Kruskal Tensor: sum of outer or Kronecker products?
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I'm reading an ACL 2014 paper: Lei, Tao, et al. "Low-Rank Tensors for Scoring Dependency Structures.", ACL 2014.
It defines the Kruskal form of a tensor as a sum of Kronecker products:
However, http://www.ima.umn.edu/industrial/2006-2007/kolda/kolda.pdf defines the Kruskal form of a tensor as a sum of outer products:
So should the Kruskal form of a tensor be defined as a sum of outer or Kronecker products?
tensor-products
asked May 16 '14 at 16:23
remindremind
78111
$endgroup$
add a comment |
$begingroup$
I'm reading an ACL 2014 paper: Lei, Tao, et al. "Low-Rank Tensors for Scoring Dependency Structures.", ACL 2014.
It defines the Kruskal form of a tensor as a sum of Kronecker products:
However, http://www.ima.umn.edu/industrial/2006-2007/kolda/kolda.pdf defines the Kruskal form of a tensor as a sum of outer products:
So should the Kruskal form of a tensor be defined as a sum of outer or Kronecker products?
tensor-products
asked May 16 '14 at 16:23
remindremind
78111
$endgroup$
$begingroup$
They're the same thing.
$endgroup$
– Qiaochu Yuan
May 16 '14 at 16:25
$begingroup$
math.stackexchange.com/q/796709/150385 says a Kronecker product of 2 vectors outputs a vector, while an outer product of 2 vectors outputs returns a matrix. Also, in the slides I pointed to the question (ima.umn.edu/industrial/2006-2007/kolda/kolda.pdf) they use a different notation for outer and for Kronecker.
$endgroup$
– remind
May 16 '14 at 16:26
$begingroup$
Okay, fine, they're slightly different. As long as you're doing operations on vectors and matrices the difference is not very serious.
$endgroup$
– Qiaochu Yuan
May 16 '14 at 16:31
$begingroup$
Thanks, so should the ACL paper have used outer product instead of Kronecker? No intent to be picky, it's just to have everything clear in my head :)
$endgroup$
– remind
May 16 '14 at 16:36
$begingroup$
The outer product is a special case of a Kronecker product, which should make things clear in this context. See en.wikipedia.org/wiki/Kronecker_product.
$endgroup$
– jhclark
Jul 16 '14 at 4:38
add a comment |
$begingroup$
I'm reading an ACL 2014 paper: Lei, Tao, et al. "Low-Rank Tensors for Scoring Dependency Structures.", ACL 2014.
It defines the Kruskal form of a tensor as a sum of Kronecker products:
However, http://www.ima.umn.edu/industrial/2006-2007/kolda/kolda.pdf defines the Kruskal form of a tensor as a sum of outer products:
So should the Kruskal form of a tensor be defined as a sum of outer or Kronecker products?
tensor-products
asked May 16 '14 at 16:23
remindremind
78111
$endgroup$
I'm reading an ACL 2014 paper: Lei, Tao, et al. "Low-Rank Tensors for Scoring Dependency Structures.", ACL 2014.
It defines the Kruskal form of a tensor as a sum of Kronecker products:
However, http://www.ima.umn.edu/industrial/2006-2007/kolda/kolda.pdf defines the Kruskal form of a tensor as a sum of outer products:
So should the Kruskal form of a tensor be defined as a sum of outer or Kronecker products?
tensor-products
tensor-products
asked May 16 '14 at 16:23
remindremind
78111
asked May 16 '14 at 16:23
remindremind
78111
asked May 16 '14 at 16:23
remindremind
78111
asked May 16 '14 at 16:23
remindremind
78111
asked May 16 '14 at 16:23
remindremind
78111
78111
$begingroup$
They're the same thing.
$endgroup$
– Qiaochu Yuan
May 16 '14 at 16:25
$begingroup$
math.stackexchange.com/q/796709/150385 says a Kronecker product of 2 vectors outputs a vector, while an outer product of 2 vectors outputs returns a matrix. Also, in the slides I pointed to the question (ima.umn.edu/industrial/2006-2007/kolda/kolda.pdf) they use a different notation for outer and for Kronecker.
$endgroup$
– remind
May 16 '14 at 16:26
$begingroup$
Okay, fine, they're slightly different. As long as you're doing operations on vectors and matrices the difference is not very serious.
