Diagonalization of 4th order tensors I have been wondering about the following problem...
Let $n$ be a positive integer and denote by $M_n^s$ the space of symmetric $n\times n$ real matrices. Now, we look at the space $\mathcal L^{sym}(M_n^s,M_n^s)$, whose elements may be seen as well as matrices (fourth order symmetric tensors), we denote this space by $\tau_n^s$.
Recall then, an element $R \in \tau^s_n$ looks like
$$R = R_{ijkl} = R_{jikl} = R_{ijlk} = R_{klij}.$$ 
The question is: Can one characterize the set of $R \in \tau ^s_n$ for which there exist a change of variables in $\mathbb R^n$ (say $D$) and a symmetric matrix $A$ such that $R_{ijkl} = \delta_{jl}A_{ik} + \delta_{ik}A_{jl} + \delta_{jk}A_{il} + \delta_{il}A_{jk}$ as bilinear form over $M_n^s$? I.e.
$$\langle R(DMD^T),DMD^T\rangle = \langle AM,M\rangle.$$
[Edit: corrected formulas.]
Here, $\langle \; , \; \rangle$ stands for the inner product in the space of matrices.
Somehow i have been not able to answer it, any ideas are welcome! 
 A: (This should be considered as empiric answer, not as a rigorous proof)
As far as I know, there is no diagonalisation for general 4-tensors.
The reason is the following.
Let us assume any combinatorial question of the following form
"One has a square 2-dimensional lattice and colours its vertices into finite number of colours. Also to the each edge corresponds the weight, which depends of the colours of the vertices of this edge. For example, we could make some of these weights being 0 and some being 1, making the edges with 1 admissible, and with 0 non-admissible.
Now we take a sum over the colourings with the weights $d\Pi q_i^{l_i}$, where $d$ is the product of all weights of edges, $q_i$ correspond to colours, $l_i$ - number of vertices coloured in $q_i$.
Calculate this sum for
a) Square m x n (or even torus m x n) [AFAIK it is called statistical sum]
b) Square m x n (or even torus m x n) with some vertices already coloured [AFAIK it is called a correlator, but im not sure]"
c) It would be nice if the answer will be some understandable function of m and n.
If you could solve this question, you could, basically, solve any 2-dimensional statistical-mechanical system, and it is very general class of questions.
For instance, having (c) will solve for you undecidable things, as tilings are the special case of this question, and there are tilings which emulate Turing machines (so if you could prove that there is a tiling with a prescribed rule for any m and n you could solve termination problem).
Ok, what I'm going to say is that there is 4-tensor, in terms of which you could easily expose the answer, and any kind of diagonalisation solves c).
Ok, let us take the vector space V with basis ${e^i}$.
Let us fix the scalar product $d_{ij}$ = weight coefficient between colours $i$ and $j$. Now let us fix the 4-tensor $S^{ijkl} = \Sigma q_i e^i \otimes e^i \otimes e^i \otimes e^i$.
Make this tensor sit in each vertex and the scalar product in each edge. We are done.
//next part is over $\mathbb{C}$
Now let us find matrix M such that $d_{ij} = M^i_k\delta_{kl}M^j_l$. Now we can contract $S$ with $M$-s in each index, and now it all reduces to the diagonalisation of this new tensor wrt the standard scalar product (for example, diagonalisation you provided solves (c) instantly, as like as diagonalisation of matrix $X$ solves the problem of finding $tr X^n$)
I consider you only look onto orthogonal operators, otherwise you break symmetricity.
But you could also consider (2, 2) tensor $S^{ijmn}d_{mk}d_{nl}$, and for these tensors you could do any linear transformations on your vector space.
So, this question in the any suitable formulation has the answer NO.
A: This paper has some very interesting information on this subject (§3).  The most relevant part I will quote here:

Given  a  tensor $t_{ij..k}$of  order $r,2≤r≤4$,  is  it  possible to find a linear change of coordinates defined by an invertible matrix $A$ such that the tensor takes a diagonal form?  A necessary condition is that the number of free parameters be preserved.  A symmetric tensor has $\left(\begin{array}{c}r\\N+r−1\end{array}\right)$ independent parameters.   In the matrix case,  we  thus have $N(N+ 1)/2$ parameters, which is smaller than the $N^2$ entries of $A$. But in the tensor case, it is clear that the number of free parameters would decrease: $O(N^r)$ in the original symmetric tensor, and $N^2$ in the decomposition.  This immediate statement proves that only a small subset of symmetric tensors of order larger than 2 is (linearly) diagonalizable.

Which corroborates the "No" found by the other answer.  OTOH, the authors go on to say that if $A$ is a linear transformation in more than $N$ dimensions, or if only an approximate diagonalization is required, then the situation is much better.
