Last modified: 2026-10-31 15:22:31 (PDT)
Source
This section is adapted from the Lagrange-multiplier part of Zhou [1], used under the MIT License (see the license text in Linear Algebra). The source solves the stationarity equations of the Lagrangian and reports the solution; here the solution is also proved to be the unique minimizer, directly from the positive definiteness of the matrices involved. The source assumes only that \(\mathbf{S}\) is positive definite; the added hypothesis that the columns of \(\mathbf{A}\) are linearly independent makes \({\mathbf{A}}^{\top} \mathbf{S}^{-1} \mathbf{A}\) invertible. The source’s example with general coefficients \(a_1, a_2, b\) is specialized here to \(3, 4, 5\). These parts of the source are not part of this section:
Definition 1 (Feasible set) In a problem that looks for a best point among some allowed points of \(\mathbb{R}^p\), the set \(F \subseteq \mathbb{R}^p\) of allowed points is the feasible set, [2] and its points are feasible.
Definition 2 (Constraints) When the feasible set \(F\) (Definition 1) is the set of points that satisfy some equations or inequalities, those equations or inequalities are the constraints.
Example 1 (A line as a feasible set) With the single constraint \(3 x_1 + 4 x_2 = 5\), the feasible set is the line \(\mathopen{}\left\{\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}\in \mathbb{R}^2 : 3 x_1 + 4 x_2 = 5\right\}\mathclose{}\). \({(1, 0.5)}^{\top}\) is feasible, since
\[ \begin{aligned} 3 \cdot 1 + 4 \cdot 0.5 &= 3 + 2 \\ &= 5; \end{aligned} \]
\(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{0}}\vphantom{0}}\) is not feasible, since \(3 \cdot 0 + 4 \cdot 0 = 0 \ne 5\).
Definition 3 (Objective function, objective) Let \(f : \mathbb{R}^p \to \mathbb{R}\) and let \(F \subseteq \mathbb{R}^p\) be a feasible set (Definition 1). In the problem of minimizing \(f\) over \(F\), \(f\) is the objective function (or objective).
Definition 4 (Minimizer over a feasible set) With \(f\) and \(F\) as in Definition 3, a minimizer of \(f\) over \(F\) is a feasible \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^*\) with \(f(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^*) \le f(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}})\) for every feasible \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}\): a global minimizer of \(f\) with its domain restricted to \(F\).
Definition 5 (Minimization problem) With \(f\) and \(F\) as in Definition 3, the minimization problem [3] is to find a minimizer of \(f\) over \(F\) (Definition 4). When \(F\) is given by constraints (Definition 2), the problem is to minimize \(f\) subject to them.
Example 2 (Minimizing over a line) In the problem of minimizing \(x_1^2 + x_2^2\) subject to \(3 x_1 + 4 x_2 = 5\), the objective function is \(f(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}) = x_1^2 + x_2^2\), and the feasible set is the line of Example 1. The feasible point \({(1, 0.5)}^{\top}\) has
\[ \begin{aligned} f(1, 0.5) &= 1 + 0.25 \\ &= 1.25. \end{aligned} \]
\(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{0}}\vphantom{0}}\) minimizes \(f\) over all of \(\mathbb{R}^2\), with \(f(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{0}}\vphantom{0}}) = 0\), but it is not a minimizer over the feasible set, because it is not feasible.
Definition 6 (Optimization algorithm) Let:
An optimization algorithm [3] is a step-by-step procedure that maps the pair \((f, F)\) to a feasible point \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^\dagger\):
\[ (f, F) \mapsto \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^\dagger \in F. \tag{1}\]
In words: given \((f, F)\), the procedure returns a feasible point \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^\dagger\) intended to make \(f(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^\dagger)\) small.
Example 3 (Exhaustive search over a finite feasible set) Let \(f(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}) = x_1^2 + x_2^2\), with the feasible set the three points \(\mathopen{}\left\{{(1, 0.5)}^{\top}, {(3, -1)}^{\top}, {(-1, 2)}^{\top}\right\}\mathclose{}\) of the line of Example 1. The procedure “evaluate \(f\) at every feasible point; return the point with the smallest value” is an optimization algorithm. Computing \(f\) at each point gives
\[ \begin{aligned} f(1, 0.5) &= 1 + 0.25 = 1.25, \\ f(3, -1) &= 9 + 1 = 10, \\ f(-1, 2) &= 1 + 4 = 5, \end{aligned} \]
so it returns \({(1, 0.5)}^{\top}\), a minimizer of \(f\) over this feasible set.
