 Solution

The linear least-squares problem is to choose $x\in {\mathbb{R}}^n$ minimizing the Euclidean norm of the residual:For a QR decomposition $A=QR$, orthogonality of $Q$ gives $\Vert Ax-b{\Vert }_2=\Vert Rx-Q^{T}b{\Vert }_2$. If $A$ has full column rank and $R=\left(\genfrac{}{}{0ex}{}{R_1}{0}\right)$ is in standard form with invertible upper-triangular $R_1\in {\mathbb{R}}^{n\times n}$, the minimizer solves $R_{1}x=(Q^{T}b{)}_{1:n}$.

Applying the Gram-Schmidt process to the columns of the given matrix yieldsIndeed, $Q$ is an orthogonal matrix and $A=QR$. Moreover,Back substitution in the leading triangular system givesso the unique least-squares solution isThe unused final transformed residual component is $1$, so the minimum residual norm is $1$.