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We consider a matrix

$$M_{\mu} = \begin{pmatrix} 1 & \mu & 1 & 0 \\ -\mu & 1 & 0 & 1 \\ -1 & 0 & 0 & 0 \\ 0 &-1 & 0 & 0 \end{pmatrix}$$

One easily checks that $\operatorname{det}(M_{\mu})=1$.

I however noticed something peculiar:

Consider a sequence of real numbers $\mu_i$ then the four eigenvalues $\lambda_1,..,\lambda_4$ of $$ A=\prod_{i=1}^n M_{\mu_i}$$ have the property that they can be chosen to satisfy $\lambda_1 = \overline{\lambda_2}$ and $\lambda_3 = 1/\lambda_1$ and consequently by the determinant $\lambda_4 = 1/\lambda_2.$

I tried to explicitly study the product and see if I can see some structure explaining all this, but so far only with limited luck.

Question: Show that the eigenvalues of $A$ for every $n$ and $\mu_i \in \mathbb R$ satisfy $$\lambda_1 = \overline{\lambda_2 } = 1/\lambda_3 = 1/\overline{\lambda_4}.$$

Update: What we know so far is that $\lambda_1 = \frac{1}{\lambda_3}$ and $\lambda_2 = \frac{1}{\lambda_4}$ and if any of the eigenvalues is non-real then we have it. However, we cannot exclude $\lambda_1>\lambda_2$ both real.

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3 Answers 3

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The explanation is pretty simple with a suitable change of basis.

Letting $$B = \begin{pmatrix} 1 & 0 & 1 & 0 \\ i & 0 & -i & 0 \\ 0 & 1 & 0 & 1 \\ 0 & i & 0 & -i \end{pmatrix}$$ we have $$B^{-1}M_{\mu}B = \begin{pmatrix} 1+i\mu & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 1-i\mu & 1 \\ 0 & 0 & -1 & 0 \end{pmatrix}$$ Letting $N_\mu = \begin{pmatrix} 1+i\mu & 1 \\ -1 & 0 \\ \end{pmatrix}$, we thus have $$B^{-1}AB = \begin{pmatrix} \prod N_{\mu_i} & 0 \\ 0 & \overline{\prod N_{\mu_i}} \end{pmatrix}$$ where the bar denotes entry-wise complex conjugation. Thus the eigenvalues of $A$ are those of $\prod N_{\mu_i}$ plus those of $\overline{\prod N_{\mu_i}}$, which are their complex conjugates. Moreover, since $N_\mu$ has determinant $1$, so does $\prod N_{\mu_i}$, so its two eigenvalues are inverses of each other.

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    $\begingroup$ may I ask how you found that transformation? $\endgroup$
    – Dreifuss
    May 25, 2021 at 6:35
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    $\begingroup$ @AmazonPrime I computed the eigenvectors of $M_\mu$ and realized that, for any $\mu$, it has two eigenvectors in the subspace $x_1=ix_0, x_3=ix_2$ and two others in the subspace $x_1=-ix_0, x_3=-ix_2$. Then it natural to take as a basis the union of bases of these two subspaces, since it makes $M_\mu$ block-diagonal. $\endgroup$ May 25, 2021 at 13:09
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EDIT: just a partial answer that does not settle the question completely.

This is a variant of symplectic matrices. Your matrices $M_\mu$ are orthogonal wrt the indefinite scalar product induced by $$ \Omega = \begin{bmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & -1 & 0 & 0\\ -1 & 0 & 0 & 0 \end{bmatrix}, $$ i.e., they satisfy the property $$ M^{*}\Omega M = \Omega. $$

One can see that if $(\lambda, v)$ is an eigenpair for $M$, then $(\frac{1}{\lambda}, \Omega v)$ is an eigenpair for $M^*$, and hence $\frac{1}{\overline{\lambda}}$ is in the spectrum of $M$: $$ \frac{1}{\lambda}\Omega v = \frac{1}{\lambda}M^*\Omega Mv = \frac{1}{\lambda}M^*\Omega \lambda v = M^*\Omega v.$$

This property, together with the fact that real matrices have eigenvalues that come in complex conjugate pairs, puts strong constraints on which eigenvalues you can get.

