# Can the eigenvalues of a real symmetric tensor be complex?

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  (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)

• 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?
– Matt
Feb 16, 2021 at 4:25
• 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. Feb 16, 2021 at 5:13
• @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. Feb 16, 2021 at 15:08
• I misread. I will delete my comments. Feb 16, 2021 at 21:27

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.)

• Isn't the original question about a complex eigenvalue, not just a complex eigenvector with real eigenvalue? Feb 16, 2021 at 14:53
• 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? Feb 16, 2021 at 14:57
• @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. Feb 16, 2021 at 15:13
• @ZachTeitler - the paper cited by the OP explicitly stresses that $x^2 =1$ is intended, not $\bar{x} x=1$. Feb 16, 2021 at 15:28
• @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. Feb 17, 2021 at 1:08

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.

• I deleted my comments. Apr 24, 2021 at 19:52