 Solution

Under the permitted singular value bounds, $m=4n$ givesFor the symmetric positive-definite matrix $H=G^{T}G$ in part (iii), its eigenvalues are the squared singular values of $G$, soApply the supplied conjugate gradient convergence bound to $H$. For $e_k=x_k-x_{\ast }$ and nonzero $e_0$, the function $(\sqrt{\kappa }-1)/(\sqrt{\kappa }+1)$ increases with $\kappa \ge 1$, givingThus the error in this weighted norm contracts at a guaranteed exponential factor of at most $1/2$ per iteration, up to the prefactor $2$. Here $\Vert e{\Vert }_H=\Vert Ge{\Vert }_2$. The Pythagorean identity in part (iii) also identifies $\Vert e_k{\Vert }_H^2$ with the excess squared linear least-squares problem objective, which consequently decreases by a bound proportional to $4^{-k}$. For the Euclidean norm, the same eigenvalue bounds give $\Vert e_k{\Vert }_2/\Vert e_0{\Vert }_2\le 62^{-k}$.

For relative accuracy $0<ϵ<1$ in the $H$-weighted norm, it is sufficient to takeCompute $c=G^{T}b$ once in $O(mn)$ operations. Each iteration applies $p\mapsto G^T(Gp)$ with two dense matrix multiplications and uses $O(n)$ further arithmetic operations; when $m=4n$ this costs $O(n^2)$. Hence the total cost isUsing finite termination of the conjugate gradient method in exact arithmetic refines this to $O(n^2\min \{n,\lceil {\log }_2(2/ϵ)\rceil \})$. Explicitly constructing $G^{T}G$ first would add $O(m{n}^2)=O(n^3)$ setup work and is unnecessary. These accuracy estimates concern relative error; an absolute tolerance replaces $2/ϵ$ by $2\Vert e_0{\Vert }_H/ϵ$ when $\Vert e_0{\Vert }_H>ϵ$. If $e_0=0$, the initial guess already solves the problem.

As in part 2(b)(v), the Gaussian assumption alone does not make the printed singular value interval a certainty. For example, the event that every entry of $G$ has absolute value below a sufficiently small positive number has positive probability and forces $\Vert G{\Vert }_F<\sqrt{n}$, contradicting the asserted lower endpoint. The displayed rate is therefore conditional on the singular-value bounds authorized by the paper. The cost and convergence derivation are exact under that condition; an unconditional probabilistic statement would need an additional failure-probability estimate.