Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 1 / 2A / iii / Solution

Rows two and three of $C$ both have zero in column one. Any rotation of these rows therefore preserves those zeros, so $\boxed{d_{21}=d_{31}=0}$. In column two, $d_{32}=c_{22}\sin {\theta }_3+c_{32}\cos {\theta }_3$. For $r_3=\sqrt{c_{22}^2+c_{32}^2}>0$, chooseThis gives $d_{32}=0$. Again choose ${\theta }_3=0$ if both relevant entries are zero. Thus the successive Givens rotations have eliminated all entries below the diagonal without recreating earlier ones.