 Solution

The columns outside indices $i,j$ are unchanged coordinate vectors. The remaining two columns have nonzero entries only in rows $i,j$, with the $2\times 2$ block $M(\theta )$. They have squared norm ${\cos }^2\theta +{\sin }^2\theta =1$ and mutual inner product zero; they are also orthogonal to all the unchanged columns. Hence $J^{T}J=I$.

Let $r=\sqrt{x_i^2+x_j^2}$. If $r>0$, choose $\theta \in [-\pi ,\pi ]$ with $\cos \theta =x_i/r$ and $\sin \theta =x_j/r$. ThenEvery other coordinate is unchanged. If $r=0$, choose $\theta =0$; the same conclusions hold. Therefore the procedure can be fixed with the nonnegative pivot conventionThis is a Givens rotation.