12
$\begingroup$

Consider the two matrices with some parameter $s \in \mathbb R$

$$A_1= \begin{pmatrix} s& -1 &0& 0 \\1&0 &0&0 \\ 0&0&1&0 \\0&0&0&1 \end{pmatrix}$$ and $$A_2= \begin{pmatrix} s& -1 &-1& 0 \\1&0 &0&0 \\ -1&0&s&-1 \\0&0&1&0 \end{pmatrix}.$$

I then noticed that the eigenvalues of arbitrary products of $A_1$ and $A_2$, i.e. e.g. $A_1A_2A_1$ and $A_1A_1A_2A_1$ etc. all have eigenvalues $\lambda_1,1/\lambda_1$ and $\lambda_2, 1/\lambda_2.$

It is clear that the product of eigenvalues is equal to one, as both matrices are in $\text{SL}(4,\mathbb R)$, but I don't see why they have to come in two pairs that multiply up to one, respectively.

$\endgroup$
2
  • 7
    $\begingroup$ Likely, $A_1$ and $A_2$ belong to a classical group $G$, defined by the identity $A^TJA=J$ for some invertible $J$ (which you must find). Then $JAJ^{-1}=A^{-T}$ tells you that the spectrum if invariant under $\lambda\mapsto\lambda^{-1}$. $\endgroup$ Jun 17, 2022 at 9:50
  • 2
    $\begingroup$ @DenisSerre thanks. Based on Carlo Beenakker's answer, this is true as $A_1^{-1} = UA_1U$. Thus $A_1^{-T}=VA_1^{-1}V=VUA_1 UV.$ Thus $J=VU.$ $\endgroup$ Jun 17, 2022 at 9:55

1 Answer 1

13
$\begingroup$

This follows from the identities $$A_1^{-1}=UA_1U^{-1},\;\;A_2^{-1}=UA_2U^{-1},$$ $$A_1^{\top}=VA_1V^{-1},\;\;A_2^{\top}=VA_2V^{-1},$$ with $$U=U^{-1}=\left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{array} \right),\;\;V=V^{-1}=\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{array} \right).$$ Hence for any string of products $M=A_1^{n_1}A_2^{n_2}A_1^{n_3}A_2^{n_4}\cdots A_1^{n_N-1}A_2^{n_N}$ of the two matrices $A_1$ and $A_2$ one has $$M^\top = VU M^{-1} (VU)^{-1}.$$ It follows that if $\lambda$ is an eigenvalue of $M$, then also $1/\lambda$ is an eigenvalue: $${\rm det}\,(\lambda-M)={\rm det}\,(\lambda-M^\top)={\rm det}\,(\lambda-VU M^{-1}(VU)^{-1})={\rm det}\,(\lambda-M^{-1}),$$ which implies that $${\rm det}\,(\lambda-M)=0\Leftrightarrow {\rm det}\,(\lambda^{-1}-M)=0.$$ The case $\lambda=0$ is excluded because $A_1$ and $A_2$ are nonsingular for any $s$.

$\endgroup$
4
  • $\begingroup$ Hm, how does it follow? I see how it would follow from identities $A_i^{-1}=VA_i^{t}V^{-1}$ for $i=1,2$ $\endgroup$ Jun 17, 2022 at 8:55
  • $\begingroup$ I added the intermediate steps $\endgroup$ Jun 17, 2022 at 9:44
  • $\begingroup$ In the second-to-last displayed equation, is the middle term really what you mean, or should it be the transpose of the matrix $A_2^{n_N} A_1^{n_{N_1}} \dots A_2^{n_4} A_1^{n_3} A_2^{n_2} A_1^{n_1}$ you get by multiplying in the opposite order? $\endgroup$
    – Will Sawin
    Jun 21, 2022 at 21:15
  • $\begingroup$ @WillSawin --- I have added some further intermediate steps. $\endgroup$ Jun 22, 2022 at 6:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.