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Let $(A,\mathcal{D}(A))$ be an infinitesimal generator of a strongly continuous semigroup $(T(t))_{t\ge 0}$ on a Banach space $X$ and define on $\mathcal{X} := X \times X$ the operator matrix $$\mathcal{A}=\left( \begin{array}{cc} 0 & A \\ A & 0 \\ \end{array} \right)$$

with domain $\mathcal{D}(\mathcal{A}) := \mathcal{D}(A) \times D(A).$

I want to know if $\mathcal{A}$ generates a strongly continuous semigroup on the product space $\mathcal{X}$.

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    $\begingroup$ I seems to me that your matrix is diagonal rather than anti-diagonal (as suggested in the title). Do you mean $\mathcal{A} = \begin{pmatrix} 0 & A \\ A & 0 \end{pmatrix}$? $\endgroup$
    – Jochen Glueck
    Commented Sep 26, 2019 at 12:16
  • $\begingroup$ yeah exactly thank you that's what I meant. (The diagonal case is direct) $\endgroup$
    – Malik Amine
    Commented Sep 26, 2019 at 12:30

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The answer is no in general.

For a counterexample, let $A$ be your favourite semigroup generator that has a sequence of eigenvalues $(\lambda_n) \subseteq \mathbb{R}$ such that $\lambda_n \to -\infty$ (for instance, let $A$ be the Dirichlet or Neumann Laplace operator on $L^2(0,1)$).

If $f_n$ is an eigenvector for $\lambda_n$, then $(f_n,-f_n) \in \mathcal{D}(\mathcal{A})$ and $\mathcal{A}(f_n,-f_n) = -\lambda_n (f_n,-f_n)$. Hence, $\mathcal{A}$ has a sequence of eigenvalues that converges to $\infty$ and thus, it cannot be a semigroup generator.

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