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 $v_1\otimes v_2\otimes\ldots\otimes v_k$ 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 $e_1^{s_1}\otimes\ldots\otimes e_k^{s_k}$ 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.