decomposable tensor

A tensor – an element in a tensor product of $k$ vector spaces $V_1\otimes V_2\otimes\ldots\otimes V_k$ – is said to be **decomposable** if it can be written in the form $v_1\otimes v_1\otimes\ldots\otimes v_k$ where $v_i \in V_i$ for $i = 1,ldots, k$.

If all $V_i$ are copies of $V$ or $V^*$ for the same $V$ then we often talk of vectors in the tensor product as tensors and the tensor product the space of tensors. If $V_i$ has a basis $\{e^s_i\}_{s= 1}^{n_i}$ then $V_1\otimes V_2\otimes\ldots\otimes V_k$ has a basis consisting of all decomposable vectors of the form $e_1^{s_1}\otimes e_k^{s_k}$ for all $(s_1,\ldots,s_k)$ such that $1\leq s_i\leq n_i$.

Let $V$ be finite-dimensional vector space. Then for a tensor $A\in V^{\otimes k}$ we say that $A$ has **decomposability rank** $r$ if

$r = min\{ h | \exists A_1,\ldots, A_h decomposable, A = A_1+\ldots+A_h \}$

Distinguish this invariant from the covariance rank of a tensor.

While the decomposability rank of a covariance rank 2 tensor $A$ is the same as the rank of the corresponding matrix, for higher covariance rank tensors we do not have general algorithms how to determine the decomposability rank.

Revised on September 2, 2011 00:02:55
by Urs Schreiber
(131.211.239.22)