Steepest descent (gradient method) for quadratic function Let the quadratic function F:Rn⟶R defined by: F(x)=12⟨Ax,x⟩−⟨c,x⟩,∀x∈Rn where A is a real symmetric positive definite (square) matrix, c∈Rn.

It holds Au⋅=c, so the gradient expresses as ∇F(x)=Ax−c=A(x−u⋅) The gradient update yields u1=u0−ρ0A(u0−u⋅) For the given choice u0=u⋅+αwm, this simplifies into u1=u⋅+αwm−αρ0λmwm By choosing the step size ρ0=1λm, the gradient descent method reaches the solution in one step!

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High School
Algebra
Algebra I
Quadratic function and equation

Kamren Franco

Answered question

2022-02-01

Steepest descent (gradient method) for quadratic function
Let the quadratic function $F : R^{n} ⟶ R$ defined by:
$F (x) = \frac{1}{2} ⟨ A x, x ⟩ - ⟨ c, x ⟩, \forall x \in R^{n}$
where A is a real symmetric positive definite (square) matrix, $c \in R^{n}$ .

Answer & Explanation

izumrledk

Beginner2022-02-02Added 15 answers

It holds

A u \cdot = c

, so the gradient expresses as

\nabla F (x) = A x - c = A (x - u^{\cdot})

The gradient update yields

u_{1} = u_{0} - ρ_{0} A (u_{0} - u^{\cdot})

For the given choice

u_{0} = u^{\cdot} + α w_{m}

, this simplifies into

u_{1} = u^{\cdot} + α w_{m} - α ρ_{0} λ_{m} w_{m}

By choosing the step size

ρ_{0} = \frac{1}{λ_{m}}

, the gradient descent method reaches the solution in one step!

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