How can I calculate the inverse of a 4th order tensor?

[Galbi]

Homework Statement

I'm looking for how to calculate inverse of the 4th order tensor. That is,
A:A^-1=A^-1:A=I⁽⁴⁾
If I know a fourth order tensor A, then how can I calculate A^-1 ?
Let's just say it is inversible.

Homework Equations

The Attempt at a Solution

 

[jambaugh]

By "fourth order" I presume you mean rank 4, but that is with regard to the original vector space. A tensor is also a "Vector" in its own vector space as well as a linear operator in several settings but you seem to be implying a double contraction wherein you are mapping rank 2 tensors to rank 2 tensors.

Your task then is to write down a basis for the rank 2 tensor space upon which the rank 4 tensor acts and express that rank 4 tensor's components in that basis in the form of a matrix. You then use the standard matrix inversion techniques to find its inverse.

Example for 2 dimensional vectors you would choose as a basis for your rank 2 space say,$e^{11},e^{12},e^{21},e^{22}$ where $e^{ij}=e^i\otimes e^j$. Then the rank 4 tensor has matrix form:
$$A=\left[\begin{array}{cccc} A_{1111} & A_{1112} & A_{1121} & A_{1122} \\ A_{1211} & A_{1212} & A_{1221} & A_{1222} \\ A_{2111} & A_{2112} & A_{2121} & A_{2122} \\ A_{2211} & A_{2212} & A_{2221} & A_{2222} \end{array}\right]$$

Invert that matrix and you have the "inverse" in the same basis.

 

[Galbi]

Thank you very much.
Since I'm a novice at tensor calculation, It's hard to understand what you're saying.

If I have 3x3x3x3 tensors, then do the components of inversion of the matrix as seen below correspond to the components of inverse of A ?

$$A=\left[\begin{array}{cccc} A_{1111} & A_{1112} & A_{1113} & A_{1121}& A_{1122}& A_{1123}& A_{1131}& A_{1132}& A_{1133} \\ A_{1211} & A_{1212} & A_{1213} & A_{1221}& A_{1222}& A_{1223}& A_{1231}& A_{1232}& A_{1233} \\ A_{1311} & A_{1312} & A_{1313} & A_{1321}& A_{1322}& A_{1323}& A_{1331}& A_{1332}& A_{1333} \\ A_{2111} & A_{2112} & A_{2113} & A_{2121}& A_{2122}& A_{2123}& A_{2131}& A_{2132}& A_{2133} \\ A_{2211} & A_{2212} & A_{2213} & A_{2221}& A_{2222}& A_{2223}& A_{2231}& A_{2232}& A_{2233} \\ A_{2311} & A_{2312} & A_{2313} & A_{2321}& A_{2322}& A_{2323}& A_{2331}& A_{2332}& A_{2333} \\ A_{3111} & A_{3112} & A_{3113} & A_{3121}& A_{3122}& A_{3123}& A_{3131}& A_{3132}& A_{3133} \\ A_{3211} & A_{3212} & A_{3213} & A_{3221}& A_{3222}& A_{3223}& A_{3231}& A_{3232}& A_{3233} \\ A_{3311} & A_{3312} & A_{3213} & A_{3321}& A_{3322}& A_{3323}& A_{3331}& A_{3332}& A_{3333} \\ \end{array}\right]$$I'm going to check this with MATLAB now.
Thank you again!

By "fourth order" I presume you mean rank 4, but that is with regard to the original vector space. A tensor is also a "Vector" in its own vector space as well as a linear operator in several settings but you seem to be implying a double contraction wherein you are mapping rank 2 tensors to rank 2 tensors.

Your task then is to write down a basis for the rank 2 tensor space upon which the rank 4 tensor acts and express that rank 4 tensor's components in that basis in the form of a matrix. You then use the standard matrix inversion techniques to find its inverse.

Example for 2 dimensional vectors you would choose as a basis for your rank 2 space say,$e^{11},e^{12},e^{21},e^{22}$ where $e^{ij}=e^i\otimes e^j$. Then the rank 4 tensor has matrix form:
$$A=\left[\begin{array}{cccc} A_{1111} & A_{1112} & A_{1121} & A_{1122} \\ A_{1211} & A_{1212} & A_{1221} & A_{1222} \\ A_{2111} & A_{2112} & A_{2121} & A_{2122} \\ A_{2211} & A_{2212} & A_{2221} & A_{2222} \end{array}\right]$$

Invert that matrix and you have the "inverse" in the same basis.

