= Solution
{c}
A real square <matrix> is orthogonal if $Q^TQ=I$. Then $\|Qx\|^2=x^TQ^TQx=x^Tx=\|x\|^2$, proving length preservation.
Define $J(i,j,\theta)$ to be identity outside rows and columns $i,j$, with block
$$
\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix}.
$$
The 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 \b[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. Thus
$$
\boxed{R=\begin{pmatrix}*&*&*\\0&*&*\\0&0&*\end{pmatrix},\qquad
Q=J(1,2)^TJ(1,3)^TJ(2,3)^T.}
$$
Inverting 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 $A=\left(\begin{smallmatrix}1&0&0\\1&1&0\\1&1&1\end{smallmatrix}\right)$ gives positive-pivot
$$
R=\begin{pmatrix}\sqrt3&2/\sqrt3&1/\sqrt3\\0&\sqrt{2/3}&1/\sqrt6\\0&0&1/\sqrt2\end{pmatrix},
$$
whose 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 is
$$
\boxed{R=\operatorname{diag}(1,1,\det A).}
$$
Without a specified pivot-sign convention, the invariant statement is a diagonal <matrix> with entries $\pm1$ and product $\det A$.
Back to article page
