Condensed SVD decomposition of an outer product

Question

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.

Answer 1

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 $\|u\|\|v\|$.