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I've been given a matrix of the form

$$\begin{pmatrix} -1 & 1/a_{21} & \cdots & 1/a_{n1} \\ a_{21} & -1 &\cdots & 1/a_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & -1 \end{pmatrix}$$ which satisfies $$ \sum_{j=1}^n a_{ij} = 0,\space\space\forall\space i $$ and I've been asked to prove whether it can have complex eigenvalues when all its coefficients are real. I've been trying to throw properties of matrices and eigevalues to it for a week but I haven't gotten anything.

For $n = 2$, I know its eigenvalues are $0$ and $-2$, its determinant is $0$ and its trace is $-2$. For $n = 3$ it also has a $0$ determinant, but this time only after applying the extra property that the sum of all the coefficients of any column must be $0$. I haven't been able to prove whether the determinant should be $0$ for any natural $n$ or not.

I don't know if there's an easy way to prove this, maybe it's possible to prove they don't even without that extra constraint. Any hint or idea I could try would be very welcome!

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  • $\begingroup$ Without the equality constraint, this is a FX matrix minus the identity, right? $\endgroup$ Jul 19, 2023 at 8:30

2 Answers 2

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At the very least, the second condition is necessary. Randomly generated counterexample:

$$ A = \pmatrix{ -1. & 0.41745274 &-1.26280132\\ 2.39548077 & -1. & -0.62360058\\ -0.79189021 & -1.60359056 &-1. } $$ the eigenvalues of this matrix are $$ 1.28354084, \quad -2.14177042\pm 0.95442081i. $$ It is notable that for the case of $n=3$, the matrix necessarily has rank $3$ so that the only non-zero eigenvalue is equal to the trace (i.e. $3$).

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  • $\begingroup$ This doesn't use the 2nd constraint that 2nd constraint that $\Sigma_{j=1}^n a_{ij} = 0 $ $\endgroup$
    – IanJ
    Dec 8, 2020 at 21:40
  • $\begingroup$ @IanJ yes, that is implied by the first sentence $\endgroup$ Dec 8, 2020 at 22:49
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With WolframAlpha and some brute force, you can see that a matrix of that form whose sum of elements of each row is zero, has only real eigenvalues for $\mathbf{n\leq 4}$. Now, for $n=5$ I have constructed this counterexample: $$\left[ \begin{array}{ccccc} -1 & 1 & \frac{1}{2} \left(3-\sqrt{21}\right) & \frac{1}{2} \left(\sqrt{21}-1\right) & -1 \\ 1 & -1 & 1 & \frac{1}{10} \left(\sqrt{21}-11\right) & \frac{1}{10} \left(1-\sqrt{21}\right) \\ \frac{1}{6} \left(-3-\sqrt{21}\right) & 1 & -1 & 1 & \frac{1}{6} \left(\sqrt{21}-3\right) \\ \frac{1}{10} \left(1+\sqrt{21}\right) & \frac{10}{\sqrt{21}-11} & 1 & -1 & 1 \\ -1 & \frac{1}{2} \left(-1-\sqrt{21}\right) & \frac{1}{2} \left(3+\sqrt{21}\right) & 1 & -1 \\ \end{array} \right].$$ You can verify that this matrix is of the desired form and each of its rows sums zero. But, this matrix has these approximate eigenvalues $$-3.42668 \pm 0.743579 i.$$ On the other hand, these matrices always have determinant zero since $0$ is always an eigenvalue with associated eigenvector $(1,1,\ldots,1)$.

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