Let $T$ be a fully symmetric tensor of rank $3$ and size $N$.

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 $\sum_i x_i^2=1$.

In the literature [1] (top of page 4) it is said that the eigenvalues and the eigenvectors can be complex. I completely fail to see this. Here is my reasoning:

Since $T$ is fully symmetric, for all $i$, $j$, $k$ we have $T_{ijk}=T_{jki}=\dots=T_{kji}$. Let me define the matrix $M$ such that: \begin{equation} M_{ij}=\sum_k^N T_{ijk}x_k. \end{equation} Due to the symmetry of $T$, my matrix $M$ is also symmetric and the first equation can now be written as: \begin{equation} \sum_{j}^NM_{ij}x_j=\lambda x_i. \end{equation} All $\lambda$ and all $x$ are real since $M$ is symmetric. Therefore $T$ cannot have complex eigenvalues.

Is this correct? if not, could we find a counter-example? (for $N=3$ or $4$ for example)

[1] Gurau - On the generalization of the Wigner semicircle law to real symmetric tensors.

realsymmetric if you've already assumed $x$ is a real vector. A complex symmetric matrix can certainly have non-real eigenvalues. $\endgroup$real$T$, so the natural example you propose doesn't qualify. Indeed, if $T$ is real then $T(e_1, e_1)$ will also be real. $\endgroup$