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For a matrix $Y(t) \in \mathbb{R}^{m \times n}$, its rank-r approximation could be represented in a factorized SVD-like form. $$ Y(t) = U(t) S(t) V^T(t), $$ where $U^{T}U = I_m$, $V^{T}V = I_n$ and $S \in \mathbb{R}^{r \times r}$.

I want to integrate the matrix $Y$ and have obtained the following iterative format: $$ U^{k+1} S^{k+1} V^{k+1} = U^{k} S^{k} V^{k} + (1 - (U^{k} S^{k} V^{k}).^2). \qquad \tag{1} $$

where $1-{}$ and $.^2$ denote elementwise operations. Then, can $U^{k+1}$, $S^{k+1}$, $V^{k+1}$ be represented by some combination of $U^{K}$, $S^{K}$, and $V^{k}$?

In fact, equation (1) could easily be rewritten." $$ Y^{k+1} = U^{k} S^{k} V^{k} + (1 - (U^{k} S^{k} V^{k}).^2). \tag{2} $$ then $$ [U^{k+1}, S^{k+1}, V^{k+1}] = svd(Y^{k+1}). $$ However, I don't want to conduct SVD decomposition at every iteration step. Is it possible to directly obtain $U^{k+1},S^{k+1},V^{k+1}$?

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  • $\begingroup$ Have you considered the dynamic low-rank approximation? See doi.org/10.1137/050639703, eq 2.8 $\endgroup$ Commented Mar 13 at 4:29
  • $\begingroup$ @StevenRoberts This is what I did. The low-rank approximation involves an iterative step, and I don't know if it is possible to update U, S, and V for equation (1) without an additional SVD decomposition. $\endgroup$
    – Owen Jun
    Commented Mar 13 at 4:42

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