$
\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?
Best Answer
$\newcommand{\tp}{^{\mathsf{T}}}$ Looking at Watkins' text, the question you're trying to answer is to find the condensed SVD of A.
Here we have $A = uσv\tp$ and you want to find $u$ and $v$ subject to the isometry conditions:
The way I'd approach this is by looking at the columns of $uσv\tp$ individually i.e. for the first column: $σu_1 v_1 = 3$ & $σu_2 v_1 = 4$.
Solving for $u_1$ & $u_2$ and using $(1)$ we get: $σ^2 v_1^2 = 25$.
Similarly, we can use the other columns to get $σ^2 v_2^2 = 100$ & $σ^2 v_3^2 = 225$.
Adding these together gives $σ^2(v_1^2 + v_2^2 + v_3^2) = 350$ and using $(2)$ we get: $σ^2 = 350$.
Finally we can solve for our $v_1,v_2,v_3$ followed by our $u_1,u_2$ and you're done - this is how you get the condensed SVD by inspection. What's left to do is to compare to Theorem $4.1.12.$