# The Inverse of a Fourth Order Tensor

Suppose that we have a fourth order tensor $${\bf{A}}$$

$${\bf{A}}=A_{ijkl} {\bf{e}}_i \otimes {\bf{e}}_j \otimes {\bf{e}}_k \otimes {\bf{e}}_l$$

in the orthonormal basis $$\{{\bf{e}}_1,{\bf{e}}_2,{\bf{e}}_3\}$$ for $$\mathbb{R}^3$$. Then we define the inverse of $${\bf{A}}$$ denoted by $${\bf{B}}$$ as follows

$${\bf{A}} : {\bf{B}} = {\bf{B}} : {\bf{A}} = {\bf{I}}$$

where $${\bf{I}}$$ is the fourth order identity tensor

\begin{align} {\bf{I}} &=I_{ijkl} {\bf{e}}_i \otimes {\bf{e}}_j \otimes {\bf{e}}_k \otimes {\bf{e}}_l\\ &=\delta_{ik} \delta_{jl} {\bf{e}}_i \otimes {\bf{e}}_j \otimes {\bf{e}}_k \otimes {\bf{e}}_l \end{align}

where $$\delta_{ij}$$ is the Kronecker's Delta

$$\delta_{ij}= \begin{cases} 1 & i=j \\ 0 & i \ne j \end{cases}$$

and $$:$$ is the double contraction defined by

$${\bf{A}} : {\bf{B}}=A_{ijmn}B_{mnkl} {\bf{e}}_i \otimes {\bf{e}}_j \otimes {\bf{e}}_k \otimes {\bf{e}}_l$$

I want to write a code to compute $${\bf{B}}$$. However, I could not find any good resource on the net that gives the elements of $${\bf{B}}$$ in terms of the elements of $${\bf{A}}$$. How should I compute $${\bf{B}} = {\bf{A}}^{-1}$$?

An application of this can be found in the theory of elasticity, where the fourth order tensor have the following symmetries

$$A_{ijkl}=A_{jikl}=A_{ijlk}=A_{klij}.$$

That would be great if you can also touch upon this.

• I would try to transform the problem to an equivalent problem stated in terms of matrices. Then you can use existing codes for inverting matrices and transform back. Since a 4th order tensor corresponds to a block matrix, it seems like this approach may work. Can you give more detail regarding the tensor-tensor product you are using to define the inverse. Jan 24 '16 at 15:18
• @K.Miller: Thanks for the attention. I added the definition to the question. :) Jan 24 '16 at 15:35
• I assume by double scalar product you mean the sum over indexes $m,n$? Jan 24 '16 at 16:17
• @K.Miller: Yes. :) Whatever terminology you like. :) Jan 24 '16 at 16:32
• There are three $4^{th}$-order isotropic tensors, but only one of them $(E_{ijkl}=\delta_{ik}\delta_{jl})$ acts like an identity with respect to the double contraction product \eqalign{E:A &= A:E = A} The other two $(F_{ijkl}=\delta_{il}\delta_{jk},\,G_{ijkl}=\delta_{ij}\delta_{kl})$, produce a trace or a transpose under the product\eqalign{F:A &= A:F = A^T \cr G:A &= A:G = {\rm tr}(A)\,I \cr}
– lynn
Jan 24 '16 at 20:24

In general

$$\sum_{m=1}^N\sum_{n=1}^N A_{ijmn}B_{mnkl} = \delta_{ik}\delta_{jl}$$

for $i,j,k,l = 1,\ldots,N$. Now suppose the tensor $\mathbf{A}$ is unfolded as the $N^2\times N^2$ matrix given by

$$A = \left[ \begin{array}{ccccccccc} A_{1111} & A_{1112} & \cdots & A_{111N} & \cdots & A_{11N1} & A_{11N2} & \cdots & A_{11NN}\\ A_{1211} & A_{1212} & \cdots & A_{121N} & \cdots & A_{12N1} & A_{12N2} & \cdots & A_{12NN}\\ \vdots & \vdots & &\vdots & & \vdots & \vdots & & \vdots\\ A_{NN11} & A_{NN12} & \cdots & A_{NN1N} & \cdots & A_{NNN1} & A_{NNN2} & \cdots & A_{NNNN}\\ \end{array} \right]$$

If the same is done for $\mathbf{B}$, then the matrix product $AB$ corresponds to the unfolded tensor product $\mathbf{A} : \mathbf{B}$. Let $E$ be the same unfolding of the tensor identity matrix $\mathbf{I}$. Then

$$\mathbf{A} : \mathbf{B} = \mathbf{I} \iff AB = E$$

Similarly,

$$\mathbf{B} : \mathbf{A} = \mathbf{I} \iff BA = E$$

In order for such a matrix $B$ to exist it follows that $AB = BA$ must hold, i.e., the product of $A$ and $B$ must commute. Now suppose $A$ is invertible. Then $B$ must satisfy the equations

$$B = A^{-1}E \quad\text{and}\quad B = EA^{-1} \iff A^{-1}E = EA^{-1} \iff EA = AE$$

Thus, in the case that $A$ is invertible, the product of $A$ and $E$ must commute in order for such a matrix $B$ to exist. If these conditions are met, then you should be able to compute a unique $B$ from the equation $AB = E$ via the LU factorization or by some other matrix factorization.

• I just noticed that if the tensor ${\bf{A}}$ has the symmetries $A_{ijkl}=A_{jikl}=A_{ijlk}=A_{klij}$ then the unfolded matrix form $A$ will be singular! :) Mar 5 '16 at 14:43