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Starting from $x_0\in {\mathbb{R}}^n$, gradient descent repeatedly computes the gradient and updatesstopping when the gradient norm, step, or objective decrease is sufficiently small.

The Hessian bounds say that $f$ is $\alpha$-strongly convex and has $\beta$-smooth gradient. With $\eta =1/\beta$,Thus the iteration count isconvergence becomes slower linearly with the condition number $\kappa =\beta /\alpha$.

Forthe Hessian matrix is $\mathrm{diag}(2,200,20000)$, soTakeThenwhose Hessian is $2I$ and whose condition number is $1$.

The maximizer defining $g^{\ast }(x)$ is ${\mathrm{prox}}_{tf}(x)$, soConsequentlyThus the proximal point algorithm for $f$ is ordinary gradient descent with step $t$ on its smooth Moreau envelope.

The gradient descent iteration isFor the error $e^{(k)}=x^{(k)}-A^{-1}b$,Part (a) shows that convergence for every initial vector is equivalent to
$|1-\alpha {\lambda }_i|<1$ for every eigenvalue of $A$. Since all eigenvalues are positive,

Forthe Fréchet derivative in direction $h$ isThus the Hilbert-space gradient isThe gradient descent update with step size $\gamma$ is consequentlywhich is exactly Landweber iteration. Its stationary points solve the normal equation and minimize the convex quadratic functional $J$.

Gradient descent is $x_{t+1}=x_t-\eta \nabla f(x_t)$. Integrating the Hessian along a segment gives the descent lemma. Applying it to the hinted $z$ and convexity yields cocoercivity: $(\nabla f(x)-\nabla f(y){)}^T(x-y)\ge \Vert \nabla f(x)-\nabla f(y){\Vert }^2/\beta$. Apply this to $\varphi =f-\alpha \Vert x{\Vert }^2/2$, whose Hessian lies between $0$ and $(\beta -\alpha )I$, and rearrange to obtain the displayed strengthened inequality. With $y=x^{\ast }$, $\nabla f(x^{\ast })=0$, and $\eta =2/(\alpha +\beta )$, expansion of the squared update gives contraction factor $((\beta -\alpha )/(\beta +\alpha ){)}^2$ per step and the stated bound.

At $d=1000$, gradient descent and full BFGS are practical candidates; a dense Newton method can also be feasible, but its extra work and indefinite curvature need attention. The number of examples $n$, data sparsity and conditioning still matter.

For dense $A$, a gradient evaluation costs $O(nd)$ arithmetic, with $O(d)$ working storage beyond the data and accumulation buffers. Gradient descent is inexpensive per iteration and, with an appropriate line search, follows a descent direction, but poor conditioning can make it slow.

Forming the Hessian matrix explicitly costs $O(n{d}^2)$ arithmetic, and a dense Newton method solve costs $O(d^3)$. A $1000$-square double-precision matrix uses about $8$ MB, so storage alone is manageable, but repeated factorization may be expensive. Because the Hessian matrix can be indefinite or singular, an unmodified Newton step need not decrease the loss; a modified Hessian, damping or a trust region is appropriate.

Full BFGS uses gradients and an $O(d^2)$ curvature approximation, avoiding explicit second derivatives. Its rank-two updates and direction calculation cost $O(d^2)$ per step and require roughly the same matrix storage. A positive-definite initial approximation with safeguarded curvature updates and a suitable line search preserves descent directions. For this scale it is often a good compromise, without guaranteeing a global minimum.^[ref]

At $d=10^6$, gradient descent remains a plausible large-scale choice; dense Newton method and full-memory BFGS are generally impractical. A dense $d\times d$ double-precision matrix alone contains $10^{12}$ entries, about $8$ TB. Dense factorization also has order $10^{18}$ arithmetic work. Full BFGS avoids factorization but retains quadratic storage and update work, so it does not solve that scaling problem.

Gradient descent uses linear working storage in $d$, though the cost of reading the data and evaluating the gradient is still $O(nd)$ for dense $A$. Sparsity or streaming can be important. Limited-memory BFGS, a distinct limited-memory variant, stores only a fixed number $m$ of recent update pairs and uses $O(md)$ storage and update work, making it a realistic alternative to full BFGS.^[ref]

The available Hessian matrix factorization also permits a matrix-free Hessian-vector product:It costs $O(nd)$ for dense data without forming $H$. A suitable iterative, safeguarded Newton method can exploit this structure, so the conclusion rules out the dense implementation rather than every possible second-order algorithm.

