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Jun 23, 2021 at 22:18 comment added aglearner Mikael, thanks! I got it. So the matrix $(A-1)^*(A-1)$ is non-negative definite, and it vanishes exactly on the kernel of $A-1$. Same for $(B-1)^*(B-1)$. Then indeed, if this equality holds for an interval, it holds for the whole $\mathbb R$. That's nice. What is also great is that this generalises to representations of any finitely generated group.
Jun 23, 2021 at 21:13 answer added Chris timeline score: 5
Jun 23, 2021 at 19:06 comment added Mikael de la Salle Hint: two matrices $A$ and $B$ have a common eigenvector with eigenvalue $1$ if and only if $\mathrm{det}((A-1)^* (A-1)+(B-1)^* (B-1))=0$.
Jun 23, 2021 at 17:59 comment added aglearner Chris, thanks, sorry, I fixed the misprint
Jun 23, 2021 at 17:58 history edited aglearner CC BY-SA 4.0
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Jun 23, 2021 at 17:56 comment added Chris You write $A_t(v_t)=B_t(v_t)=0$ for $v_t\neq 0$ but $A_t, B_t$ are invertible, thus their kernel is zero.
Jun 23, 2021 at 17:32 history asked aglearner CC BY-SA 4.0