Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 1 / 2A (Vectors and Matrices)

Left multiplication by the given Givens rotation mixes rows two and three, so $b_{31}=a_{21}\sin {\theta }_1+a_{31}\cos {\theta }_1$. If $r_1=\sqrt{a_{21}^2+a_{31}^2}>0$, chooseEquivalently ${\theta }_1=atan2(-a_{31},a_{21})$ modulo $2\pi$. Substitution gives $b_{31}=0$ and $b_{21}=r_1$. If $a_{21}=a_{31}=0$, take ${\theta }_1=0$; the desired zero is already present. This also covers singular matrices without division by zero.

The second Givens rotation leaves row three unchanged, so $\boxed{c_{31}=b_{31}=0}$. Its other relevant entry is $c_{21}=b_{11}\sin {\theta }_2+b_{21}\cos {\theta }_2$. For $r_2=\sqrt{b_{11}^2+b_{21}^2}>0$, takeThen $c_{21}=0$ and $c_{11}=r_2$. If both entries vanish, choose ${\theta }_2=0$. The first column now has zeros below its diagonal.

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.

Each plane rotation satisfies $R_j^T{R}_j=I$. Hence $U=R_3{R}_2{R}_1$ is an orthogonal matrix and $D=UA$ is an upper triangular matrix. Multiplication by $U^T$ gives the QR decompositionThe zero-pair cases already handled make this construction valid for every real $3\times 3$ matrix, including rank-deficient matrices. Orthogonality follows from the product identity, not from any invertibility assumption on $A$.