A real tensor $T$ of order $r$ in dimension $n$ is a collection of real coefficients $T_{i_1i_2\dots i_r}$, $1\leq i_1, i_2, \dots, i_r\leq n$. A tensor $T$ is symmetric if the coefficients $T_{i_1i_2\dots i_r}$ are invariant under permutation of indices.
Do tensors of the form $n_in_jn_kn_l$ span all 4th order symmetric tensors?
This question came up when I was studying the elastic behavior of Bravais lattices. I believe the answer to be negative in 3D because some symmetric 4th order tensors are isotropic and I don't see how a (finite) sum of $n_in_jn_kn_l$ can be isotropic. I am mostly interested in this case (4th order tensors in 3D). But I wouldn't mind learning about other dimensions (say 2D) or orders.
By the way, this is true for 2nd order tensors in any dimension: matrices $n_in_j$ span all symmetric matrices because they are diagonalizable. For symmetric 4th order tensors, diagonalizability ensures that they are spanned by $m_{ij}m_{kl}$ but I don't see how to further break these into a sum of $n_in_jn_kn_l$, or if this is possible at all.
Tensors of the form $n_in_jn_kn_l$ (tensor products of four vectors, also known as rank-1 tensors) can produce all fully symmetric tensors by linear combination, but they are not fully symmetric themselves. However, the symmetrizations of $n_in_jn_kn_l$ do span all fully symmetric tensors while being symmetric themselves. Note that linear combination is necessary, as not every fully symmetric tensor is the symmetrization of a rank-1 tensor.