I am trying to understand rank of a $d \times d \times d$ tensor. The way that I understand the $d \times d$ case is that a rank $r$, $d \times d$ tensor is a tensor that can be written as the sum of $r$ rank 1, $d \times d$ tensors, and each rank 1 tensor can be written as the outer product of two vectors. I understand that a rank $r$, $d \times d \times d$ tensor is one that can be written as the sum of $r$, $d \times d \times d$ rank 1 tensors, but I don't understand how to show that $d \times d \times d$ tensor is rank $1$. For example, how would one show that the following $2 \times 2 \times 2$ tensor is rank 1?

\begin{equation*} \left( \begin{array}{cc|cc} a_1 b_1 c_1 & a_1 b_1 c_2 & a_2 b_1 c_1 & a_2 b_1 c_2 \\ a_1 b_2 c_1 & a_1 b_2 c_2 & a_2 b_2 c_1 & a_2 b_2 c_2 \end{array} \right) \end{equation*}

Is there a way to write this as an outer product?

Thank you in advance.

  • $\begingroup$ Maybe this article would be helpful. As mentioned here, "Every tensor can be expressed as a sum of rank $1$ tensors. The rank of a general tensor $T$ is defined to be the minimum number of rank $1$ tensors with which it is possible to express $T$ as a sum (Bourbaki 1989, II, §7)". "A tensor of rank one is a tensor that can be written as a tensor product of the form $T=a\otimes b\otimes\cdots\otimes d$ where $a, b, \dots, d$ are nonzero and in $V$ or $V^*$" $\endgroup$
    – Vlad
    Aug 5, 2015 at 11:08
  • $\begingroup$ I think my question is how to go from the tensor given in indices, as in my question statement, to a tensor given by the tensor product of the three vectors, as in the definition you've quoted. The definitions are equivalent, my question is how to go between them. In the matrix case it's obvious to me, but in the $d \times d \times d$ it is not. $\endgroup$
    – daniel
    Aug 5, 2015 at 11:43

1 Answer 1


We could write your tensor as $$ (a_1,a_2) \otimes (b_1,b_2) \otimes (c_1, c_2) $$ Or, depending on your notation, perhaps $$ (b_1,b_2) \otimes (c_1, c_2) \otimes (a_1,a_2) $$ One way to check that this tensor is rank one is to note that one matrix is a multiple of the other, and that each matrix is rank one.

  • 2
    $\begingroup$ "one matrix is a multiple of the other", here saying that a $2 \times 2 \times 2$ tensor is basically two $2 \times 2$ matrices stacked on top of each other, right? You have to make sure that the matrix in question is rank 1 as well. $\endgroup$
    – Calle
    Aug 5, 2015 at 18:45
  • $\begingroup$ @omnomnomnom say $a = (a_1, a_2), ..., c = (c_1, c_2)$. Do we get to the tensor product $a \otimes b \otimes c$ from the array in my original post by fixing a basis and then expanding $a \otimes b \otimes c$ to $\sum a_i b_j c_k (x_i \otimes x_j \otimes x_k)$? Is that the idea? Thanks. $\endgroup$
    – daniel
    Aug 5, 2015 at 19:40
  • $\begingroup$ @daniel yes! ${}{}$ $\endgroup$ Aug 5, 2015 at 19:58
  • $\begingroup$ I'm still a little confused then why rank 1 tensors are said to be vectors. In what sense is the tensor above a vector? But I understand my original question a bit better, thanks. $\endgroup$
    – daniel
    Aug 6, 2015 at 7:12
  • 1
    $\begingroup$ @daniel that's because there are two unrelated ways in which people use the term "rank" for tensors, and people who say that "rank 1 tensors are vectors" mean the other definition. For the alternate definition, see here and here. $\endgroup$ Aug 6, 2015 at 8:45

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