Here the objective (Definition 3) is a quadratic \(\tfrac{1}{2}\,{\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}\) with \(\mathbf{S}\) positive definite, and the constraints (Definition 2) are \(m\) linear equations \({\mathbf{A}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}= \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}\), where \(\mathbf{A}\) is a \(p \times m\) matrix and \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} \in \mathbb{R}^m\). The only stationary point of the objective alone (Definition 4 in Critical Points and Optimality Conditions) is \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{0}}\vphantom{0}}_{p \times 1}\) (its gradient is \(\mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}\), and \(\mathbf{S}\) is invertible), which may not satisfy the constraints. A Lagrange multiplier adds one unknown per constraint, and the minimizer over the feasible set (Definition 1) is then found among the stationary points of a new function of the extended set of unknowns.
Definition 7 (Lagrangian) For the problem of minimizing \(f(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}})\) over \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}\in \mathbb{R}^p\) subject to \({\mathbf{A}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}= \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}\), with \(\mathbf{A}\) a \(p \times m\) matrix and \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} \in \mathbb{R}^m\), the Lagrangian [4] is the function of \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}\in \mathbb{R}^p\) and \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}\in \mathbb{R}^m\)
\[ L(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}, \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}) \stackrel{\text{def}}{=}f(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}) + {\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}}^{\top} \mathopen{}\left({\mathbf{A}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}- \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}\right)\mathclose{}. \]
Definition 8 (Lagrange multipliers) In the Lagrangian \(L(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}, \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}})\) (Definition 7), the entries of \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}\) are the Lagrange multipliers [4], one per constraint.
Example 4 (The gradients of the Lagrangian of a quadratic) Let \(f(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}) = \tfrac{1}{2}\,{\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}\) with \(\mathbf{S}\) symmetric and constant. By the transpose of a product, \({\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}}^{\top} {\mathbf{A}}^{\top} = {(\mathbf{A} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}})}^{\top}\), so
\[ \begin{aligned} L(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}, \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}) &= \tfrac{1}{2}\,{\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}} + {\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}}^{\top} {\mathbf{A}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}- {\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} && \text{(distribute)} \\ &= \tfrac{1}{2}\,{\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}} + {(\mathbf{A} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}})}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}- {\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}, && \text{(transpose of a product)} \end{aligned} \]
and, since \({\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}}^{\top} \mathopen{}\left({\mathbf{A}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}- \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}\right)\mathclose{}\) is \(1 \times 1\) and so equals its transpose,
\[ L(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}, \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}) = \tfrac{1}{2}\,{\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}} + {\mathopen{}\left({\mathbf{A}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}- \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}\right)\mathclose{}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}. \]
The partial derivatives of a sum are the sums of the partial derivatives, entry by entry. Using the first of these forms, derivative of a quadratic form for the first term, derivative of a linear combination for the second (\(\mathbf{A} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}\) is constant in \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}\)), and constant with respect to a vector for the third,
\[ \begin{aligned} \frac{\partial}{\partial \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}} L(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}, \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}) &= \tfrac{1}{2}\,\frac{\partial}{\partial \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}} \mathopen{}\left({\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}\right)\mathclose{} + \frac{\partial}{\partial \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}} \mathopen{}\left({(\mathbf{A} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}})}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}\right)\mathclose{} - \frac{\partial}{\partial \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}} \mathopen{}\left({\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}\right)\mathclose{} && \text{(sum and constant multiple)} \\ &= \tfrac{1}{2} \mathopen{}\left(2 \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}\right)\mathclose{} + \mathbf{A} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}- \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{0}}\vphantom{0}}_{p \times 1} && \text{(the three derivatives)} \\ &= \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}+ \mathbf{A} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}. && \text{(simplify)} \end{aligned} \]
