Skip to content
nonconvexcard

Gradient descent with exact line search on a quadratic

ProofAdvanced⏱ 1 hourOptimizationline searchkantorovichquadraticrate

Problem

For f(x)=12x⊤Axf(x) = \frac12x^\top Ax with μI⪯A⪯LI\mu I\preceq A\preceq LI, gradient descent with exact line search chooses ηt=arg⁡min⁡ηf(xt−η∇f(xt))\eta_t = \arg\min_\eta f(x_t-\eta\nabla f(x_t)).

  1. Derive the closed form for ηt\eta_t.
  2. Show that the rate is f(xt+1)≤(κ−1κ+1)2f(xt)f(x_{t+1}) \le \left(\frac{\kappa-1}{\kappa+1}\right)^2f(x_t) (the Kantorovich bound).
  3. Explain why exact line search does not remove the κ\kappa dependence.

Write a complete argument. Every inequality you use should be one you could state precisely.

Solution

Opening this before you have a real attempt on paper is the fastest way to feel like you understand something you cannot reproduce.

Related problems

All problems
SolvedProblemFieldTypeDifficultyTime
The exact step-size threshold on a quadraticOptimizationProofBeginner15 min
Linear convergence under strong convexityOptimizationProofIntermediate30 min
Measure the covariance estimation rate runHigh-Dim StatisticsCodingIntermediate1 hour
Non-convex gradient descent reaches an approximate stationary pointOptimizationProofIntermediate30 min
SGD on convex Lipschitz objectivesSGDProofIntermediate30 min