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Is there a Command in Mathematica that fourth rank tensor can be represented as 6 by 6 matrix?

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  • $\begingroup$ ArrayReshape, e.g., ArrayReshape[m, {6,6}] (if Times@@TensorDimensions[m] ==36)? $\endgroup$
    – kglr
    Commented Jul 3, 2018 at 8:09
  • $\begingroup$ @kglr, thanks a lot! $\endgroup$
    – ABCDEMMM
    Commented Jul 3, 2018 at 8:13

1 Answer 1

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aa = Array[a, {2, 2, 3, 3}];
TensorRank[aa]

4

aa // MatrixForm // TeXForm

$\tiny\left( \begin{array}{cc} \left( \begin{array}{ccc} a(1,1,1,1) & a(1,1,1,2) & a(1,1,1,3) \\ a(1,1,2,1) & a(1,1,2,2) & a(1,1,2,3) \\ a(1,1,3,1) & a(1,1,3,2) & a(1,1,3,3) \\ \end{array} \right) & \left( \begin{array}{ccc} a(1,2,1,1) & a(1,2,1,2) & a(1,2,1,3) \\ a(1,2,2,1) & a(1,2,2,2) & a(1,2,2,3) \\ a(1,2,3,1) & a(1,2,3,2) & a(1,2,3,3) \\ \end{array} \right) \\ \left( \begin{array}{ccc} a(2,1,1,1) & a(2,1,1,2) & a(2,1,1,3) \\ a(2,1,2,1) & a(2,1,2,2) & a(2,1,2,3) \\ a(2,1,3,1) & a(2,1,3,2) & a(2,1,3,3) \\ \end{array} \right) & \left( \begin{array}{ccc} a(2,2,1,1) & a(2,2,1,2) & a(2,2,1,3) \\ a(2,2,2,1) & a(2,2,2,2) & a(2,2,2,3) \\ a(2,2,3,1) & a(2,2,3,2) & a(2,2,3,3) \\ \end{array} \right) \\ \end{array} \right)$

ArrayReshape[aa, {6, 6}] // MatrixForm // TeXForm

$\tiny\left( \begin{array}{cccccc} a(1,1,1,1) & a(1,1,1,2) & a(1,1,1,3) & a(1,1,2,1) & a(1,1,2,2) & a(1,1,2,3) \\ a(1,1,3,1) & a(1,1,3,2) & a(1,1,3,3) & a(1,2,1,1) & a(1,2,1,2) & a(1,2,1,3) \\ a(1,2,2,1) & a(1,2,2,2) & a(1,2,2,3) & a(1,2,3,1) & a(1,2,3,2) & a(1,2,3,3) \\ a(2,1,1,1) & a(2,1,1,2) & a(2,1,1,3) & a(2,1,2,1) & a(2,1,2,2) & a(2,1,2,3) \\ a(2,1,3,1) & a(2,1,3,2) & a(2,1,3,3) & a(2,2,1,1) & a(2,2,1,2) & a(2,2,1,3) \\ a(2,2,2,1) & a(2,2,2,2) & a(2,2,2,3) & a(2,2,3,1) & a(2,2,3,2) & a(2,2,3,3) \\ \end{array} \right)$

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  • $\begingroup$ very helpful for me! $\endgroup$
    – ABCDEMMM
    Commented Jul 3, 2018 at 8:19

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