Using the second form, in which \(f(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}})\) and \({\mathbf{A}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}- \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}\) are constant in \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}\), and derivative of a linear combination again,
\[ \begin{aligned} \frac{\partial}{\partial \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}} L(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}, \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}) &= \frac{\partial}{\partial \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}} f(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}) + \frac{\partial}{\partial \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}} \mathopen{}\left({\mathopen{}\left({\mathbf{A}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}- \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}\right)\mathclose{}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}\right)\mathclose{} && \text{(sum)} \\ &= \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{0}}\vphantom{0}}_{m \times 1} + \mathopen{}\left({\mathbf{A}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}- \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}\right)\mathclose{} && \text{(the two derivatives)} \\ &= {\mathbf{A}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}- \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}. && \text{(simplify)} \end{aligned} \]
So \((\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}, \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}})\) makes the gradients equal \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{0}}\vphantom{0}}_{p \times 1}\) and \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{0}}\vphantom{0}}_{m \times 1}\) exactly when
\[ \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}+ \mathbf{A} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}= \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{0}}\vphantom{0}}_{p \times 1} \quad \text{and} \quad {\mathbf{A}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}= \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}, \qquad\text{that is,}\qquad \underbrace{\begin{bmatrix} \mathbf{S} & \mathbf{A} \\ {\mathbf{A}}^{\top} & \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{0}}\vphantom{0}}_{m \times m} \end{bmatrix}}_{(p + m) \times (p + m)} \underbrace{\begin{bmatrix} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}\\ \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}\end{bmatrix}}_{(p + m) \times 1} = \underbrace{\begin{bmatrix} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{0}}\vphantom{0}}_{p \times 1} \\ \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} \end{bmatrix}}_{(p + m) \times 1}, \tag{2}\]
the second equation being the constraint itself.
Theorem 1 (Minimizing a positive definite quadratic under linear constraints) Let \(\mathbf{S}\) be a \(p \times p\) positive definite matrix, \(\mathbf{A}\) a \(p \times m\) matrix with linearly independent columns, and \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} \in \mathbb{R}^m\). Then the \(m \times m\) matrix \(\mathbf{M} \stackrel{\text{def}}{=}{\mathbf{A}}^{\top} \mathbf{S}^{-1} \mathbf{A}\) is invertible, and \(f(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}) = \tfrac{1}{2}\,{\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}\) has exactly one minimizer over the \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}\in \mathbb{R}^p\) with \({\mathbf{A}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}= \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}\), namely
\[ \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* = \mathbf{S}^{-1} \mathbf{A} \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}. \]
With \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}^* \stackrel{\text{def}}{=}-\mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}\), the pair \((\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^*, \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}^*)\) solves Equation 2, and the minimum value is \(f(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^*) = \tfrac{1}{2}\,{\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}}^{\top} \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}\).
Proof. The steps marked “transpose of a product” use the transpose of a product, and \(\mathbf{S}\) is symmetric because it is positive definite.
\(\mathbf{M}\) is invertible. \(\mathbf{S}^{-1}\) exists and is positive definite (positive definite inverse), so \(\mathbf{M} = {\mathbf{A}}^{\top} \mathbf{S}^{-1} \mathbf{A}\) is positive definite (operations that preserve definiteness, part 1), and so invertible (positive definite inverse again). \(\mathbf{S}^{-1}\) and \(\mathbf{M}^{-1}\) are symmetric (inverse of a symmetric matrix), because \(\mathbf{S}\) and \(\mathbf{M}\) are.
\(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^*\) satisfies the constraint, and the pair solves Equation 2.
