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Let $A \in \mathbb{R}^{n \times m}$, by Singular Values Decomposition we know that $A = U\Sigma V^\top$, where $U \in \mathbb{R}^{n \times n} and V \in \mathbb{R}^{m \times m} and \Sigma \in \mathbb{R}^{n\times m}$.

If rank(A) = $r = number of \{i|\Sigma_{i,i} \neq 0\}$ How to prove that $A = \tilde{U}\tilde{\Sigma}\tilde{V}^\top$, where $\tilde{U} \in \mathbb{R}^{n\times r}, \tilde{V} \in \mathbb{R}^{m\times r}, and \tilde{\Sigma}=Diag(\sigma_1,...,\sigma_r) \in \mathbb{R}^{r\times r}$, i.e. the reduced SVD

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  • $\begingroup$ What is there to prove? in both cases using the definition of matrix multiplication we get $A = \sum_{i=1}^r \sigma_i u_i v_i^\top$, where $u_i,v_i$ are the columns of $U,V$. $\endgroup$
    – Theorem
    Oct 24, 2020 at 7:41

1 Answer 1

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Write it under block matrix form: $$U \Sigma V=(\tilde{U}|\tilde{U'})\begin{pmatrix}\tilde{\Sigma}&0\\ 0&0\end{pmatrix}\begin{pmatrix}\tilde{V}^\top\\\tilde{V'}^\top\end{pmatrix}$$

and expand it as if it were numbers (but without commutation!). Please note that, fortunately, all dimensions are compatible...

$$U \Sigma V=\tilde{U} \tilde{\Sigma} \tilde{V}^\top+\underbrace{\tilde{U} O \tilde{V'}^\top+ \tilde{U'} O \tilde{V}^\top +\tilde{U'} O \tilde{V'}^\top}_O.$$

Remark: the solution proposed by @Theorem is of the same vein.

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