What if we apply resolvent analysis to the linear system identified by DMD?
The first two steps remain the same
Step 1: Rank-r truncation $$ X=U_r\Sigma_r V_r^H $$
Step 2: System matrix - we make the distinction that $A_d=\exp(A\Delta t)$ is for discrete-time system $$ A_d = X^1 V_r\Sigma_r^{-1}U_r^H \equiv BU_r^H $$ and the reduced system matrix $$ \tilde{A}_d = U_r^H A_d U_r = U_r^H B $$
For next step, let's now consider the more general Jordan form, i.e., when $\tilde{A}_d$ is not necessarily diagonalizable,
Using exact DMD formulation,
One can verify that the primal and adjoint DMD modes are orthogonal to each other $$ \begin{align} \Psi^H\Phi = (U_rS)^H (BW) J_d^{-1} = S^H \underbrace{U_r^H B}_{\tilde{A}_d} W J_d^{-1} = J_d \underbrace{S^H W}_{I} J_d^{-1} = I \end{align} $$
Now that one obtains a linear system from DMD, it is natural to apply resolvent analysis to extract out the gain characteristics over the spectrum. Due to the special structure in DMD operator, the resolvent can be computed efficiently.
We consider the following system with arbitrary forcing $$ \dot{x} = Ax+f $$ with DMD modes $\Phi$ and adjoint modes $\Psi$.
Understanding that the dynamics is happening in a reduced subspace spanned by the DMD modes, represent the states and forcing as $$ x=\Phi a,\quad f=\Phi b $$ Then $$ \Phi \dot{a} = A\Phi a + \Phi b = \Phi J a + \Phi b $$ and left-multiply by $\Psi^H$ $$ \dot{a} = J a + b $$
As usual, for a harmonic forcing $$ b = \hat{b}\exp(i\omega t) $$ and by Fourier Transform $$ \hat{a} = (i\omega I-J)^{-1} \hat{b} $$ In the full-order space $$ \hat{x}=\Phi\hat{a},\quad \hat{f}=\Phi\hat{b} $$
Note that $$ \norm{\Phi a} = \sqrt{a^H\Phi^H\Phi a} = \sqrt{a^H F^HF a} = \norm{F a} $$ where we used Cholesky decomposition $\Phi^H\Phi=F^HF$; $F$ is upper-triangular and invertible.
We want to maximize the gain $$ G = \max_{\hat{f}} \frac{\norm{\hat{x}}}{\norm{\hat{f}}} $$ The ratio of norms can be simplified $$ \frac{\norm{\hat{x}}}{\norm{\hat{f}}} = \frac{\norm{\Phi\hat{a}}}{\norm{\Phi\hat{b}}} = \frac{\norm{F\hat{a}}}{\norm{F\hat{b}}} = \frac{\norm{F(i\omega I-J)^{-1}\hat{b}}}{\norm{F\hat{b}}} = \frac{\norm{F(i\omega I-J)^{-1}F^{-1}(F\hat{b})}}{\norm{F\hat{b}}} $$ Then $$ G = \max_{F\hat{b}} \frac{\norm{F(i\omega I-J)^{-1}F^{-1}(F\hat{b})}}{\norm{F\hat{b}}} $$
Should not be surprising now, the optimal solution is given by the SVD $$ F(i\omega I-J)^{-1}F^{-1} = U(\omega)\Sigma(\omega)V(\omega)^H $$ by taking the maximum singular value $\sigma_1(\omega)$ and the associated singular vectors $u_1(\omega)$ and $v_1(\omega)$
But be careful as we optimized over $F\hat{b}$, we need to convert back to the reduced-order system $\hat{b}$, then the full-order system $\hat{f}$. The input and output modes are, respectively $$ V_f(\omega) = \Phi F^{-1} V(\omega),\quad U_f(\omega) = \Phi F^{-1} U(\omega) $$
Nevertheless, the accurate characterization of the input and output modes using data-driven DMD really relies on the data quality
For highly non-normal cases