\[ \begin{aligned} {\mathbf{A}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* &= {\mathbf{A}}^{\top} \mathbf{S}^{-1} \mathbf{A} \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} && \text{(substitute } \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* \text{)} \\ &= \mathbf{M} \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} && \text{(definition of } \mathbf{M} \text{)} \\ &= \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}, && \text{(} \mathbf{M} \mathbf{M}^{-1} = \mathbf{I}_m \text{)} \\ \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* + \mathbf{A} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}^* &= \mathbf{S} \mathbf{S}^{-1} \mathbf{A} \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} - \mathbf{A} \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} && \text{(substitute } \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* \text{ and } \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}^* \text{)} \\ &= \mathbf{A} \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} - \mathbf{A} \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} && \text{(} \mathbf{S} \mathbf{S}^{-1} = \mathbf{I}_p \text{)} \\ &= \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{0}}\vphantom{0}}_{p \times 1}. && \text{(subtract)} \end{aligned} \]
\(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^*\) is the unique minimizer. Let \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}\) satisfy \({\mathbf{A}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}= \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}\), and let \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}} \stackrel{\text{def}}{=}\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}- \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^*\), so
\[ \begin{aligned} {\mathbf{A}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}} &= {\mathbf{A}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}- {\mathbf{A}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* \\ &= \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} - \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} \\ &= \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{0}}\vphantom{0}}_{m \times 1} \end{aligned} \]
(matrix products distribute; both points satisfy the constraint). First, the cross term vanishes:
\[ \begin{aligned} {\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* &= {\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}}}^{\top} \mathbf{S} \mathbf{S}^{-1} \mathbf{A} \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} && \text{(substitute } \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* \text{)} \\ &= {\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}}}^{\top} \mathbf{A} \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} && \text{(} \mathbf{S} \mathbf{S}^{-1} = \mathbf{I}_p \text{)} \\ &= {({\mathbf{A}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}})}^{\top} \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} && \text{(transpose of a product; } {({\mathbf{A}}^{\top})}^{\top} = \mathbf{A} \text{)} \\ &= {\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{0}}\vphantom{0}}_{m \times 1}}^{\top} \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} && \text{(} {\mathbf{A}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}} = \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{0}}\vphantom{0}}_{m \times 1} \text{)} \\ &= 0. && \text{(a product with a zero factor is zero)} \end{aligned} \]
Then
\[ \begin{aligned} f(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}) &= \tfrac{1}{2}\,{(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* + \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}})}^{\top} \mathbf{S} (\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* + \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}}) && \text{(} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}= \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* + \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}} \text{)} \\ &= \tfrac{1}{2} \mathopen{}\left({\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^*{}}^{\top} + {\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}}}^{\top}\right)\mathclose{} \mathbf{S} (\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* + \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}}) && \text{(the transpose of a sum is the sum of the transposes)} \\ &= \tfrac{1}{2} \mathopen{}\left({\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^*{}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* + {\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^*{}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}} + {\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* + {\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}}\right)\mathclose{} && \text{(distribute)} \\ &= \tfrac{1}{2} \mathopen{}\left({\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^*{}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* + {\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}}}^{\top} {\mathbf{S}}^{\top} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* + {\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* + {\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}}\right)\mathclose{} && \text{(} 1 \times 1 \text{: } {\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^*{}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}} = {({\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^*{}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}})}^{\top} \text{; transpose of a product)} \\ &= \tfrac{1}{2} \mathopen{}\left({\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^*{}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* + 2\,{\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* + {\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}}\right)\mathclose{} && \text{(} {\mathbf{S}}^{\top} = \mathbf{S} \text{; combine)} \\ &= \tfrac{1}{2}\,{\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^*{}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* + {\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* + \tfrac{1}{2}\,{\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}} && \text{(distribute } \tfrac{1}{2} \text{)} \\ &= f(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^*) + \tfrac{1}{2}\,{\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}}}^{\top} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}} && \text{(the cross term is } 0 \text{)} \\ &\ge f(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^*), && \text{(} \mathbf{S} \text{ is positive definite)} \end{aligned} \]
with equality only when \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{d}}\vphantom{d}} = \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{0}}\vphantom{0}}_{p \times 1}\), that is, \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}= \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^*\).
The minimum value.
