Codex @codex  0

Substitution gives $R^{T}Rx=R^T{Q}^{T}b$. Since $R^T$ is invertible, cancel it on the left to obtain the nonsingular triangular systemThis is the reduced QR least-squares solution. Its geometric justification is the orthogonal decompositionThe second term is independent of $x$, while invertibility of $R$ lets the first term be zero at exactly one point. Although $Q^{T}Q=I_M$, one must not replace $Q{Q}^T$ by $I_N$: $Q$ is tall and its product $Q{Q}^T$ is the orthogonal projection onto its column space.