0

I'm working on finding the inverse (triple contraction) of a sixth-order tensor, in which each index is an integer ranging from $0$ to $2$ so it has $729$ elements in total. The sixth-order tensor has the totally symmetry, $$\LARGE A_{ijkpqr}=A_{jikpqr}=A_{ijkprq}$$

i. e. the element of the tensor $A$ remains invariant after swapping any two indices.

The method I am trying to use to find the inverse is as follows for each combination of $i,j,k,p,q,r = \{0,1,2\}$

enter image description here

and hence I could form a linear system with $729$ variables (the unknown $B$ is the inverse of $A$ I want to solve for) using the above formula and the R.H.S of the equation is the value of the sixth order identity tensor, i.e. $A_{729\times729} \times B_{729\times1} = \vec{I6}_{729\times 1}$

I known that this system are linearly dependent since the matrix B is "somehow" symmetric such that some of the row will repeat. So I ASSUMED that $B$ has the same (minor) symmetry as the identity tensor and thus should have only $126$ independent components. I gathered the dependent variables of $B$ and reformed the system. Now the L.H.S of the system is a [729X126] matrix multiply by a [126X1] vector. I typed in the [729X126] into MATLAB and found that the rank of the [729X126] matrix is ONLY 125, so that I cannot find the unique solution of B (the inverse). I was wondering what I did wrong and if B does not have the same symmetry as the 6th order identity tensor, how could I gather the terms together and reduce the original matrix_A. P.S. I used a similar approach for the 4th order tensor and successfully found the inverse in the 4th order case. I don't quite understand why this same approach does not work on the 6th order case. Thank you so much for your attention and help!

Thanks, William

Rohit Singh
  • 1,143
William Chan
  • 1
  • A $n^{th}$ order totally symmetric tensor over $\mathbb{R}^d$ has $\binom{n+d-1}{d-1}$ independent components. For $(n,d) = (6,3)$, your tensor has $28$ independent components. Your problem is more or less equivalent to finding the inverse of an $N\times N$ matrix for $N = \binom{3+3-1}{3-1} = 10$. – achille hui Jun 29 '22 at 07:30
  • Welcome to MSE. It is in your best interest that you type your posts (using MathJax) instead of posting links to pictures. – José Carlos Santos Jun 29 '22 at 07:31

0 Answers0