Uniqueness of Tensor Decompositions (Aren't Matrix Decompositions a Special Case?)
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It seems that higher-order tensors (of order 3 or higher) generally have unique decompositions under relatively mild conditions. For example, Kruskal proved that if an order-3 Tensor $T$ can be decomposed as the outer product of matrices $A, B, C$, with Kruskal ranks $k_A, k_B, k_C$ respectively, then as long as the rank $R$ of $T$ satisfies
$$R le frac{k_A + k_B + k_C - 2}{2},$$
then the decomposition is unique.
My question is: aren't matrix decompositions a special case of tensor decompositions? In other words, I can think of a matrix $M$ as a tensor whose third dimension is of size 1. If I was to apply the same result to matrix decompositions, I would get that that the matrix decomposition is unique as long as its rank is:
$$R le frac{k_A + k_B + 1 - 2}{2} = frac{k_A + k_B - 1}{2},$$
(where we have assumed that $C$ is just a matrix of ones), but we know that this is generally not true. You can take a matrix, and rotate its component matrices, and get a different decomposition (the so-called "rotation problem", so matrix decompositions are not unique.
How do we reconcile these facts?
matrices tensors matrix-decomposition tensor-rank tensor-decomposition
asked Jun 1 '18 at 12:59
Curious StudentCurious Student
797
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add a comment |
$begingroup$
It seems that higher-order tensors (of order 3 or higher) generally have unique decompositions under relatively mild conditions. For example, Kruskal proved that if an order-3 Tensor $T$ can be decomposed as the outer product of matrices $A, B, C$, with Kruskal ranks $k_A, k_B, k_C$ respectively, then as long as the rank $R$ of $T$ satisfies
$$R le frac{k_A + k_B + k_C - 2}{2},$$
then the decomposition is unique.
My question is: aren't matrix decompositions a special case of tensor decompositions? In other words, I can think of a matrix $M$ as a tensor whose third dimension is of size 1. If I was to apply the same result to matrix decompositions, I would get that that the matrix decomposition is unique as long as its rank is:
$$R le frac{k_A + k_B + 1 - 2}{2} = frac{k_A + k_B - 1}{2},$$
(where we have assumed that $C$ is just a matrix of ones), but we know that this is generally not true. You can take a matrix, and rotate its component matrices, and get a different decomposition (the so-called "rotation problem", so matrix decompositions are not unique.
How do we reconcile these facts?
matrices tensors matrix-decomposition tensor-rank tensor-decomposition
asked Jun 1 '18 at 12:59
Curious StudentCurious Student
797
$endgroup$
add a comment |
$begingroup$
It seems that higher-order tensors (of order 3 or higher) generally have unique decompositions under relatively mild conditions. For example, Kruskal proved that if an order-3 Tensor $T$ can be decomposed as the outer product of matrices $A, B, C$, with Kruskal ranks $k_A, k_B, k_C$ respectively, then as long as the rank $R$ of $T$ satisfies
$$R le frac{k_A + k_B + k_C - 2}{2},$$
then the decomposition is unique.
My question is: aren't matrix decompositions a special case of tensor decompositions? In other words, I can think of a matrix $M$ as a tensor whose third dimension is of size 1. If I was to apply the same result to matrix decompositions, I would get that that the matrix decomposition is unique as long as its rank is:
$$R le frac{k_A + k_B + 1 - 2}{2} = frac{k_A + k_B - 1}{2},$$
(where we have assumed that $C$ is just a matrix of ones), but we know that this is generally not true. You can take a matrix, and rotate its component matrices, and get a different decomposition (the so-called "rotation problem", so matrix decompositions are not unique.
How do we reconcile these facts?
matrices tensors matrix-decomposition tensor-rank tensor-decomposition
asked Jun 1 '18 at 12:59
Curious StudentCurious Student
797
$endgroup$
It seems that higher-order tensors (of order 3 or higher) generally have unique decompositions under relatively mild conditions. For example, Kruskal proved that if an order-3 Tensor $T$ can be decomposed as the outer product of matrices $A, B, C$, with Kruskal ranks $k_A, k_B, k_C$ respectively, then as long as the rank $R$ of $T$ satisfies
$$R le frac{k_A + k_B + k_C - 2}{2},$$
then the decomposition is unique.
My question is: aren't matrix decompositions a special case of tensor decompositions? In other words, I can think of a matrix $M$ as a tensor whose third dimension is of size 1. If I was to apply the same result to matrix decompositions, I would get that that the matrix decomposition is unique as long as its rank is:
$$R le frac{k_A + k_B + 1 - 2}{2} = frac{k_A + k_B - 1}{2},$$
(where we have assumed that $C$ is just a matrix of ones), but we know that this is generally not true. You can take a matrix, and rotate its component matrices, and get a different decomposition (the so-called "rotation problem", so matrix decompositions are not unique.
How do we reconcile these facts?
matrices tensors matrix-decomposition tensor-rank tensor-decomposition
matrices tensors matrix-decomposition tensor-rank tensor-decomposition
asked Jun 1 '18 at 12:59
Curious StudentCurious Student
797
asked Jun 1 '18 at 12:59
Curious StudentCurious Student
797
asked Jun 1 '18 at 12:59
Curious StudentCurious Student
797
asked Jun 1 '18 at 12:59
Curious StudentCurious Student
797
asked Jun 1 '18 at 12:59
Curious StudentCurious Student
797
797
add a comment |
add a comment |
1 Answer
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Kruskal rank can not be higher than the canonical rank, so you can never have:
2*R <= 2*R - 1
which means that the decomposition is never unique in the general case.
edited Jan 16 at 20:58
Yassine Zniyed
31
answered Jan 16 at 19:45
Yassine ZNIYEDYassine ZNIYED
261
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Kruskal rank can not be higher than the canonical rank, so you can never have:
2*R <= 2*R - 1
which means that the decomposition is never unique in the general case.
edited Jan 16 at 20:58
Yassine Zniyed
31
answered Jan 16 at 19:45
Yassine ZNIYEDYassine ZNIYED
261
$endgroup$
add a comment |
$begingroup$
Kruskal rank can not be higher than the canonical rank, so you can never have:
2*R <= 2*R - 1
which means that the decomposition is never unique in the general case.
edited Jan 16 at 20:58
Yassine Zniyed
31
answered Jan 16 at 19:45
Yassine ZNIYEDYassine ZNIYED
261
$endgroup$
add a comment |
$begingroup$
Kruskal rank can not be higher than the canonical rank, so you can never have:
2*R <= 2*R - 1
which means that the decomposition is never unique in the general case.
edited Jan 16 at 20:58
Yassine Zniyed
31
answered Jan 16 at 19:45
Yassine ZNIYEDYassine ZNIYED
261
$endgroup$
Kruskal rank can not be higher than the canonical rank, so you can never have:
2*R <= 2*R - 1
which means that the decomposition is never unique in the general case.
edited Jan 16 at 20:58
Yassine Zniyed
31
answered Jan 16 at 19:45
Yassine ZNIYEDYassine ZNIYED
261
edited Jan 16 at 20:58
Yassine Zniyed
31
edited Jan 16 at 20:58
Yassine Zniyed
31
edited Jan 16 at 20:58
Yassine Zniyed
31
31
answered Jan 16 at 19:45
Yassine ZNIYEDYassine ZNIYED
261
answered Jan 16 at 19:45
Yassine ZNIYEDYassine ZNIYED
261
answered Jan 16 at 19:45
Yassine ZNIYEDYassine ZNIYED
261
261
add a comment |
add a comment |
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