There is also the possibility that the eigenvalues come in the form $\lambda_1,\lambda_2, \frac1{\lambda_1}, \frac1{\lambda_2}$, for $\lambda_1,\lambda_2\in \mathbb{R}$. This case actually happens: for instance, $M_2M_{-2}$ has four real eigenvalues, although in this case I get $\lambda_1 = \lambda_2$.

Four purely imaginary eigenvalues can also appear: for instance $M_{-2}M_{1/2}$ has eigenvalues $\{\pm 2i, \pm \frac12 i\}$.

EDIT: on further thought, I still cannot exclude that the eigenvalues are in the form suggested by Sascha in a comment; this is not a counterexample as the real eigenvalues have multiplicity 2. Sorry :(

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    $\begingroup$ could one not also have $1/\lambda_4=\lambda_1 >\lambda_2 = 1/\lambda_3$ in this case?-as an example. $\endgroup$
    – Sascha
    May 24, 2021 at 13:55
  • $\begingroup$ Incidentally, despite Wikipedia, for some of us this one is our favourite symplectic form. (Personally, I like it because it means that the upper triangular matrices in the symplectic group form a Borel subgroup.) $\endgroup$
    – LSpice
    May 24, 2021 at 14:00
  • $\begingroup$ I agree, I fail to see how one gets that the eigenvalues come in pairs (up to complex conjugation) $\endgroup$
    – Dreifuss
    May 24, 2021 at 14:08
  • $\begingroup$ @AmazonPrime Added a quick proof in the answer. $\endgroup$ May 24, 2021 at 14:51
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    $\begingroup$ @Sascha Good point. I thought I had solved this but on second thought I haven't. :( $\endgroup$ May 24, 2021 at 15:10
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Your matrix $M_\mu$ is symplectic: $M_\mu^T\Omega M_\mu=\Omega$ where $$\Omega=\begin{pmatrix} 0_2 & Y \\ -Y & 0_2 \end{pmatrix},\qquad Y=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$$ Then every product $A$ is still symplectic. This implies $A^T\Omega=\Omega A^{-1}$. because the spectrum of $A^T$, equal to that of $A$, is the inverse of that of $A^{-1}$, this shows that the spectrum is stable under $z\mapsto 1/z$. Because the matrix is real, the spectrum is invariant under $z\mapsto \bar z$ as well.

There remains to eliminate the possibility of a spectrum $\lambda_1,\lambda_2,\lambda_1^{-1},\lambda_2^{-1}$ with both $\lambda_j$ real... TODO

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  • $\begingroup$ thank you, as Federico Poloni showed in an example, it can happen that the eigenvalues are real, but in that case they should be degenerate. Why this happens seems to be however not clear so far. $\endgroup$
    – Dreifuss
    May 24, 2021 at 15:42
  • $\begingroup$ From this result, we know that if $\rho e^{i\theta}$ is an eigenvalue, then $\rho e^{-i\theta}$, $\rho^{-1} e^{i\theta}$ and $\rho^{-1} e^{-i\theta}$ are also eigenvalues. Then by continuity, if the eigenvalues are real, they are twice degenerated. $\endgroup$
    – Adam
    May 24, 2021 at 23:55
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    $\begingroup$ @Adam If this proof worked, then you could also use the same argument to prove that every real eigenvalue $\lambda$ of a real matrix $M$ has even multiplicity (when an obvious counterexample is $M = \begin{bmatrix}1 & 0\\ 0 & 3\end{bmatrix}$, for instance). Instead, what the structure tells us is that a small (structure-preserving) perturbation $M+\delta M$ must still have real eigenvalues, not to violate the symmetry constraints. $\endgroup$ May 25, 2021 at 6:17

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