# Eigenvalues of large symmetric random tensors

I am studying the eigenvalues of large random tensors and realise that very little is known about it. I was wondering what is already known and what could be potential leads to find their limiting distribution.

### Random symmetric tensor

Let $$\mathbf{T}$$ be a real fully symmetric tensor of order $$3$$ and size $$N$$. Its components can be represented as $$T_{ijk}\in \mathbb{R}$$ for all $$1\leq i,j,k\leq N$$. By symmetric I mean that if I permute any indices the value stays the same: $$$$T_{ijk}=T_{ikj}=\dots=T_{jki}$$$$

### Eigenvalues

One possible definition of eigenvalues: let $$x\in \mathbb{C}^N$$ and $$\lambda\in\mathbb{C}$$ such that: $$$$\sum_{jk}^NT_{ijk}x_kx_j=\lambda x_i \label{eq1}$$$$ with the constraint that $$\sum_i x_i^2=1$$.

It has been shown (in [1] for example) that for a tensor of order 3 and size $$N$$ there are $$2^{N}-1$$ eigenvalues $$\lambda$$. Perhaps an intuitive idea on why is to see that the eigenvalue equation is composed of $$N$$ quadratic equations. Each equation possesses two roots, hence $$2^N$$ possible combinations of solutions, minus the normalisation constraint.

It would be interesting to find the limiting distribution of eigenvalues of random tensors for large $$N$$. The eigenvalues are real only if their eigenvectors are real. I conjecture that the purely real eigenvalues scale with $$\sqrt{N}$$ whereas the general eigenvalues scale with $$N$$. That is to say, I would expect the real eigenvalues to be distributed on a closed non trivial interval when the elements of the tensor have a variance scaled with $$1/\sqrt{N}$$.

1. Reducing it to a non-linear matrix problem:

We define the matrix $$M$$ such that: $$M_{ij}=\sum_{k}T_{ijk}x_k$$. Therefore the eigenvalue problem would boil down to find the eigenvalues of $$M$$:

\begin{align} \sum_{jk}^NT_{ijk}x_kx_j&=\lambda x_i\\ \implies \sum_{j}M_{ij}x_j&=\lambda x_i \end{align}

Where the elements of $$M$$ now also depend on the eigenvectors.

2. Choosing an appropriate basis to uncouple the equations:

If we fix the index $$i$$, we can represent $$T$$ as a collection of $$N$$ matrices: $${M}^{(1)},{M}^{(2)}\dots{M}^{(N)}$$ where $$T_{ijk}={M_{jk}}^{(i)}$$.

Since $$T$$ is real and fully symmetric, $${M}^{(i)}$$ is also real symmetric $$(1\leq i\leq N)$$. There will also share some elements in common and not be fully independent from one another. For example $$T_{ijk}=T_{kij}\implies M_{ij}^{(1)}=M_{1j}^{(i)}$$. So matrix $$M^{(1)}$$ will be composed of the first column of all the other matrices. What is the relation between the eigenvalues of $$M^{(i)}$$ and $$M^{(j)}$$? Are they directly linked to the eigenvalues of $$T$$?

Here were some of the first ideas that came naturally to my mind. If you know a paper or some ideas on how to tackle this problem this would be very interesting.

[1]: Cartwright, D., & Sturmfels, B. (2013). The number of eigenvalues of a tensor. Linear algebra and its applications, 438(2), 942-952.

• Looking at the eigenvalues as the roots of random polynomials is something very new to me, thank you for your reference. On the other hand I studied the work of Razvan Gurau and the spectral density he derives in his paper does not seem to be directly linked with the eigenvalues. I computed the eigenvalues numerically for low ranks and size $\approx 20$ and the distribution and choice of scaling is noticeably different from his spectral density...