= Solution
The first column is five times $(4,0,3,0)^\mathsf T/5$. Its <orthogonal> companion in the first-and-third-coordinate plane is $(3,0,-4,0)^\mathsf T/5$, and the second column has positive residual seven times that vector. The remaining coordinate directions can therefore be read off immediately rather than performing a general four-dimensional orthogonalization. One obtains
$$
\boxed{Q=\begin{pmatrix}4/5&3/5&0&0\\0&0&1&0\\3/5&-4/5&0&0\\0&0&0&1\end{pmatrix},\qquad R=\begin{pmatrix}5&1&5&2\\0&7&10&-11\\0&0&9&6\\0&0&0&1\end{pmatrix}.}
$$
The columns of $Q$ are visibly <orthonormal>. Taking their inner products with each column of the given <matrix> gives the displayed $R$, which also verifies $QR=A$ and positive diagonal entries.
Back to article page
