# Condensed SVD decomposition of an outer product

Let $$A = uv^{T} \in \mathbb{R}^{m \times n}$$. Find the (condensed) SVD decomposition of $$A$$.

Theorem (Condensed SVD decomposition)

Let $$A \in \mathbb{R}^{n \times m}$$ be a non-zero matrix of rank $$r$$. Then there exists $$\hat{U} \in \mathbb{R}^{n \times r}, \hat{\Sigma} \in \mathbb{R}^{r \times r} , \hat{V} \in \mathbb{R}^{r \times m}$$ such that $$\hat{U}$$ and $$\hat{V}$$ are isometries , $$\hat{\Sigma}$$ is a diagonal matrix with main diagonal entries $$\sigma_{1} \ge \dots \ge \sigma_r > 0$$ and $$A = \hat{U} \hat{\Sigma}\hat{V}^{T}.$$

### Attempt at solution

Here the rank of $$A$$ is $$1$$ since each column is linearly dependent on $$u$$. So by the outer product form of the SVD decomposition , we get $$A = \sum_{j=1}^{r} \sigma_j u_j v_{j}^{T} = \sigma_1 u_1 v_j^{T} = \hat{U}\hat{\Sigma}\hat{V}^{T}.$$ It follows from the theorem above that $$A = \underbrace{\begin{pmatrix} u_1 \dots u_m\end{pmatrix}}_{\widehat{U}} \underbrace{\begin{pmatrix} \sigma_1 \end{pmatrix}}_{\widehat{\Sigma}} \underbrace{\begin{pmatrix} v_1 \\ \vdots \\ v_n\end{pmatrix}}_{\widehat{V}^{T}}.$$ To satisfy the initial equality ,we set $$\sigma_1 =1$$.

Could someone confirm my reasoning ? I feel like I am missing a step or something is wrong.

• Vectors in the DVD are normalized. You'll adjust $\sigma$ to scale the product. Jun 27, 2021 at 0:35

When $$r=1$$, then $$\hat{U}$$ is a single column and $$\hat{V}^\top$$ is a single row. You might try $$A=uv^\top = u \hat{\Sigma} v^\top$$ where $$\hat{\Sigma}$$ is just the $$1\times 1$$ matrix $$1$$, but the problem is that $$u$$ and $$v$$ aren't isometries because their columns are not normalized.
However, there is an easy fix. The SVD of $$A=uv^\top$$ is $$\hat{U} \hat{\Sigma} \hat{V}^\top$$ where $$\hat{U} = u/\|u\|$$, $$\hat{V} = v/\|v\|$$, and the single entry of $$\hat{\Sigma}$$ is $$\|u\|\|v\|$$.