Suppose we have a real analytic family of matrices $A_t, B_t\in GL_n(\mathbb C)$, with $t\in \mathbb R$. Suppose that for $t\in (0,1)$ there is a common non-zero eigenvector $v_t\in \mathbb C^n\setminus 0$ such that $A_t(v_t)=B_t(v_t)=v_t$. Is it true that such a common (non-zero) vector exists for all $t\in \mathbb R$?
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$\begingroup$ 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. $\endgroup$– ChrisCommented Jun 23, 2021 at 17:56
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$\begingroup$ Chris, thanks, sorry, I fixed the misprint $\endgroup$– aglearnerCommented Jun 23, 2021 at 17:59
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3$\begingroup$ Hint: two matrices $A$ and $B$ have a common eigenvector with eigenvalue $1$ if and only if $\mathrm{det}((A-1)^* (A-1)+(B-1)^* (B-1))=0$. $\endgroup$– Mikael de la SalleCommented Jun 23, 2021 at 19:06
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2$\begingroup$ 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. $\endgroup$– aglearnerCommented Jun 23, 2021 at 22:18
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1 Answer
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Define the sets $$M = \{ (A,B,l)\in GL_n(\mathbb{C})\times GL_n(\mathbb{C})\times\mathbb{P}^{n-1}(\mathbb{C}) : Aw=Bw=w \text{ for all } w\in l \},$$ $$N = \{ (A,B)\in GL_n(\mathbb{C})\times GL_n(\mathbb{C}):\exists v\in\mathbb{C}^n\setminus\{0\}, Av=Bv=v\}, $$ where a point of $\mathbb{P}^{n-1}(\mathbb{C})$ is thought of as a line in $\mathbb{C}^n$. The set $M$ is closed and analytic (actually algebraic). Let $$\pi: GL_n(\mathbb{C})\times GL_n(\mathbb{C})\times\mathbb{P}^{n-1}(\mathbb{C})\to GL_n(\mathbb{C})\times GL_n(\mathbb{C})$$ be the projection. Then $\pi(M)=N$, and since $\pi$ is proper, the set $N$ is closed and analytic.
