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Consider $N \times N$ matrices

$$A = \begin{bmatrix} 0 & 0 & \cdots & 0 & 1 \\ 1 & 0 & 0 & & 0 \\ \vdots & 1 & 0 & \ddots & \vdots \\ 0 & & \ddots & \ddots & 0 \\ 0 & 0 & \cdots &1 & 0 \\ \end{bmatrix}$$ and

$$B=\operatorname{diag}( \cos(2\pi\cdot 0/N),...,\cos(2\pi\cdot (N-1)/N)).$$

Does anybody know why the eigenvalues of $i(A+A^T)+2B$ are invariant under 90° rotations?- Numerics seem to imply this. What I mean by this is that if $\lambda$ is an eigenvalue, then also $e^{i \frac{\pi}{2}} \lambda.$

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  • $\begingroup$ This property of the eigenvalues does not seem to hold for $N=2$, $N=3$ or $N=4$ (I haven't tried any higher). Of course, for $N=2$ or $N=3$, this can only be true if all eigenvalues are zero (but they aren't). For $N=4$, I had some hope, but alas, no: The eigenvalues are $i\sqrt{3} $, $-i\sqrt{3} $, $0$, $0$. $\endgroup$
    – Michael Engelhardt
    Commented Jul 31, 2022 at 2:36
  • $\begingroup$ @MichaelEngelhardt sorry there was a 2 missing in front of $B$, now it should be fine. $\endgroup$
    – Sascha
    Commented Jul 31, 2022 at 13:46
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    $\begingroup$ Related? mathoverflow.net/questions/426279 $\endgroup$
    – Fred Hucht
    Commented Jul 31, 2022 at 14:33
  • $\begingroup$ The new factor 2 in front of $B$ makes things rather more pathological: For $N=2$ and $N=4$, the matrices now are deficient, consisting of $2\times 2$ Jordan blocks, all with eigenvalue 0; so yes, in a trivial sense, you could say that, if 0 is an eigenvalue, also $e^{i\pi /2} 0$ is an eigenvalue. For $N=3$, the eigenvalues are nonzero, and the claimed property doesn't (and can't) hold. $\endgroup$
    – Michael Engelhardt
    Commented Jul 31, 2022 at 14:54
  • $\begingroup$ @FredHucht - if it weren't for that factor $i$ in front of $A+A^T $, which completely changes things ... but if we leave out the factor $i$, the matrix is Hermitean, and the eigenvalues real, so the claimed property again can't arise ... $\endgroup$
    – Michael Engelhardt
    Commented Jul 31, 2022 at 15:00

1 Answer 1

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Assume $N$ is even (this is false when N is odd). Let $X=2B, Y=A+A^T$. Let $$P = \begin{bmatrix} 1 & 1 & 1 & \cdots & 1 \\ 1 & \zeta & \zeta^2 & \cdots & \zeta^{N-1} \\ 1 & \zeta^2 & \zeta^4 & \cdots & \zeta^{2(N-1)} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & \zeta^{N-1} & \zeta^{2(N-1)} & \cdots & \zeta^{(N-1)^2} \\ \end{bmatrix}$$ $$Q = \text{diag}(1, -1, 1, -1, \ldots, 1, -1)$$ where $\zeta=e^{\frac{2i\pi}{N}}$.

One can easily check that $$PXP^{-1}=Y$$ $$PYP^{-1}=X$$ $$QXQ^{-1}=X$$ $$QYQ^{-1}=-Y$$ from which it follows that $$P(X+iY)P^{-1}=Y+iX=i(X-iY)$$ $$Q(X+iY)Q^{-1}=X-iY$$ This shows that $X+iY$ is conjugate to $i(X+iY)$, so its eigenvalues are invariant under multiplication by $i$.

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