$endgroup$
– Qiaochu Yuan
May 16 '14 at 16:31
$begingroup$
Thanks, so should the ACL paper have used outer product instead of Kronecker? No intent to be picky, it's just to have everything clear in my head :)
$endgroup$
– remind
May 16 '14 at 16:36
$begingroup$
The outer product is a special case of a Kronecker product, which should make things clear in this context. See en.wikipedia.org/wiki/Kronecker_product.
$endgroup$
– jhclark
Jul 16 '14 at 4:38
add a comment |
$begingroup$
They're the same thing.
$endgroup$
– Qiaochu Yuan
May 16 '14 at 16:25
$begingroup$
math.stackexchange.com/q/796709/150385 says a Kronecker product of 2 vectors outputs a vector, while an outer product of 2 vectors outputs returns a matrix. Also, in the slides I pointed to the question (ima.umn.edu/industrial/2006-2007/kolda/kolda.pdf) they use a different notation for outer and for Kronecker.
$endgroup$
– remind
May 16 '14 at 16:26
$begingroup$
Okay, fine, they're slightly different. As long as you're doing operations on vectors and matrices the difference is not very serious.
$endgroup$
– Qiaochu Yuan
May 16 '14 at 16:31
$begingroup$
Thanks, so should the ACL paper have used outer product instead of Kronecker? No intent to be picky, it's just to have everything clear in my head :)
$endgroup$
– remind
May 16 '14 at 16:36
$begingroup$
The outer product is a special case of a Kronecker product, which should make things clear in this context. See en.wikipedia.org/wiki/Kronecker_product.
$endgroup$
– jhclark
Jul 16 '14 at 4:38
$begingroup$
They're the same thing.
$endgroup$
– Qiaochu Yuan
May 16 '14 at 16:25
$begingroup$
They're the same thing.
$endgroup$
– Qiaochu Yuan
May 16 '14 at 16:25
$begingroup$
math.stackexchange.com/q/796709/150385 says a Kronecker product of 2 vectors outputs a vector, while an outer product of 2 vectors outputs returns a matrix. Also, in the slides I pointed to the question (ima.umn.edu/industrial/2006-2007/kolda/kolda.pdf) they use a different notation for outer and for Kronecker.
$endgroup$
– remind
May 16 '14 at 16:26
$begingroup$
math.stackexchange.com/q/796709/150385 says a Kronecker product of 2 vectors outputs a vector, while an outer product of 2 vectors outputs returns a matrix. Also, in the slides I pointed to the question (ima.umn.edu/industrial/2006-2007/kolda/kolda.pdf) they use a different notation for outer and for Kronecker.
$endgroup$
– remind
May 16 '14 at 16:26
$begingroup$
Okay, fine, they're slightly different. As long as you're doing operations on vectors and matrices the difference is not very serious.
$endgroup$
– Qiaochu Yuan
May 16 '14 at 16:31
$begingroup$
Okay, fine, they're slightly different. As long as you're doing operations on vectors and matrices the difference is not very serious.
$endgroup$
– Qiaochu Yuan
May 16 '14 at 16:31
$begingroup$
Thanks, so should the ACL paper have used outer product instead of Kronecker? No intent to be picky, it's just to have everything clear in my head :)
$endgroup$
– remind
May 16 '14 at 16:36
$begingroup$
Thanks, so should the ACL paper have used outer product instead of Kronecker? No intent to be picky, it's just to have everything clear in my head :)
$endgroup$
– remind
May 16 '14 at 16:36
$begingroup$
The outer product is a special case of a Kronecker product, which should make things clear in this context. See en.wikipedia.org/wiki/Kronecker_product.
$endgroup$
– jhclark
Jul 16 '14 at 4:38
$begingroup$
The outer product is a special case of a Kronecker product, which should make things clear in this context. See en.wikipedia.org/wiki/Kronecker_product.
$endgroup$
– jhclark
Jul 16 '14 at 4:38
add a comment |
1 Answer
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The former is using $otimes$ as an outer product, not a kronecker product. However there is a nice relationship between outer products and kronecker products. That is, if we define tensor vectorization in reverse lexicographic ordering (so that we have no inconsistencies with regular matrix vectorization) and we denote the vector outer product using $circ$, then
$
vec big( circ_{i=1}^p mathbf{a}_{i} big)
=bigotimes_{i=p}^1 mathbf{a}_i
$, where $mathbf{a}_1,..., mathbf{a}_p $ are vectors of any size.
