# Eigenvalues come in pairs

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.

• 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 $\lambda\mapsto\lambda^{-1}$. Jun 17, 2022 at 9:50
• @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 $J=VU.$ Jun 17, 2022 at 9:55

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 $$s$$.
• Hm, how does it follow? I see how it would follow from identities $A_i^{-1}=VA_i^{t}V^{-1}$ for $i=1,2$ Jun 17, 2022 at 8:55
• 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? Jun 21, 2022 at 21:15