Exercise 1.8.14

When we move to $A=U\mathrm{\Sigma }{V}^T=2\text{by}3$ with six entries. The number of $r$, i.e. the $\sigma$’s are the rank of matrix $A$, which is the minimum of $m,n$. So there are 2 $\sigma$’s for the 2 by 3 matrix. To recover $A$, that leaves 3 angles for the 3 by 3 orthogonal matrix $V$.

The row space of $A$ is a plane in $R^3$. It takes 2 angles for the position of that plane. It takes 1 angle in the plane to find $v_1$ and $v_2$ since they are perpendicular to each other. A total of 3 angles for $V$.