Two ways of obtaining Dynamic Mode Decomposition modes - are they equivalent?

Question

In this lecture prof. Kutz gives the Dynamic Mode Decomposition modes:

$\Phi = X' V_r \Sigma_r^{-1} W $

which are the eigenvectors of the linear propagator matrix. This results from splitting the full data matrix $X$ into $X' = X(:,2:\text{end})$ and $X_1 = X(:,1:\text{end}-1)$. $U_r$, $\Sigma_r$ and $V_r$ are matrices resulting from the SVD of the data matrix $X_1$ and rank-$r$ truncation afterwards. $W$ is the approximation to the eigenvectors of the linear propagator matrix.

However, in the textbook by prof. Kutz Data-driven modeling and scientific computation, this formula is given for the DMD modes:

$\Phi = U_r W$

and this is also the formulation that was originally given by Peter Schmid.

I don't see them being equivalent equations, but perhaps I overlook something. To my understanding, they would be equivalent if in the first equation we used the rank-$r$ approximation matrix $X_{1 r} \approx X_1$ instead of $X'$:

$X_{1 r} V_r \Sigma_r^{-1} W = U_r \Sigma_r V_r^T V_r \Sigma_r^{-1} W = U_r W$

since $V_r$ is both orthogonal and orthonormal, and $\Sigma_r$ is a diagonal matrix.

Could you please explain the logic/intuition behind both approaches and if and how these two approaches differ from each other?

Answer 1

There's two issues here. First, are we using X(:,2:end) or X(:,1:end−1)? I have no idea.

Second, are we using the rank r approximation or the full matrix? Turns out it doesn't matter. The missing bit is actually 0. Let $USV^T$ be the full SVD of $X_1$. Let $S_r$, $U_r$, and $V_r$ be identical to $S$, $U$, and $V$ except that columns $r+1$ and onwards are set to 0. So, $U_rS_rV_r^T$ is a truncated SVD, but the individual matrices all stay at their original (non-truncated) dimensions. This allows me to show how they interact with one another. Let $S_{r+1}$, $U_{r+1}$ and $V_{r+1}$ be the remainders: they equal $S$, $U$ and $V$ respectively but with the first $r$ columns set to 0. By looking back at the outer product form of the SVD:

$$\sum_r u_r s_r v_r^T$$

... you'll hopefully see that

$$X_1 = U_{r} S_{r}V_{r}^T + U_{r+1} S_{r+1}V_{r+1}^T= X_{1r} + U_{r+1} S_{r+1}V_{r+1}^T$$.

The key to seeing why

$$X_{1}V_{r} = X_{1r}V_{r}$$

is that the remainder term, ending in $V_{r+1}^T$, will completely cancel with $V_r$.