In :
%matplotlib inline
import numpy as np;
import matplotlib
import matplotlib.pyplot as plt

np.random.seed(1337)

kwargs = {'linewidth' : 3.5}
font = {'weight' : 'normal', 'size'   : 24}
matplotlib.rc('font', **font)

def error_plot(ys, yscale='log'):
plt.figure(figsize=(8, 8))
plt.xlabel('Step')
plt.ylabel('Error')
plt.yscale(yscale)
plt.plot(range(len(ys)), ys, **kwargs)

$\LaTeX \text{ commands here} \newcommand{\R}{\mathbb{R}} \newcommand{\im}{\text{im}\,} \newcommand{\norm}{||#1||} \newcommand{\inner}{\langle #1 \rangle} \newcommand{\span}{\mathrm{span}} \newcommand{\proj}{\mathrm{proj}} \newcommand{\OPT}{\mathrm{OPT}}$

Georgia Tech, CS 4540

L8: Duality, Convexity, & Postive Semidefiniteness¶

Jake Abernethy & Benjamin Bray

Thursday, September 13, 2018

Recall: Linear Program in Standard Form¶

\begin{align} \max_{x \in \R^n} &\quad c^T x \\ \text{such that} &\quad Ax \leq b & \text{(only $\leq$ constraints)} \\ &\quad x \geq 0 & \text{(variables nonnegative)} \end{align}
• Decision variables $x \in \R^n$
• Linear objective $c^T x$ for $c \in \R^n$
• Constraint matrix $A \in \R^{m \times n}$ and vector $b \in \R^m$. Each row corresponds to a constraint.

Recall: Activity Analysis Problem¶

A company has $m$ different resources which can be used to manufacture $n$ different products.

• $x = (x_1,\dots,x_n) \in \R^n$ is the amount of each product manufactured
• $r = (r_1,\dots,r_m) \in \R^m$ is the supply of each resource
• $c = (c_1,\dots,c_n) \in \R^n$ is the profit generated by each product
• $A = [a_{ij}] \in \R^{m \times n}$, where $a_{ij}$ is the amount of resource $i$ needed to make product $j$
$$\max_{x \in \R^n} \quad c^T x \quad \text{s.t.} \quad A x \leq r \quad\text{and}\quad x \geq 0$$

Problem: Interpreting the Dual¶

Problem: Write down the dual linear program for the activity analysis problem. What "units" do the dual variables need have? How can they be interpreted?

If it helps, assume the amount of each resource available is measured in kilograms, and our profit is measured in euros €. Maybe we're manufacturing rope, so the amount of each product is measured in meters.

Solution: Interpreting the Dual (1)¶

Using our rope factory example,

• $r_i$ is measured in $(kg)$
• $x_j$ is measured in $(m)$
• $c_j$ is measured in $(€/m)$
• $a_{ij}$ is measured in $(kg/m)$

Solution: Interpreting the Dual (2)¶

The dual program has the constraint $y^T A \geq c$, which has terms like $y_i a_{ij} \geq c_j$. The units are

$$\bigg( ? \bigg) \times \bigg( \frac{kg}{m} \bigg) \geq \bigg( \frac{€}{m} \bigg)$$
• For this to make sense, $y_i$ needs to have units $(€/kg)$ or euros per kilogram.
• Then the dual objective $y^T r$ has units $(€/kg)\times (kg) = (€)$.

Solution: Interpreting the Dual (3)¶

We decided that $y_i$ has units $(€/kg)$, and $y^T r$ has units $(€)$.

• Imagine that we also have the option of selling our resources directly to a buyer, for a price of $y_i$ euros per kilogram, rather than turning them into rope.
• This is only worth it to us if the resources used to make each rope are more valuable than the ropes themselves. This is the dual constraint!
• The buyer knows this, so he'll try to minimize the amount he pays for our raw materials by choosing as small $y_i$ as possible.
\begin{align} \min_{y \in \R^m} &\quad y^T r \\ \text{such that} &\quad y^T A \geq c^T & \text{(dominates objective)} \\ &\quad y \geq 0 & \text{(nonnegative combinations)} \end{align}

Farkas Lemma (v2) and Strong Duality¶

Lemma. (Farkas) Let $A \in \R^{m \times n}$ and $b \in \R^m$. Then exactly one of the following two statements is true:

1. There exists $x \in \R^n$ such that $Ax \geq b$ and $x \geq 0$.
2. There exists $y \in \R^m$ such that $y^T A \leq 0$ and $b^T y > 0$ and $y \geq 0$

Problem: Strong Duality¶

Prove that strong duality holds using the Farkas Lemma (v2). That is, the following two are equal:

\begin{align} \max_{x \in \R^n} &\quad c^T x & \text{such that} &\quad A x \leq b, \quad x \geq 0 \\ = \min_{y \in \R^m} &\quad y^T b & \text{such that} &\quad y^T A \geq c^T, \quad y \geq 0 \\ \end{align}

Positive Semidefinite Matrices¶

Definition: A matrix $M \in \R^{n\times n}$ is positive semidefinite (PSD) if $x^\top M x \geq 0\; \forall x \in \R^n$