These methods seek a local solution and require convergence checks appropriate to nonconvex optimization. The formula does not even ensure a finite minimizer exists: with one positive coefficient and nonzero $\mathbf{a}$, the positive exponential loss approaches zero along an unbounded direction without attaining it. Dimension alone cannot supply a convergence guarantee.

Use mathematical induction on the iteration. Initially $x_0=b=1\cdot b$. If $x_n={\lambda }_{n}b$, then $A{x}_n={\lambda }_{n}b/2$, $x_n^{T}b={\lambda }_n$, and $x_n^{T}A{x}_n={\lambda }_n^2/2$. ThereforeA gradient descent step with any scalar step length ${\alpha }_n$ consequently has the formThus every iterate remains on the same eigenline, withThis proves that gradient descent preserves an eigenline for a cubic product. The statement is invariance, not convergence to a minimum: along the negative direction of this line the cubic polynomial objective is unbounded below.

Gradient descent uses only the current gradient and a suitable step size; its additional storage is order $d$. It is attractive for very large $d$ or when low per-iteration cost matters most. For a smooth strongly convex function, standard step choices give linear convergence, with a rate sensitive to the condition number. Mere convexity gives a global-minimum interpretation but does not alone imply a linear rate; for nonconvex objectives the usual guarantee concerns stationary points.

BFGS uses successive gradients to update an approximation to inverse curvature, incorporating information that can correct poor scaling. Full BFGS uses order $d^2$ storage and update work. With suitable line searches and local smoothness and positive-curvature assumptions near a nondegenerate minimum, it has superlinear convergence. This often favours it in moderate dimensions when gradients are not extremely cheap relative to the extra work.

Neither gradient descent nor BFGS requires an explicitly evaluated Hessian matrix. Thus an expensive Hessian matrix is compatible with both, rather than a reason to reject BFGS. If an accurate Hessian matrix and its factorization are exceptionally cheap, Newton's method in optimization becomes an additional reasonable option. Convexity helps identify a stationary point as a global minimum; without it, both methods need suitable descent and curvature safeguards. The BFGS update learns curvature from the Quasi-Newton secant equation.^[ref]

For squared loss $E={\sum }_k(t_k-z_k{)}^2/2$, let $e_k=(t_k-z_k)g_k^{\prime }(x_k)$ and $e_j=g_j^{\prime }(x_j){\sum }_k{w}_{jk}{e}_k$. The chain rule gives $\partial E/\partial w_{jk}=-e_k{z}_j$ and $\partial E/\partial w_{ij}=-e_j{z}_i$. A gradient descent step of size $\eta$ uses $\Delta w_{jk}=\eta e_k{z}_j$ and $\Delta w_{ij}=\eta e_j{z}_i$, with every derivative evaluated at the same pre-update weights.

For a symmetric positive-definite matrix $A$, solving $A{x}_{\ast }=b$ is equivalent to minimizing the quadratic form with linear termWriting the error vector $e=x-x_{\ast }$ and expanding gives the exact energy norm identityThe cross term vanishes because $A{x}_{\ast }=b$. With $\eta =Se$ and $B=S^{-1}A{S}^{-1}$, the same quantity is ${\eta }^{T}B\eta /2$. Symmetric positive-definite preconditioning changes coordinates without changing the energy assigned to a corresponding error vector.

In an orthonormal eigenbasis of $A$, the level set $e^{T}Ae=t>0$ has semiaxis lengths $\sqrt{t/{\lambda }_i}$. Its longest-to-shortest axis ratio is $\sqrt{{\kappa }_2(A)}$. A large spectral condition number of a positive-definite matrix therefore appears geometrically as a thin ellipsoid. Good scaling makes the transformed ellipsoid rounder, allowing a simple search direction to improve all coordinates together.

The negative gradient of $F$ is the linear-system residual $r$. Relative to the norm $\Vert v{\Vert }_M=\sqrt{v^{T}Mv}$, a steepest direction is $z=M^{-1}r$: the Cauchy-Schwarz inequality applied to $r^{T}v=(S^{-1}r{)}^T(Sv)$ shows that the greatest decrease per unit $M$-length occurs along this direction. Equivalently, ordinary gradient descent in $y=Sx$ becomes preconditioned steepest descent in $x$.^[ref]