# Anti-diagonal matrix operator

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 $$\mathcal{X}$$.

• I seems to me that your matrix is diagonal rather than anti-diagonal (as suggested in the title). Do you mean $\mathcal{A} = \begin{pmatrix} 0 & A \\ A & 0 \end{pmatrix}$? Sep 26 '19 at 12:16
• yeah exactly thank you that's what I meant. (The diagonal case is direct) Sep 26 '19 at 12:30

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.