How does the outer product work in matrix approximation using SVD?

SVD factorizes a matrix $$A \in \mathbb{R}^{mxn}$$ such that $$A = U\Sigma V^T$$, quoting from Deisenroth et al:

Instead of doing the full SVD factorization, we will now investigate how the SVD allows us to represent a matrix $$A$$ as a sum of simpler (low-rank) matrices $$A_i$$, which lends itself to a matrix approximation scheme that is cheaper to compute than the full SVD. We construct a rank-1 matrix $$A_i \in \mathbb{R}^{mxn}$$ as: $$A_i := u_iv_i^T,$$ which is formed by the outer product of the ith orthogonal column vector of $$U$$ and $$V$$.

Each $$A_i$$ is then multiplied by its corresponding $$\sigma _i$$ in $$\Sigma$$, and all the rank-1 matrices are summed up to the chosen rank approximation.

I need to understand the intuition behind using the outer product, I know that each rank-1 matrix is weighted by its singular value (which is sorted in a descending order, so that the last few matrices in the summation have minimal weight), but I can't grasp how the outer product does the trick. So, any pointers?

Write the singular value decomposition as $$\begin{equation*} A = U\Sigma V^\top = \begin{bmatrix} u_1 & u_2 & \cdots & u_m\end{bmatrix}\begin{bmatrix}\tilde{\Sigma} & 0_{r\times (n-r)} \\ 0_{(m-r)\times r} & 0_{(m-r)\times(n-r)} \end{bmatrix}\begin{bmatrix} v_1^\top \\ v_2^\top \\ \vdots \\ v_n^\top \end{bmatrix} \end{equation*}$$ where $$\tilde{\Sigma} = \begin{bmatrix}\sigma_1 & 0 & \cdots & 0 \\ 0 & \sigma_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \sigma_r\end{bmatrix}$$. Then multiplying out the above expression yields $$\begin{equation*} A = \sum_{i=1}^r \sigma_i u_i v_i^\top. \end{equation*}$$ You can now drop terms from the above dyadic sum in order to approximate $$A$$ by its leading principal components. In fact, by the Eckart–Young–Mirsky theorem, we can show that the best rank-$$k$$ approximation of $$A$$ (in terms of the spectral norm and the Frobenius norm) is given by $$\begin{equation*} \hat{A}^* = \arg\min_{\text{rank}(\hat{A}) \le k}\|\hat{A} - A\| = \sum_{i=1}^k\sigma_i u_iv_i^\top. \end{equation*}$$ See here for a proof of this optimality.
• Thank you brenderson for the answer, I want to clarify that I am asking why the summation of the rank-1 matrices approximates the original matrix, I think my question is not clear. I am aware of Eckart–Young–Mirsky theorem, but my question is about the intuition behind $A_i = \sigma _i u_i v_i^T$
• The sum of all $r$ nonzero rank-1 matrices is an exact decomposition of the matrix $A$. There is no approximation when summing up all $r$ of these outer products. The approximation comes from deleting the $r-k$ terms with smallest influence, as determined by the size of their corresponding singular values. Mar 22 '20 at 7:57