# Let $A \in \mathbb{R}^{n \times m},$ by Singular Values Decomposition we know that $A = U\Sigma V^\top$, how to prove the reduced SVD?

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

• 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 $U,V$. Oct 24, 2020 at 7:41

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