In the dynamic mode decomposition, we consider a sequence of data vectors $\{z_0, \dots, z_m\}$ where $z_k \in \mathbb{R}^n$ for all $n$. We assume that the data satisfies the linear relationship $z_{k+1} =Az_k$ for some matrix $A$. We define the $n \times m$ matrices $X = [z_0 \: \cdots \: z_{m-1}]$ and $Y = [z_1 \: \cdots \: z_m]$, which satisfies $Y = AX$, so we let $A = YX^+$.
We use the SVD to compute a rank $r$ approximation $X_r = U_r \Sigma_r V_r^*$, and we define the matrices $$A_r = YX_r^+ = YV_r \Sigma_r^{-1}U_r^*$$ $$\tilde{A}_r = U_r^*A_rU_r = U_r^*YV_r \Sigma_r^{-1}$$
then $A_r$ and $\tilde A_r$ have the same non-zero eigenvalues, and we assume that $A_r$ will be a reasonable substitute for $A$.
This is nice and all, but I haven't found much information on what assumptions we need about the data for $A_r$ to be a reasonable substitute for $A$. In fact, $\|X^+-X_r^+\| = \frac{1}{\sigma_s}$ in the spectral norm, where $\sigma_s$ is the smallest nonzero singular value of $X$. Moreover, we could define $X_r = LR^*$ for any full rank $n \times r$ and $m \times r$ matrices $L$ and $R$ as our "low rank approximation" of $X$. Then we may define $A_r$ and $\tilde A_r$ analogously, and they will also have the same nonzero eigenvalues. However, in this case $A_r$ is clearly not a suitable substitute for $A$ for arbitrary $L$ and $R$.
So what are the important properties that our data vectors $\{z_0, \dots, z_m\}$ should have so that $A_r$ would be a reasonable substitute for $A$ in the dynamic mode decomposition? Similarly, what are the properties that our low rank approximation should have?
