ≥ 2 + 2− 1 = 3
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
Lower bounds to border rank
Unfortunately the same technique cannot be applied to border rank
W1 may be singular in the limit as ε → 0 so that E •3 W−11 may not be
infinitesimal anymore
Remedy: compute the QR factorization of W and multiply W by QT
which has Euclidean norm 1 for any ε. One has
W ′ = QTW =
∗ ∗ ∗ ∗ ∗ ∗ ∗0 ∗ ∗ ∗ ∗ ∗ ∗0 0 ∗ ∗ ∗ ∗ ∗0 0 0 ∗ ∗ ∗ ∗
Remark
There exists an m × n × k subtensor A such that
brk(A) ≤ brk(A)− (p − k)
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
Theorem
brk(A) ≥ minT
minS
(p − k + brk(A •3 TS))
T: k × p submatrix of Ip, S: p × p nonsingular
Corollary
If any basis of the linear space spanned by A1, . . . ,Ap is made upby matrices of rank at most q then brk(A) ≥ p + q − 1
Corollary (Generalization to higher dimension)
Let A = [A1, . . . ,Anh]. If any basis of the linear space spanned by
A1, . . . ,Anhis made up by tensors of rank at most q then
brk(A) ≥ nh + q − 1
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
Theorem
brk(A) ≥ minT
minS
(p − k + brk(A •3 TS))
T: k × p submatrix of Ip, S: p × p nonsingular
Corollary
If any basis of the linear space spanned by A1, . . . ,Ap is made upby matrices of rank at most q then brk(A) ≥ p + q − 1
Corollary (Generalization to higher dimension)
Let A = [A1, . . . ,Anh]. If any basis of the linear space spanned by
A1, . . . ,Anhis made up by tensors of rank at most q then
brk(A) ≥ nh + q − 1
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
Theorem
brk(A) ≥ minT
minS
(p − k + brk(A •3 TS))
T: k × p submatrix of Ip, S: p × p nonsingular
Corollary
If any basis of the linear space spanned by A1, . . . ,Ap is made upby matrices of rank at most q then brk(A) ≥ p + q − 1
Corollary (Generalization to higher dimension)
Let A = [A1, . . . ,Anh]. If any basis of the linear space spanned by
A1, . . . ,Anhis made up by tensors of rank at most q then
brk(A) ≥ nh + q − 1
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
Example
The tensor A ∈ Fn2×n2×n2whose 3-slabs span the linear space
I ⊗ S =
S. . .
S
, S =
s1,1 . . . s1,n...
. . ....
sn,1 . . . sn,n
is associated with n × n matrix multiplication.
All the matrices in any basis of the linear space have rank at leastn. Therefore
brk(A) ≥ n2 + n − 1
The bound can be improved to
brk(A) ≥ n2 + 2n − 2
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
Example
The tensor A ∈ Fn2×n2×n2whose 3-slabs span the linear space
I ⊗ S =
S. . .
S
, S =
s1,1 . . . s1,n...
. . ....
sn,1 . . . sn,n
is associated with n × n matrix multiplication.
All the matrices in any basis of the linear space have rank at leastn. Therefore
brk(A) ≥ n2 + n − 1
The bound can be improved to
brk(A) ≥ n2 + 2n − 2
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
Example
The tensor A ∈ F4×4×3 whose 3-slabs span the linear space s1 s2 0 0s3 s4 0 0
0 0 s1 s2
is associated with the multiplication of a 2× 2 triangular matrixand a full 2× 2 matrix
For any basis of the linear space there exist two matrices whichform a tensor of rank at least 4.
rk(A) ≥ 4− 2 + 4 = 6
Since there exists a canonical approximate decomposition of length5 of A then brk(A) ≤ 5 < 6 = rk(A)
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
Border rankLower bounds
Example
The tensor A ∈ F4×4×3 whose 3-slabs span the linear space s1 s2 0 0s3 s4 0 0
0 0 s1 s2
is associated with the multiplication of a 2× 2 triangular matrixand a full 2× 2 matrix
For any basis of the linear space there exist two matrices whichform a tensor of rank at least 4.
rk(A) ≥ 4− 2 + 4 = 6
Since there exists a canonical approximate decomposition of length5 of A then brk(A) ≤ 5 < 6 = rk(A)
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
The direct sum conjectureSome tensors of unknown rank
The direct sum conjecture
Let A ∈ Fm×n×p, B ∈ Fm′×n′×p′ . Consider the direct sum of Aand B
C = A⊕ B ∈ F(m+m′)×(n+n′)×(p+p′)
Direct sum conjecture, Strassen 1973
rk(C) = rk(A) + rk(B)
The complexity of two disjoint sets of bilinear forms is the sum ofthe complexities of each set.
Fact
There are cases where brk(C) < brk(A) + brk(B)
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
The direct sum conjectureSome tensors of unknown rank
The direct sum conjecture
Let A ∈ Fm×n×p, B ∈ Fm′×n′×p′ . Consider the direct sum of Aand B
C = A⊕ B ∈ F(m+m′)×(n+n′)×(p+p′)
Direct sum conjecture, Strassen 1973
rk(C) = rk(A) + rk(B)
The complexity of two disjoint sets of bilinear forms is the sum ofthe complexities of each set.
Fact
There are cases where brk(C) < brk(A) + brk(B)
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
The direct sum conjectureSome tensors of unknown rank
Example (A. Schoenhage 81)
brk(A) = 10 < 9 + 4
A =
s11 s12 s13s21 s22 s23s21 s22 s23
tt
tt
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
The direct sum conjectureSome tensors of unknown rank
. . .
. 1 1 ε ε
. 1 1 ε ε
ε ε2
ε ε2
ε ε2
ε ε2
1 1 11 . . −ε −ε1 . . −ε −ε
−ε−ε
−ε−ε
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
The direct sum conjectureSome tensors of unknown rank
Matrix multiplication and open problems
2× 2 matrix product:
A = I2 ⊗[
s11 s12s21 s22
]=
s11 s12s21 s22
s11 s12
s21 s22
rk(A) = brk(A) = 7 [Strassen 69, Landsberg 05]
3× 3 matrix product A = I3 ⊗
s11 s12 s13
s21 s22 s23s31 s32 s33
19 ≤ rk(A) ≤ 23 [Laderman 76], [Blaser 03]
13 ≤ brk(A) ≤ 22 [Schonhage 81]
Dario A. Bini The role of tensor rank
PreliminariesProperties of tensor rank
Open problems
The direct sum conjectureSome tensors of unknown rank
Matrix multiplication and open problems
4× 4 matrix product A = I4 ⊗
s11 s12 s13 s14s21 s22 s23 s24s31 s32 s33 s34s41 s42 s43 s44
34 ≤ rk(A) ≤ 49
22 ≤ brk(A) ≤ 49
Dario A. Bini The role of tensor rank
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