So you could define the kruskal tensor using the kronecker product, but only via its vectorization.
answered Aug 15 '18 at 15:44
Carlos LlosaCarlos Llosa
12
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$begingroup$
The former is using $otimes$ as an outer product, not a kronecker product. However there is a nice relationship between outer products and kronecker products. That is, if we define tensor vectorization in reverse lexicographic ordering (so that we have no inconsistencies with regular matrix vectorization) and we denote the vector outer product using $circ$, then
$
vec big( circ_{i=1}^p mathbf{a}_{i} big)
=bigotimes_{i=p}^1 mathbf{a}_i
$, where $mathbf{a}_1,..., mathbf{a}_p $ are vectors of any size.
So you could define the kruskal tensor using the kronecker product, but only via its vectorization.
answered Aug 15 '18 at 15:44
Carlos LlosaCarlos Llosa
12
$endgroup$
add a comment |
$begingroup$
The former is using $otimes$ as an outer product, not a kronecker product. However there is a nice relationship between outer products and kronecker products. That is, if we define tensor vectorization in reverse lexicographic ordering (so that we have no inconsistencies with regular matrix vectorization) and we denote the vector outer product using $circ$, then
$
vec big( circ_{i=1}^p mathbf{a}_{i} big)
=bigotimes_{i=p}^1 mathbf{a}_i
$, where $mathbf{a}_1,..., mathbf{a}_p $ are vectors of any size.
So you could define the kruskal tensor using the kronecker product, but only via its vectorization.
answered Aug 15 '18 at 15:44
Carlos LlosaCarlos Llosa
12
$endgroup$
add a comment |
$begingroup$
The former is using $otimes$ as an outer product, not a kronecker product. However there is a nice relationship between outer products and kronecker products. That is, if we define tensor vectorization in reverse lexicographic ordering (so that we have no inconsistencies with regular matrix vectorization) and we denote the vector outer product using $circ$, then
$
vec big( circ_{i=1}^p mathbf{a}_{i} big)
=bigotimes_{i=p}^1 mathbf{a}_i
$, where $mathbf{a}_1,..., mathbf{a}_p $ are vectors of any size.
So you could define the kruskal tensor using the kronecker product, but only via its vectorization.
answered Aug 15 '18 at 15:44
Carlos LlosaCarlos Llosa
12
$endgroup$
The former is using $otimes$ as an outer product, not a kronecker product. However there is a nice relationship between outer products and kronecker products. That is, if we define tensor vectorization in reverse lexicographic ordering (so that we have no inconsistencies with regular matrix vectorization) and we denote the vector outer product using $circ$, then
$
vec big( circ_{i=1}^p mathbf{a}_{i} big)
=bigotimes_{i=p}^1 mathbf{a}_i
$, where $mathbf{a}_1,..., mathbf{a}_p $ are vectors of any size.
So you could define the kruskal tensor using the kronecker product, but only via its vectorization.
answered Aug 15 '18 at 15:44
Carlos LlosaCarlos Llosa
12
edited Sep 25 '18 at 2:03
answered Aug 15 '18 at 15:44
Carlos LlosaCarlos Llosa
12
answered Aug 15 '18 at 15:44
Carlos LlosaCarlos Llosa
12
answered Aug 15 '18 at 15:44
Carlos LlosaCarlos Llosa
12
12
add a comment |
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$begingroup$
They're the same thing.
$endgroup$
– Qiaochu Yuan
May 16 '14 at 16:25
$begingroup$
math.stackexchange.com/q/796709/150385 says a Kronecker product of 2 vectors outputs a vector, while an outer product of 2 vectors outputs returns a matrix. Also, in the slides I pointed to the question (ima.umn.edu/industrial/2006-2007/kolda/kolda.pdf) they use a different notation for outer and for Kronecker.
$endgroup$
– remind
May 16 '14 at 16:26
$begingroup$
Okay, fine, they're slightly different. As long as you're doing operations on vectors and matrices the difference is not very serious.
$endgroup$
– Qiaochu Yuan
May 16 '14 at 16:31
$begingroup$
Thanks, so should the ACL paper have used outer product instead of Kronecker? No intent to be picky, it's just to have everything clear in my head :)
$endgroup$
– remind
May 16 '14 at 16:36
$begingroup$
The outer product is a special case of a Kronecker product, which should make things clear in this context. See en.wikipedia.org/wiki/Kronecker_product.
$endgroup$
– jhclark
Jul 16 '14 at 4:38