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I would like to apply a linear transformation to a tensor. My linear transformation is encoded by a matrix, for example

M = Table[m[i, j], {i, 4}, {j, 4}]

Take some vectors

V = Table[v[i], {i, 4}]

W = Table[w[i], {i, 4}]

I apply the linear transformation to vectors with the dot operator

M . V

M . W

Now I would like to apply the same linear transformation to tensors, in a compatible ways. For a tensor of order two, the linear transformation is encoded by the tensor product matrix

TensorProduct[M,M]

But how do I apply it to a tensor? The dot operator doesn't give the right result. For example, imagine I want to apply my linear transformation to the following tensor

TensorProduct[V,W]

I would like to have that

TensorProduct[M,M] . TensorProduct[V,W] === TensorProduct[M . V, M . W]

But this is False. The dot operator is not the correct way. What is the right way? I would like to apply this linear transformation also to other tensors, not just tensors of the form TensorProduct[V,W]

Moreover, I would like to apply the linear transformation also to tensors of higher order, not just tensors of order 2.

Thank you for any help!

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    – bbgodfrey
    Commented Jan 16, 2022 at 16:10

2 Answers 2

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If I understand your question correctly, you're looking to rotate a tensor of arbitrary rank using a transformation matrix. This can be done using the tensor product:

\begin{equation} \hat{B}_{i_1 i_2 ... i_ n} = \sum_{j_1 = 1}^d \sum_{j_1 = 1}^d ... \sum_{j_n = 1}^d A_{i_1 j_1} A_{i_2 j_2} ... A_{i_n j_n} B_{j_1 j_2 ... j_n} \ . \end{equation}

There's several implementations of this in this question. Here's one example:

rayleighProduct[transformationMatrix_, tensor_] := Block[{n, it, t1},
  n = TensorRank[tensor];
  it = RotateLeft[Range[n]];
  t1 = tensor;
  Do[t1 = TensorTranspose[transformationMatrix . t1, it], {i, n}];
  t1]

Some examples:

M = Table[m[i, j], {i, 4}, {j, 4}];
V = Table[v[i], {i, 4}];
Q = Table[q[i, j], {i, 4}, {j, 4}];

M.V == rayleighProduct[M, V]
M.Q.Transpose[M] == rayleighProduct[M, Q]
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    $\begingroup$ Thank you very much! I didn't know the name Rayleigh Product. I tested it. The following command Simplify[rayleighProduct[M, TensorProduct[V, W]] - TensorProduct[M . V, M . W]] Gives zero. It also works for tensors of order 3: ` Simplify[rayleighProduct[M, TensorProduct[V, W, X]] - TensorProduct[M . V, M . W, M . X]] ` this also gives zero. So I think this is what I need! $\endgroup$
    – LeibnizGW
    Commented Jan 16, 2022 at 16:44
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Assume you have 2 vectors V and W and construct the tensor product "t1=TesorProduct[V,W]". If you apply a linear transformation: M to the vectors V,W: M.V,M.W. How do you then get "t2=TensorProduct[M.V,M.W]" from t1?

Well the simple answer is:

t2=M.TensorProduct[V, W].Transpose[M]

You can check this by a simple example:

n = 4;
M = Table[m[i, j], {i, n}, {j, n}];
V = Table[v[i], {i, n}];
W = Table[w[i], {i, n}];

M. TensorProduct[V, W].Transpose[M] == 
  TensorProduct[M . V, M . W] // Simplify

( True *)

Update

The same can be written using "TensorContract" in a form that can easily be generalized to more dimensions:

TensorProduct[M . V, M . W] == 
  TensorContract[
   TensorProduct[M, V, M, W], {{2, 3}, {5, 6}}] // Expand

(* True *)

For more dimensions we may have e.g:

X = Table[w[i], {i, n}];
TensorProduct[M . V, M . W, M . X] == 
  TensorContract[
   TensorProduct[M, V, M, W, M, 
    X], {{2, 3}, {5, 6}, {8, 9}}] // Expand 

(* True*)
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  • $\begingroup$ Thanks! Yes, I know the transpose on the other side can work for tensors of order 2. But How can I generalize this for tensors of order 3 or higher? George's answer works for tensors of any order. $\endgroup$
    – LeibnizGW
    Commented Jan 16, 2022 at 17:36
  • $\begingroup$ With the right notation, this can easily be generalized for higher tensors. Look my update. $\endgroup$
    – Daniel Huber
    Commented Jan 16, 2022 at 18:18
  • $\begingroup$ Unfortunately, using TensorProduct and TensorContract like this becomes quickly super slow... $\endgroup$
    – Henrik Schumacher
    Commented Jan 16, 2022 at 18:21
  • $\begingroup$ I see, I had never used TensorContract before. But there is still an issue with this solution. You have to imagine that I don't know V and W, I only have one tensor (in this case, TensorProduct[V,W]). So, what I need to do is to permute the factors with reference to your solution. The right order is TensorContract[TensorProduct[M, M, V, W], {{2, 5}, {4, 6}}] $\endgroup$
    – LeibnizGW
    Commented Jan 16, 2022 at 21:05
  • $\begingroup$ With this order, I can now write the equivalent ` TensorContract[ TensorProduct[M, M, TensorProduct[V, W]], {{2, 5}, {4, 6}}]` This contains only things I know. $\endgroup$
    – LeibnizGW
    Commented Jan 16, 2022 at 21:06

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