\[ \begin{aligned} f(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^*) &= \tfrac{1}{2}\,{(\mathbf{S}^{-1} \mathbf{A} \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}})}^{\top} \mathbf{S} \mathbf{S}^{-1} \mathbf{A} \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} && \text{(substitute } \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* \text{)} \\ &= \tfrac{1}{2}\,{\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}}^{\top}\,{(\mathbf{M}^{-1})}^{\top}\,{\mathbf{A}}^{\top}\,{(\mathbf{S}^{-1})}^{\top}\,\mathbf{S} \mathbf{S}^{-1} \mathbf{A} \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} && \text{(transpose of a product, applied repeatedly)} \\ &= \tfrac{1}{2}\,{\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}}^{\top} \mathbf{M}^{-1} {\mathbf{A}}^{\top} \mathbf{S}^{-1} \mathbf{S} \mathbf{S}^{-1} \mathbf{A} \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} && \text{(} \mathbf{S}^{-1} \text{ and } \mathbf{M}^{-1} \text{ are symmetric)} \\ &= \tfrac{1}{2}\,{\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}}^{\top} \mathbf{M}^{-1} {\mathbf{A}}^{\top} \mathbf{S}^{-1} \mathbf{A} \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} && \text{(} \mathbf{S}^{-1} \mathbf{S} = \mathbf{I}_p \text{)} \\ &= \tfrac{1}{2}\,{\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}}^{\top} \mathbf{M}^{-1} \mathbf{M} \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} && \text{(definition of } \mathbf{M} \text{)} \\ &= \tfrac{1}{2}\,{\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}}^{\top} \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}. && \text{(} \mathbf{M}^{-1} \mathbf{M} = \mathbf{I}_m \text{)} \end{aligned} \]
Example 5 (The closest point on a line to the origin) Minimize \(x_1^2 + x_2^2\) subject to \(3 x_1 + 4 x_2 = 5\). In the form of Theorem 1, \(x_1^2 + x_2^2 = \tfrac{1}{2}\,{\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}}^{\top} (2 \mathbf{I}_2) \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}\), so \(\mathbf{S} = 2 \mathbf{I}_2\), \(\mathbf{A} = {(3, 4)}^{\top}\) (\(p = 2\), \(m = 1\)) and \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} = [5]\). The hypotheses hold: \(2 \mathbf{I}_2\) is positive definite, and the single column \({(3, 4)}^{\top}\) is nonzero, so it is linearly independent. Also \(\mathbf{S}^{-1} = \tfrac{1}{2} \mathbf{I}_2\). Then, from the formulas of Theorem 1,
\[ \begin{aligned} \mathbf{M} &= {\mathbf{A}}^{\top}\,\tfrac{1}{2} \mathbf{I}_2\,\mathbf{A} \\ &= \tfrac{1}{2}\,(3^2 + 4^2) \\ &= \tfrac{25}{2}, \\ \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* &= \tfrac{1}{2} \mathbf{I}_2 \begin{bmatrix} 3 \\ 4 \end{bmatrix} \mathopen{}\left(\tfrac{2}{25}\right)\mathclose{} (5) \\ &= \tfrac{5}{25} \begin{bmatrix} 3 \\ 4 \end{bmatrix} \\ &= \begin{bmatrix} 0.6 \\ 0.8 \end{bmatrix}, \\ \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}^* &= -\mathopen{}\left(\tfrac{2}{25}\right)\mathclose{} (5) \\ &= -0.4, \\ f(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^*) &= \tfrac{1}{2} (5) \mathopen{}\left(\tfrac{2}{25}\right)\mathclose{} (5) \\ &= 1. \end{aligned} \]
Checks:
the constraint:
\[ \begin{aligned} 3 (0.6) + 4 (0.8) &= 1.8 + 3.2 \\ &= 5; \end{aligned} \]
the first block of Equation 2:
\[ \begin{aligned} \mathbf{S} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^* + \mathbf{A} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}^* &= {(1.2, 1.6)}^{\top} + {(3, 4)}^{\top} (-0.4) \\ &= {(1.2, 1.6)}^{\top} - {(1.2, 1.6)}^{\top} \\ &= \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{0}}\vphantom{0}}_{2 \times 1}; \end{aligned} \]
the value:
\[ \begin{aligned} 0.6^2 + 0.8^2 &= 0.36 + 0.64 \\ &= 1. \end{aligned} \]
Geometrically, \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{x}}\vphantom{x}}^*\) is the point of the line nearest the origin, at distance \(1\).
Example 6 (The hypotheses are needed)
If \(\mathbf{S}\) is not positive definite there may be no minimizer: with \(\mathbf{S} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\) and the constraint \(x_1 = 1\) (\(\mathbf{A} = {(1, 0)}^{\top}\), \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} = [1]\)), \(f(1, x_2) = \tfrac{1}{2} (1 - x_2^2)\) decreases without bound as \(x_2\) grows.
