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I feel like different sources use the term "rank" differently, which is perhaps leading to my confusion.

When I think of rank I think of number of linearly independent columns/rows, number of pivots in RREF, etc... So a vector always has a rank of 1. A $n \times m$ matrix has $rank \leq min(m,n)$.

For tensors, I have always been under the impression that: a scalar is a 0-th order tensor; a vector is a 1st order tensor; a matrix is a 2nd order tensor.

When reading this https://www.sandia.gov/~tgkolda/pubs/pubfiles/TensorReview.pdf, I came across the term "rank-one tensor." See page 463 (page number is located on top right), in the sentence below Fig. 3.1, it states "The CP decomposition factorizes a tensor into a sum of component rank-one tensors." In this context, is rank-one tensor just a vector? That's what it seems like from studying Eq. (3.1). Does the term "rank" here have anything to do with the number of linearly independent columns or is it just describing the order of the tensor, i.e., the number of dimensions of the tensor? For example, a rank-2 tensor would then be a matrix and a rank-0 tensor would be a scalar?

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  • $\begingroup$ The order and the rank of a tensor are different concepts> Wikipedia has an article on tensor rank decomposition (a.k.a. canonical polyadic decomposition) at en.wikipedia.org/wiki/Tensor_rank_decomposition $\endgroup$ – gandalf61 Jan 28 at 17:22
  • $\begingroup$ Thanks. I read this before. I have also seen "rank" used in contexts such as "Write the rank-3 approximation of X, where X is a rank-3 tensor." Does this mean representing X as another rank-3 tensor, or what does rank-3 approximation mean in this kind of context? $\endgroup$ – Iamanon Jan 28 at 17:33

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