Linear Least Square Optimization with Linear Equality Constraints

Question

What is the exact solution $x_{n \times 1}$ of the following constrained optimization problem \begin{align*} &\min \|A x - b\|^2 \\ s.t.& C x = 0 \end{align*} where $A$ is full column rank $m \times n$ matrix ($m>n$); $b$ is $m \times 1$ matrix; $C$ is full row rank $1 \times n$ matrix?

Answer 1

The Lagragian is the following:

$$ L = \sum \limits_{i = 1}^{m}(\sum \limits_{j = 1}^{2}a_{ij}x_{j} - b_{i})^{2} + \lambda \sum \limits_{j = 1}^{n}c_{i}x_{i} $$

$$ \frac{\partial L}{\partial x_{k}} = 2 \sum \limits_{j = 1}^{n}(\sum \limits_{i = 1}^{n}a_{ij}a_{ik})x_{j} - 2 n \sum \limits_{i = 1}^{m}b_{i} a_{ik} + \lambda c_{k} = 0 $$

There are $n + 1$ equations and $n + 1$ unknowns.

Answer 2

The problem is given by:

$$ \begin{alignat*}{3} \arg \min_{x} & \quad & \frac{1}{2} \left\| A x - b \right\|_{2}^{2} \\ \text{subject to} & \quad & C x = \boldsymbol{0} \end{alignat*} $$

The Lagrangian is given by:

$$ L \left( x, \nu \right) = \frac{1}{2} \left\| A x - b \right\|_{2}^{2} + {\nu}^{T} C x $$

From KKT Conditions the optimal values of $ \hat{x}, \hat{\nu} $ obeys:

$$ \begin{bmatrix} {A}^{T} A & {C}^{T} \\ C & 0 \end{bmatrix} \begin{bmatrix} \hat{x} \\ \hat{\nu} \end{bmatrix} = \begin{bmatrix} {A}^{T} b \\ \boldsymbol{0} \end{bmatrix} $$

Now all needed is to solve the above with any Linear System Solver.

Answer 3

$ \def\LR#1{\left(#1\right)} \def\qiq{\quad\implies\quad} $Since the constraint is linear it has an explicit solution in terms of the nullspace projector
$$\eqalign{ P &= I-C^+C \qiq P^T=P=P^2 \\ }$$ $$\eqalign{ Cx &= 0 \qiq x = Pu }$$ where a new unconstrained vector variable $u$ has been introduced.

In terms of $u,\,$ the objective function is likewise unconstrained $$\eqalign{ \phi &= \|APu-b\|^2 \\ }$$ Thus it can be immediately solved using a pseudoinverse $$\eqalign{ u &= \LR{AP}^+b \qiq x = P\LR{AP}^+b = \LR{I-C^+C}\LR{A-AC^+C}^+b \\ }$$ ${\bf NB\!:}\;$ This solution remains valid when $C$ is an arbitrary matrix, not simply a vector.