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Register a new userRecovering orthogonal tensors under arbitrarily strong, but locally correlated, noise
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Added by
Oscar Mickelin
Publication date
2021
fields
Informatics Engineering
and research's language is
English
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We consider the problem of recovering an orthogonally decomposable tensor with a subset of elements distorted by noise with arbitrarily large magnitude. We focus on the particular case where each mode in the decomposition is corrupted by noise vectors with components that are correlated locally, i.e., with nearby components. We show that this deterministic tensor completion problem has the unusual property that it can be solved in polynomial time if the rank of the tensor is sufficiently large. This is the polar opposite of the low-rank assumptions of typical low-rank tensor and matrix completion settings. We show that our problem can be solved through a system of coupled Sylvester-like equations and show how to accelerate their solution by an alternating solver. This enables recovery even with a substantial number of missing entries, for instance for $n$-dimensional tensors of rank $n$ with up to $40%$ missing entries.
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We study the problem of finding orthogonal low-rank approximations of symmetric tensors. In the case of matrices, the approximation is a truncated singular value decomposition which is then symmetric. Moreover, for rank-one approximations of tensors of any dimension, a classical result proven by Banach in 1938 shows that the optimal approximation can always be chosen to be symmetric. In contrast to these results, this article shows that the corresponding statement is no longer true for orthogonal approximations of higher rank. Specifically, for any of the four common notions of tensor orthogonality used in the literature, we show that optimal orthogonal approximations of rank greater than one cannot always be chosen to be symmetric.
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A third order real tensor is mapped to a special f-diagonal tensor by going through Discrete Fourier Transform (DFT), standard matrix SVD and inverse DFT. We call such an f-diagonal tensor an s-diagonal tensor. An f-diagonal tensor is an s-diagonal tensor if and only if it is mapped to itself in the above process. The third order tensor space is partitioned to orthogonal equivalence classes. Each orthogonal equivalence class has a unique s-diagonal tensor. Two s-diagonal tensors are equal if they are orthogonally equivalent. Third order tensors in an orthogonal equivalence class have the same tensor tubal rank and T-singular values. Four meaningful necessary conditions for s-diagonal tensors are presented. Then we present a set of sufficient and necessary conditions for s-diagonal tensors. Such conditions involve a special complex number. In the cases that the dimension of the third mode of the considered tensor is $2, 3$ and $4$, we present direct sufficient and necessary conditions which do not involve such a complex number.
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