Linear Constraints

$\underset{x \in \R^n}{\lim} f(x)$ subject to $\underbrace{Ax=b}_{\text{undetermined system}}$

• effectively an “unconstrained problem”

• geometric introduction Lagrange Multipliers

• Connection to QR decomposition

Matrices $A=m \times n$, $b=m \times 1$

• $m$ constants; $n$ variables
• $rank(A)=m$ (full-rank)

  • $f:\R^n \to \R$ continuous differentiable once or twice
  • Feasible Set

    • $F=\{x \mid Ax=b\}=\phi \Leftrightarrow b\in Range(A)$

    

Equivalent representation of feasible set

$F=\{x \mid Ax=b\}=\{\bar{x}+Zp \mid p \in \R^{n-m}\}$,

$\begin{aligned}\text{where } & \bar{x}\text{ is any “particular” solution for }A\bar{x}=b,\\ & Z \text{ is any basis for }Null(A) (Z \text{ is }n\times (n-m) \text{ s.t. }AZ=0),\\& p\text{ along the null basis}\end{aligned}$

Equivalent Reduced Problem

$\underset{p \in \R^{n-m}}{\lim} f(\bar{x}+Zp) \equiv \underset{x \in \R^{n}}{\lim}{f(x)\mid Ax=b}$

• $\bar{x}+Zp^* \equiv x^*$

Example:

$$\lim \frac{1}{2}(x_1^2+x_2^2) \text{ s.t. } x_1+x_2=1$$

eliminate e.g., $x_2=1-x_1$ substitute

$\underset{x_1 \in \R}{\lim}\frac{1}{2}(x_1^2+(1-x_1)^2)$

$\begin{aligned}\text{set derivative = 0 ⇒}x_1-(1-x_1) &=0 \\ 2x_1-1 &= 0\\ x_1 &=\frac{1}{2} \end{aligned}$

$x_2 = \frac{1}{2} ⇒ (x_1,x_2)=(\frac{1}{2},\frac{1}{2})$

Optimality Conditions for “Reduced” Problem

$$\underset{p \in \R^{n-m}}{\lim} f(\bar{x}+Zp) \equiv f_Z(p)$$

1-st order necessary optimal condition $\nabla_p f_Z(p)=0$

$0=\underbrace{\nabla_p f_Z(p)}_{(n-m)\times 1}=\underbrace{Z^T}_{(n-m)\times n}\underbrace{\nabla f(\bar{x}+Zp)}_{n\times1}$

⇒ $p^*$ is a solution only if $\nabla_p f_Z(p)=0 \Leftrightarrow Z^T\nabla f(\bar{x}+Zp^*)=0$

let $x^* := \bar{x}+Zp^*$ implies

$Z^T\nabla f(x^*)=0 \Leftrightarrow \nabla f(x^*) \in Null(Z^T)$

Four Fundamental Subspaces Associated with $A$ and $Z$

$range(A^T)\oplus Null(A)=\R^n$

$Null(Z^T)\oplus range(Z)=\R^n$

$\begin{aligned}\nabla f(x^*) \in Null(Z^T) &\Leftrightarrow \nabla f(x^*) \in range(A^T) \\ &\Leftrightarrow \exist y \in \R^m\text{ s.t. } \nabla f(x^*)=A^Ty\end{aligned}$

First-order Necessary Condition for

$\underset{x\in \R^n}{\lim}{f(x)\mid Ax=y}$(p)

• A point $x^*$ is a local minimizer (stationary) of ($p$) only if there exists a vector $y \in \R^m$ s.t.:

  $\left \{\begin{aligned}&\nabla f(x^)=A^Ty &, &\text{ optimality} \\ &Ax^=b&,&\text{ feasibility}\end{aligned}\right.$

  *Note: $y$ sometimes referred to as “Lagrange Multipliers”

Lagrangian Function

$$L(x,y)=f(x)-y^T(Ax-b)$$

find a stationary pair $(x,y)$ of $L$:

$0=\nabla_x L(x,y)=\nabla f(x)-A^Ty$

$0=\nabla_y L(x,y)=-(Ax-b)$

** $\Leftrightarrow Z^T \nabla f(x^)=0 \Leftrightarrow \nabla f(x^*)^Tp=0, p\in Null(A)$

Example Least-Norm Solutions for Underdetermined

Linear-Systems

$\lim \lVert x\rVert_2$ subject to $Ax=b$ (underdetermined)

Take $f(x)=\frac{1}{2}\lVert x\rVert_2^2$, $\nabla f(x)=x$

First-order optimality: $(x,y)$ satisfies

$X_{LS}=(A^TA)^{-1}A^Tb$ (over-determined) normal equation

Example: min-norm to $x_1+x_2+\cdots+x_n=1$

$m=1$

$x_1+x_2+\cdots+x_n=\underbrace{\begin{bmatrix}1&1&\cdots&1\end{bmatrix}}_A\underbrace{\begin{bmatrix}x_1 \\ x_2\\ \vdots \\ x_n\end{bmatrix}}_x=\underbrace{1}_b$

$A=e^T\equiv \begin{bmatrix}1&1&\cdots&1\end{bmatrix}$

$x=e(e^Te)^{-1}=e(w)^{-1}=\frac{1}{w}e=\begin{bmatrix}\frac{1}{w} \\ \vdots \\ \frac{1}{w}\end{bmatrix}$

Second-order Necessary Conditions $x^*$ is a local minimizer only if

$\left \{ \begin{aligned}\nabla^2 f_Z(x^)\ge 0 \\ Ax^=b\end{aligned}\right.$