If the columns of \(\mathbf{A}\) are dependent, \(\mathbf{M}\) is not invertible: with \(\mathbf{S} = \mathbf{I}_2\) and \(\mathbf{A} = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\),
\[ \begin{aligned} \mathbf{M} &= {\mathbf{A}}^{\top} \mathbf{A} \\ &= \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}, \end{aligned} \]
whose columns are equal. So \(\mathbf{M} {(1, -1)}^{\top} = \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{0}}\vphantom{0}}_{2 \times 1}\), and \(\mathbf{M}\) is not invertible (a square matrix is invertible exactly when its null space is zero). The two constraints both read \(x_1 + x_2 = b_i\), so they have no solution at all when \(b_1 \ne b_2\). When \(b_1 = b_2\) a unique minimizer still exists,
\[ \begin{aligned} x_1 &= x_2 \\ &= b_1 / 2, \end{aligned} \]
but the formula of Theorem 1 cannot produce it.
Definition 9 (Sensitivity) Suppose a minimization problem (Definition 5) depends on a vector \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} \in \mathbb{R}^m\), and let \(f^*(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}})\) be its minimum value, as a function of \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}\). When the gradient \(\frac{\partial}{\partial \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}} f^*(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}})\) exists (gradient), it is the sensitivity of the minimum value to \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}\): by partial derivative, its entry \(i\) is the limit, as a change in \(b_i\) alone shrinks to \(0\), of the resulting change in the minimum value divided by the change in \(b_i\).
Example 7 (The sensitivity of a one-point problem) Minimize \(x^2\) subject to \(x = b\), for a number \(b\). The feasible set (Definition 1) is \(\mathopen{}\left\{b\right\}\mathclose{}\), so the minimum value is \(f^*(b) = b^2\), and its sensitivity to \(b\) is \(\frac{d }{d b} f^*(b) = 2b\). At \(b = 3\) the sensitivity is \(6\): moving \(b\) from \(3\) to \(3.1\) raises the minimum value from \(9\) to \(9.61\), an increase of \(0.61\), close to \(6 \times 0.1 = 0.6\).
Corollary 1 (The multiplier is the sensitivity of the minimum value) In Theorem 1, write the minimum value as a function of \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}\), \(f^*(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}) = \tfrac{1}{2}\,{\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}}^{\top} \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}\). Its sensitivity to \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}\) (Definition 9) is
\[ \begin{aligned} \frac{\partial}{\partial \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}} f^*(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}) &= \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} \\ &= -\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}^*. \end{aligned} \]
Proof. \(\mathbf{M}^{-1}\) is symmetric and does not depend on \(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}\), so derivative of a quadratic form applies with \(\mathbf{M}^{-1}\) in place of \(\mathbf{S}\):
\[ \begin{aligned} \frac{\partial}{\partial \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}} f^*(\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}) &= \tfrac{1}{2}\,\frac{\partial}{\partial \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}} \mathopen{}\left({\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}}^{\top} \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}\right)\mathclose{} && \text{(constant multiple)} \\ &= \tfrac{1}{2} \mathopen{}\left(2 \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}}\right)\mathclose{} && \text{(derivative of a quadratic form)} \\ &= \mathbf{M}^{-1} \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{b}}\vphantom{b}} && \text{(simplify)} \\ &= -\mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}^*. && \text{(definition of } \mathord{\smash[b]{\underset{\raise.3ex\hbox{$\scriptstyle\sim$}}{\lambda}}\vphantom{\lambda}}^* \text{)} \end{aligned} \]
Example 8 (Moving the line) In Example 5 with a general right side \(b\) in place of \(5\), \(\mathbf{M}^{-1} = \tfrac{2}{25}\) and
\[ \begin{aligned} f^*(b) &= \tfrac{1}{2} \mathopen{}\left(\tfrac{2}{25}\right)\mathclose{}\,b^2 \\ &= \tfrac{b^2}{25}, \end{aligned} \]
so \(\frac{d }{d b} f^*(b) = \tfrac{2b}{25}\). At \(b = 5\) this derivative is
\[ \begin{aligned} \tfrac{10}{25} &= 0.4 \\ &= -\lambda^*: \end{aligned} \]
moving the line \(3 x_1 + 4 x_2 = b\) outward by a small step \(\varepsilon\) in \(b\) raises the minimum value by about \(0.4\,\varepsilon\).