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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.

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  • $\begingroup$ Thanks. Based on your example, if $T$ only has real components and is fully symmetric, then my argument with $M$ still holds and complex eigenvectors and eigenvalues are still impossible, no? $\endgroup$
    – Matt
    Commented Feb 16, 2021 at 4:25
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    $\begingroup$ Unless I'm misunderstanding something, $M$ is only real symmetric if you've already assumed $x$ is a real vector. A complex symmetric matrix can certainly have non-real eigenvalues. $\endgroup$
    – lambda
    Commented Feb 16, 2021 at 5:13
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    $\begingroup$ @Malkoun, although it's not explicitly stated in the body, the title question indicates that the question is about 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. $\endgroup$
    – LSpice
    Commented Feb 16, 2021 at 15:08
  • $\begingroup$ I misread. I will delete my comments. $\endgroup$
    – Malkoun
    Commented Feb 16, 2021 at 21:27

2 Answers 2

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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: it assumes the eigenvector is real, which as this counter example shows need not be the case.)

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    $\begingroup$ Isn't the original question about a complex eigenvalue, not just a complex eigenvector with real eigenvalue? $\endgroup$
    – LSpice
    Commented Feb 16, 2021 at 14:53
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    $\begingroup$ well, the argument provided by the OP implies both real eigenvalues and real eigenvectors for any real symmetric $T$; this counterexample invalidates that, doesn't it? $\endgroup$
    – Carlo Beenakker
    Commented Feb 16, 2021 at 14:57
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    $\begingroup$ @LSpice --- in that "cheap" example, the condition $\sum_i x_i^2=1$ forces $\lambda=\pm 1$; and it would also force $t=\pm 1$ in Zach Teitler's example. $\endgroup$
    – Carlo Beenakker
    Commented Feb 16, 2021 at 15:13
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    $\begingroup$ @ZachTeitler - the paper cited by the OP explicitly stresses that $x^2 =1$ is intended, not $\bar{x} x=1$. $\endgroup$
    – Michael Engelhardt
    Commented Feb 16, 2021 at 15:28
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    $\begingroup$ @ZachTeitler - OP is explicitly interested in the paper he cites, and there the normalization happens to be $x^2 =1$. We may deplore it, but that's what the OP is looking for. $\endgroup$
    – Michael Engelhardt
    Commented Feb 17, 2021 at 1:08
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Let us take $n=2$. Let $T_{112} = T_{121} = T_{211} = 1$, $T_{222} = \frac{43}{9}$ and $T_{ijk} = 0$ otherwise. Consider the vector

$\mathbf{x} = \left( \begin{array}{c} \frac{5}{4} \\ i \frac{3}{4} \end{array} \right)$.

Then, unless I made a calculation mistake, $\mathbf{x}$ is an eigenvector of $T$, whose sum of the squares of its components is $1$, and with eigenvalue

$\lambda = i\frac{3}{2},$

which is complex and not real.

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  • $\begingroup$ I deleted my comments. $\endgroup$
    – Zach Teitler
    Commented Apr 24, 2021 at 19:52

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