I'm learning tensor algebra.
So I have a question, Given a mod 3 tensor $\chi \in \mathbf{R}^{d_1,d_2,d_2}$, find vector $x \in \mathbf{R}^{d_1}$ so that $\chi \times_1 x$ is symmetric/antisymmetric?
So here is my attempt:
We want $(\chi \times_1 x )_{ij}=(\chi \times_1 x )_{ji}$
$(\chi \times_1 x)_{ij}= \sum_k \chi_{kij}x[k]$, here $x[k]$ means the $k^{th} $ entry of $x$. it need to be equals to $(\chi \times_1 x)_{ij}= \sum_k \chi_{kji}x[k]$.
But like making $\chi_{kji}=\chi_{kij}$ will work but that is restriction on the tensor and not the vector...
Obviously the zero vector will always work. However in general non-trivial solutions do not necessarily exist: if $d_1=1$, then the matrix part needs to be symmetric/anti-symmetric already. When $d_1>1$ the problem boils down to linear dependence of the slices of the tensor.
For clearer notation, let $\mathfrak X = (X_1,\ldots, X_K)$ be the 3d tensor and $v=(v_1, \ldots, v_K)$ be the vector. Let $A^+=\frac{1}{2}(A+A^T)$ and $A^-=\frac{1}{2}(A-A^T)$ denote the symmetric/anti-symmetric part of a matrix. Then we observe that
$$\begin{aligned} \sum_k X_k v_k \overset{!}{=} +\Big(\sum_k X_k v_k\Big)^T = +\sum_{k} X_k^T v_k \iff \sum_k X_k^-v_k = 0\\ \sum_k X_k v_k \overset{!}{=} -\Big(\sum_k X_k v_k\Big)^T = -\sum_{k} X_k^T v_k \iff \sum_k X_k^+v_k = 0 \end{aligned}$$
Hence, a non-trivial solution exists if and only if the symmetric/anti-symmetric parts of the slices $X_k$ are linearly dependent. Note that the dimension of the space of symmetric/anti-symmetric matrices is $\frac{1}{2}n(n+1)$ and $\frac{1}{2}n(n-1)$ respectively. Thus, if $d_1 > \frac{1}{2}n(n\pm 1)$ then a non-trivial solution always exists.