 Solution

A real square matrix is orthogonal if $Q^{T}Q=I$. Then $\Vert Qx{\Vert }^2=x^T{Q}^{T}Qx=x^{T}x=\Vert x{\Vert }^2$, proving length preservation.

Define $J(i,j,\theta )$ to be identity outside rows and columns $i,j$, with blockThe block's transpose times itself is identity because ${\cos }^2\theta +{\sin }^2\theta =1$ and the off-diagonal terms cancel. Thus $J$ is orthogonal, with determinant one.

For $r=\sqrt{x_i^2+x_j^2}>0$, choose $\cos \theta =x_i/r$, $\sin \theta =x_j/r$. Then $y_i=r$, $y_j=0$ and every other coordinate is unchanged. If $r=0$, choose $\theta =0$. This fixes the useful nonnegative pivot convention.

Apply the rotations to the current matrix in the displayed order from right to left. First $J(1,2)$ eliminates entry $(2,1)$; then $J(1,3)$ eliminates $(3,1)$ without changing row two. Finally $J(2,3)$ eliminates $(3,2)$ while preserving the zeros in column one. ThusInverting the product gives $A=QR$. Even if $A$ starts lower triangular, the row rotations generally create nonzero entries above the diagonal; there is no further universal simplification. For example $A=\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 1& 1& 1\end{array}\right)$ gives positive-pivotwhose three entries above the diagonal are all nonzero.

If $A$ is orthogonal, $R$ is both upper triangular and orthogonal. Orthogonality of its columns successively forces all off-diagonal entries to vanish and all diagonal entries to have absolute value one. The first two pivots are positive under our convention, and the rotations have determinant one. Therefore the precise result isWithout a specified pivot-sign convention, the invariant statement is a diagonal matrix with entries $\pm 1$ and product $\det A$.