$ \newcommand{\tp}{^\mathsf{T}} \newcommand{\R}{\mathbb{R}} \newcommand{\RR}[2]{\mathbb{R}^{#1 \times #2}} \newcommand{\diag}{\mathrm{diag}} \newcommand{\set}[1]{\left\{ #1 \right\}} \newcommand{\m}[1]{\begin{pmatrix} #1 \end{pmatrix}} $The Problem: We are given the matrix $$ A := \begin{pmatrix} 3 & 6 & 9 \\ 4 & 8 & 12 \end{pmatrix} $$ From this, we want to find - via inspection, and not any detailed calculations - the singular value decomposition of $A$. (Specifically, the reduced or condensed form, in the sense later elaborated upon.)
Context: This comes up as problem $4.1.14$ in Watkins' text Fundamentals of Matrix Computations. In the text at this point, we've not come up with any of the tools typically associated with SVDs. For instance, it has not yet been developed at this point that the singular values of $A$ are tied to the eigenvalues of $A\tp A$, for instance (which has made searching for help quite difficult). We are just given a few basic theorems and definitions...
Relevant Theorems & Definitions:
Definition (Orthogonal): We say a matrix $A \in \RR n n$ is orthogonal if $A\tp A = I$, or equivalently $A\tp = A^{-1}$.
Definition (Isometry): We say a matrix $A \in \RR n m$ is an isometry if $A \tp A = I$.
Theorem $4.1.1$: (the usual SVD theorem)
Let $A \in \RR n m$ be nonzero with rank $r$. Then we may write $$ A = U \Sigma V \tp $$ for $U \in \RR n n, V \in \RR m m$ orthogonal, and where $\Sigma = \diag(\sigma_1,\cdots,\sigma_r,0,\cdots,0)$ with $\sigma_1 \ge \cdots \ge \sigma_r > 0$.
Theorem $4.1.3$: (a geometric interpretation)
Let $A \in \RR n m$ be nonzero of rank $r$. Then $\R^m$ has an orthonormal basis $\set{v_i}_{i=1}^m$ and $\R^n$ has orthogonal basis $\set{u_i}_{i=1}^n$, and there are $\sigma_1 \ge \cdots \ge \sigma_r > 0$ such that $$ Av_i = \begin{cases} \sigma_1 u_i & i = 1,\cdots,r \\ 0 & \text{otherwise} \end{cases} \qquad A\tp u_i = \begin{cases} \sigma_1 v_i & i = 1,\cdots,r \\ 0 & \text{otherwise} \end{cases} $$
Theorem $4.1.10$: (the condensed SVD theorem)
Let $A \in \RR n m$ be nonzero and of rank $r$. Then $\exists U \in \RR n r, V \in \RR m r$ which are isometrics, and $\Sigma \in \RR r r$ where $\Sigma = \diag(\sigma_1, \cdots, \sigma_r)$ with $\sigma_1 \ge \cdots \ge \sigma_r > 0$, all such that $$ A = U \Sigma V \tp $$
Theorem $4.1.12$: (a possibly relevant alternate SVD form?)
Let $A \in \RR n m$ be nonzero of rank $r$, with $\set{\sigma_i}_{i=1}^r$ the singular values of $A$ (ordered in the usual way). Each $\sigma_i$ is associated with right singular vectors $\set{v_i}_{i=1}^r$ and left singular vectors $\set{u_i}_{i=1}^r$. Then we may write $$ A = \sum_{\ell = 1}^r \sigma_\ell u_\ell v_\ell \tp $$
Questions:
How in the world should the SVD of $A$ (as given at the outset) have an SVD that is obvious on inspection? (Again, keep in mind the limited tools I'm working with; essentially only the above.) How should I see it or look for it? Clearly, $\mathrm{rank}(A) = 1$, so we only have one singular value. I can set up the decomposition (though not determine the values): it takes the form $$ A = U \Sigma V \tp = \m{u_1 \\ u_2 } \m{\sigma_1} \m{v_1 & v_2 & v_3} $$ but what can I glean beyond this?