Can we construct a larger matrix $M$ such that its eigenvalues are the same as the eigenvalues of a tensor $T$ of order 3?
Let $\mathbf{T}$ be a fully symmetric tensor of order $3$ and size $N$. Its components can be represented as $T_{ijk}$ for all $1\leq i,j,k\leq N$. By symmetric I mean that if I permute any indices the value stays the same: \begin{equation} T_{ijk}=T_{ikj}=\dots=T_{jki} \end{equation}
Using the following definition of eigenvalues, let $x\in \mathbb{C}^N$ and $\lambda\in\mathbb{C}$ such that: \begin{equation} \sum_{jk}^NT_{ijk}x_kx_j=\lambda x_i \label{eq1} \end{equation} with the constraint that $\sum_i x_i^2=1$.
It has been shown (in [1] for example) that the number of eigenvalues$^*$ $\lambda$ of $T$ is $2^N-1$. I was wondering if is has been shown whether it is possible to construct a matrix $M$ of size $(2^N-1)\times( 2^N-1)$ such that the $( 2^N-1)$ eigenvalues of $M$ are the same as the eigenvalues $\lambda$ of $T$.
A few remarks:
- Although $T$ is fully symmetric, its eigenvalues can be complex.
- for every eigenvector $\mathbf{x}$ with its eigenvalue $\lambda$, we can see that $-\mathbf{x}$ is an eigenvector for $-\lambda$. The formula $2^N-1$ does not take this into account.
- $^*$: How I make intuitively sense of this result is that there are $N$ coupled quadratic equations with $x_i$ ($1\leq i\leq N$) variables. Therefore each $x_i$ has $2$ solutions and thus there are up to $2^N$ possible eigenvectors $\mathbf{x}\in\mathbb{C}^N$. Adding the normalization constraint would give $2^N-1$
I have not found any work related to this question. Perhaps it is possible to show that it is impossible or at least very hard to find such matrix? The motivation behind this would be that then we can use the spectral theorem and all the already well known properties of matrices in order to study the eigenvalues of $T$.
[1]: Cartwright, D., & Sturmfels, B. (2013). The number of eigenvalues of a tensor. Linear algebra and its applications, 438(2), 942-952.