 

[Galbi]

Dear jambaugh,

It works! Thank you very much!

By "fourth order" I presume you mean rank 4, but that is with regard to the original vector space. A tensor is also a "Vector" in its own vector space as well as a linear operator in several settings but you seem to be implying a double contraction wherein you are mapping rank 2 tensors to rank 2 tensors.

Your task then is to write down a basis for the rank 2 tensor space upon which the rank 4 tensor acts and express that rank 4 tensor's components in that basis in the form of a matrix. You then use the standard matrix inversion techniques to find its inverse.

Example for 2 dimensional vectors you would choose as a basis for your rank 2 space say,$e^{11},e^{12},e^{21},e^{22}$ where $e^{ij}=e^i\otimes e^j$. Then the rank 4 tensor has matrix form:
$$A=\left[\begin{array}{cccc} A_{1111} & A_{1112} & A_{1121} & A_{1122} \\ A_{1211} & A_{1212} & A_{1221} & A_{1222} \\ A_{2111} & A_{2112} & A_{2121} & A_{2122} \\ A_{2211} & A_{2212} & A_{2221} & A_{2222} \end{array}\right]$$

Invert that matrix and you have the "inverse" in the same basis.

 

[Charles6]

By "fourth order" I presume you mean rank 4, but that is with regard to the original vector space. A tensor is also a "Vector" in its own vector space as well as a linear operator in several settings but you seem to be implying a double contraction wherein you are mapping rank 2 tensors to rank 2 tensors.

Your task then is to write down a basis for the rank 2 tensor space upon which the rank 4 tensor acts and express that rank 4 tensor's components in that basis in the form of a matrix. You then use the standard matrix inversion techniques to find its inverse.

Example for 2 dimensional vectors you would choose as a basis for your rank 2 space say,$e^{11},e^{12},e^{21},e^{22}$ where $e^{ij}=e^i\otimes e^j$. Then the rank 4 tensor has matrix form:
$$A=\left[\begin{array}{cccc} A_{1111} & A_{1112} & A_{1121} & A_{1122} \\ A_{1211} & A_{1212} & A_{1221} & A_{1222} \\ A_{2111} & A_{2112} & A_{2121} & A_{2122} \\ A_{2211} & A_{2212} & A_{2221} & A_{2222} \end{array}\right]$$

Invert that matrix and you have the "inverse" in the same basis.

Hi jambaugh,

Your anwser is quite inspiring, though I do encounter other questions when applying your solution.

It's quite common in elasticity for a fourth order tensor to possesses minor symmetry, i.e. $$A_{ijkl}=A_{ijlk}=A_{jikl}$$
In this case, the matrix you mentioned would be singular, while the corresponding tensor (e.g. a modulus tensor) do have an inverse (e.g. a compliance tensor).

Could you shed some light on calculating the inverse of a modulus tensor?

 

[jambaugh]

Note that by construction the strain tensor also has this index symmetry and so too the stress tensor.
I suggest you conceptualize the pairs as a single multi-index:
$$Y_{(ij)} = \sum_{(kl)} A_{(ij),(kl)} X_{(kl)}$$
One needs to be careful with definitions here as there may be factor of two differences in how you express the tensor since I am here suggesting only summing over redundant indices once.
The summation over a symmetric pair here would count e.g. (12) and (21) as one value.
In this format there are six unique terms in the sum: (11)(12)(13)(22)(23)(33) (for 3 dimensions)
So A is effectively expressed as a 6x6 matrix instead of as a 9x9. It is here where true singularities which forbid inverting the defined mapping would occur.

The factor of two I mentioned occurs where equal indices occur once in both 9x9 and 6x6 representations while the unequal ones occur only once in the 6x6 but twice in the 9x9.

I'm not sure of the application and the nature of the singularity you mention. I've just played a lot with irreducible linear representations of GL(N) which is what "tensors" ultimately are. But I hope this observation is helpful.