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](https://arxiv.org/abs/2004.02660).



Can the eigenvalues of a real symmetric tensor be complex?
Here is a counter example of a complex eigenvector: $N=3$, the nonzero elements of $T$ are $T\_{111}=2$, $T\_{122}=T\_{212}=T\_{221}=1$, $T\_{133}=T\_{313}=T\_{331}=1$. Eigenvectors with eigenvalue $2$ are $x=(1,iz,z)$, for any $z\in\mathbb{C}$.


(The flaw in the argument of the OP is indicated by [@lambda](https://mathoverflow.net/questions/384098/can-the-eigenvalues-of-a-real-symmetric-tensor-be-complex#comment977922_384098): it *assumes* the eigenvector is real, which as this counter example shows need not be the case.)


