Let $\Omega$ be an open set in $\mathbb{R^n}$. We consider the product Hilbert space $H=H^1_0(\Omega)\times L^2(\Omega)$ with the norm $$|(f,g)|^2=\int_\Omega (|\nabla f|^2+|f|^2+|g|^2 ) dx$$ We consider the operator $A:D(A)\to H$ defined by
$$A\begin{pmatrix} f \\ g \end{pmatrix} =\begin{pmatrix} 0 & I \\ \Delta-I & 0 \end{pmatrix}\begin{pmatrix} f \\ g \end{pmatrix}=\begin{pmatrix} g \\ \Delta f-f \end{pmatrix}$$ with domain $$D(A)=\left\{ \begin{pmatrix} f \\ g \end{pmatrix}\in H, \ \ \ \Delta f\in L^2(\Omega) \ \ and \ \ \ g\in H^1_0(\Omega) \right\}$$ How can we find the spectrum of this operator in the case $\Omega$ is bounded and in the case $\Omega=\mathbb{R^n}$. If I am not wrong, we can see that $A$ is a skew adjoint operator.
Because nobody is answering, I thought I'd offer a start. Your equation for the resolvent of $A$ is $$ (\lambda I -A)\left(\begin{array}{c}f \\ g\end{array}\right)= \left(\begin{array}{cc}\lambda & -I \\ I-\Delta & \lambda\end{array}\right) \left(\begin{array}{c}f \\ g\end{array}\right) =\left(\begin{array}{c}h \\ k\end{array}\right). $$ Equivalently, $$ \begin{align} \lambda f - g = h & \implies g=\lambda f - h\\ (I-\Delta)f +\lambda g = k & \implies (I-\Delta)f+\lambda^{2}f=k+\lambda h \end{align} $$ Given $h$, $k$, this amounts to solving for $f$ such that $$ -\Delta f+(1+\lambda^{2})f = k+\lambda h, $$ and then setting $g=\lambda f-h$. All of this comes down to knowing when $-1-\lambda^{2}\in\sigma(-\Delta)$. So it seems that $$ \sigma(A) = \pm i\sqrt{\sigma(-\Delta)+1}. $$ I'm being a little sloppy about the details.