[proofplan] We exploit the fact that $Q$ is orthogonal to replace $\|Ax - b\|^2$ by $\|Rx - Q^\topb\|^2$. The block structure of $R$ then decomposes the residual into two independent components: $\|R_1x - c\|^2$ (controllable by choice of $x$) and $\|d\|^2$ (a fixed irreducible error). When $A$ has full column rank, $R_1$ is invertible and the minimiser is obtained by backward substitution. [/proofplan] [step:Use orthogonality of $Q$ to simplify the residual norm] Since $Q$ is orthogonal, $\|Qy\| = \|y\|$ for all $y$. Therefore \begin{align*} \|Ax - b\|^2 = \|QRx - b\|^2 = \|Q(Rx - Q^\topb)\|^2 = \|Rx - Q^\topb\|^2. \end{align*} [/step] [step:Decompose the residual using the block structure of $R$] Write $R = \begin{pmatrix} R_1 \\ 0 \end{pmatrix}$ with $R_1 \in \mathbb{R}^{n \times n}$ upper triangular and $Q^\topb = \begin{pmatrix} c \\ d \end{pmatrix}$ with $c \in \mathbb{R}^n$, $d \in \mathbb{R}^{m-n}$. Then \begin{align*} \|Rx - Q^\topb\|^2 = \left\|\begin{pmatrix} R_1x - c \\ -d \end{pmatrix}\right\|^2 = \|R_1x - c\|^2 + \|d\|^2. \end{align*} The term $\|d\|^2$ is independent of $x$, so the minimum over $x$ is achieved by minimising $\|R_1x - c\|^2$. The minimum residual norm is therefore $\|d\|$. [/step] [step:Solve for the unique minimiser when $A$ has full column rank] If $A$ has full column rank, then $R_1$ is non-singular: the diagonal entries of $R_1$ are nonzero because $\operatorname{rank}(A) = \operatorname{rank}(R) = \operatorname{rank}(R_1)$. Setting $R_1x^* = c$ gives the unique solution $x^* = R_1^{-1}c$, which is computed by backward substitution. [/step]