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Let $\Omega_i\subseteq\mathbb R^d$ be bounded and open, $A_i$ denote the weak Laplacian with domain $\mathcal D(A):=\{u\in H_0^1(\Omega_i):\Delta u\in L^2(\Omega_i)\}$ on $L^2(\Omega_i)$ and $$T_i(t)f:=\sum_{n\in\mathbb N}e^{-\lambda^{i}_nt}\langle f,e_n^{i}\rangle_{L^2(\Omega_i)}e^{i}\;\;\;\text{for }f\in L^2(\Omega_i)\text{ and }t\ge0,$$ where $(e^i_n)_{n\in\mathbb N}$ and $(\lambda^{i}_n)_{n\in\mathbb N}$ are the eigenvectors and eigenvalues of $-A_i$, respectively.

We know that $(T_i(t))_{t\ge0}$ is a strongly continuous contraction semigroup on $L^2(\Omega_i)$ with generator $A_i$. In this paper it is claimed below equation 1.131 on p. 64 that if $U:L^2(\Omega_1)\to L^2(\Omega_2)$ satisfies $$\forall t>0:UT_1(t)=T_2(t)U\tag1,$$ then "$\Omega_1$ and $\Omega_2$ are isospectral", which I guess means that $\lambda^1_n=\lambda^2_n$ for all $n\in\mathbb N$.

How can we show that?

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    $\begingroup$ Well, if two operators are similar, they have the same eigenvalues. That's standard linear algebra, isn't it? $\endgroup$
    – Jochen Glueck
    Commented Feb 20, 2021 at 19:03
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    $\begingroup$ By the way, one should probably add that the linear operator $U$ in your question is bijective (in the paper you linked, it is even assumed to be a unitary). $\endgroup$
    – Jochen Glueck
    Commented Feb 20, 2021 at 19:05
  • $\begingroup$ @JochenGlueck Thank you for your comment. How do you define "similar" for operators and why does it follow from $(1)$ that $A_1$ and $A_2$ are similar? $\endgroup$
    – 0xbadf00d
    Commented Feb 20, 2021 at 19:38
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    $\begingroup$ Alternatively, one can argue directly via the generators (which is probably a bit clearer since it avoids the spectral mapping theorem): an unbounded operator $A_1$ on $X_1$ is said to be similar to an unbounded operator $A_2$ on $X_2$ iff there exists a a continuous linear bijection $V: X_1 \to X_2$ such that $V \mathcal{D}(A_1) = \mathcal{D}(A_2)$ and $A_1 = V^{-1} A_2 V$. In your situation, it follows immediately from (1) and from the definition of the generator of a $C_0$-semigroup that $A_1$ and $A_2$ are similar. $\endgroup$
    – Jochen Glueck
    Commented Feb 20, 2021 at 22:07
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    $\begingroup$ @JochenGlueck Thank you. Using the generators, it is really trivial. And I think we don't need the unitary assumption; $U$ can be any bijective bounded linear operator between Banach spaces. $\endgroup$
    – 0xbadf00d
    Commented Feb 21, 2021 at 10:46

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