Commonly, we work with matrices $M$ that are symmetric (that is, $M = M^\top$). In this case, the following are equivalent:

1. $M$ is PSD
2. The eigenvalues of $M$ are all $\geq 0$
3. The matrix $M$ can be written as $B^\top B$ for some $B \in \R^{n \times n}$

Problem¶

Show that (3) $\implies$ (1) $\implies$ (2)

Solution¶

• (3) $\implies$ (1): If $M = B^\top B$ then for any $x$ we have $$x^\top M x = x^\top (B^\top B) x = (x^\top B^\top) (B x) = (B x)^\top (B x) = \|B x\|_2^2 \geq 0$$
• (1) $\implies$ (2): If $M$ is PSD, then for any vector $x$ we have $x^\top M x \geq 0$. Let's take an vector $v$ of $M$, with eigenvalue $\lambda$, hence $M v = \lambda v$. We need to show that $\lambda \geq 0$. Notice that $v \ne 0$, hence, if we multiply on the left hand side by $v$, we use the fact that $M$ is PSD to get $$0 \leq v^\top M v = v^\top(\lambda v) = \lambda v^\top v = \lambda \| v\|^2.$$ The fact that $0 \leq \lambda \| v \|^2$ can only occur if $\lambda \geq 0$.

PSD matrices¶

We often use the notation $M \succcurlyeq 0$ to denote that $M$ is PSD. The notation $M \succcurlyeq N$ is equivalent to $M - N \succcurlyeq 0$, i.e. $M - N$ is PSD.

Problem A¶

Let $M \in \R^{n \times n}$ be an arbitrary symmetric matrix. Let $P \in \R$ be any nonsingular matrix. Show that $$M \succcurlyeq 0 \Longleftrightarrow P^\top M P \succcurlyeq 0$$

Problem B¶

Show that the set $S_+^n := \{ M \in \R^{n\times n}: M = M^\top, M \succcurlyeq 0 \}$ forms a convex cone.

Solutions¶

• (Problem A)
• First assume that $M \succcurlyeq 0$. Then $x^\top M x \geq 0$ for all $x$. Now take an arbitrary vector $y$, and observe that $y^\top (P^\top M P) y = (P y)^\top M (P y)$. Of course, $P y$ is just some arbitrary vector, hence $(P y)^\top M (P y) \geq 0$ as desired
• For the second direction, assume that for all $y$ we have $y^\top (P^\top M P) y \geq 0$. We need to show that for all $x$, $x^\top M x \geq 0$. Notice that $P$ is invertible, hence we can set $y = P^{-1} x$, and observe that $$0 \leq (P^{-1} x)^\top (P^\top M P) (P^{-1} x) = x^\top (P^{-1})^\top P^\top M P P^{-1} x = x^\top M x$$ as desired.
• (Problem B) Let us quickly check convexity. Let $N,M \in S_+^n$, and observe that for any $x$ we have $x^\top M x \geq 0$ and $x^\top N x \geq 0$. Hence, if we take a convex combination $\theta M + (1-\theta)$, with $\theta \in [0,1]$ we have $$x^\top (\theta M + (1-\theta) N) x = \theta x^\top M x + (1-\theta) x^\top N x$$ which is nonnegative since both terms are nonnegative

Convex Functions¶

A function $f : \R^d \to \R$ is convex if, for any $x,y \in \text{dom}(f)$ and any $\theta \in [0,1]$, we have $$f(\theta x + (1-\theta)y) \leq \theta f(x) + (1-\theta) f(y)$$

First-order Condition When $f$ is differentiable, then $f$ is convex if and only if $$f(y) \geq f(x) + \nabla f(x)^\top(y - x) \quad \text{ for any } x,y \in \text{dom}(f)$$

Problem¶

First-order Condition When $f$ is differentiable, then $f$ is convex if and only if $$f(y) \geq f(x) + \nabla f(x)^\top(y - x) \quad \text{ for any } x,y \in \text{dom}(f)$$

Use the first order condition to show that, for any convex and differentiable $f$, we have $$(\nabla f(y) - \nabla f(x))^\top(y - x) \geq 0 \text{ for any } x,y \in \text{dom}(f)$$

Let's apply the first order condition twice, once at $x$ and once at $y$: \begin{eqnarray*} f(y) & \geq & f(x) + \nabla f(x)^\top(y - x) \\ f(x) & \geq & f(y) + \nabla f(y)^\top(x - y). \end{eqnarray*} Add these two inequalities together, and subtract $f(x) + f(y)$ from both sides and you are done!
Second-order Condition When $f$ is twice differentiable, then $f$ is convex if and only if the hessian satisfies $\nabla^2f(x) \succcurlyeq 0$.
Find conditions under which the following function is convex $f(x) = \frac 1 2 x^\top A x$ for a symmetric matrix $A \in \R^{d \times d}$
As we computed earlier in lecture, for $f(x) = \frac 1 2 x^\top A x$, the hessian is $\nabla^2 f(x) = A$. We know that a twice-differentiable function is convex if its hessian is positive semi-definite. Therefore, $f$ is convex if and only if $A$ is positive semi-definite.