$X$ – finite dimensional real vector space
$U\subset X$ open
$F: U\to \mathbf{R}$ differentiable function
$S\subset U$ a smooth submanifold, which can be represented as a zero set of a differentiable map $G: U\to Y$, whre $Y$ is a real vector space and such that $d G_x$ is surjective for each $x\in S$.
We want to minimize $F(x)$ for $x\in S$. It won’t work to set $d F_x = 0$ and solving for $x$ as $x$ will not be a critical point of $F$ in general. The Lagrange multipliers are used to define another function $L$ such that solving $d L_x = 0$ gives extrema of the constrained extremization problem.
Theorem (Loomis-Sternberg 3.12.2) Suppose $F$ has a maximum on $S$ at $x$. Then there is a function(al) $l$ in $Y^*$ such that $x$ is a critical point of the function $F - l\circ G$.
The proof uses implicit function theorem and the usual extremization arguments.
To get to a more familiar form of Lagrange multipliers, one uses the local coordinates $(x_1,\ldots,x_n)$ on $U$ and sets $Y = \mathbf{R}^m$, so that $G = (g^1,\ldots, g^n)$. Now $l: Y\to\mathbf{R}$ will be of the form $l(y_1,\ldots,y_m) = \sum_{i = 1}^m \lambda_i y_i$ and $F - l\circ G = F - \sum_{i = 1}^m \lambda_i g^i$ and $d (F - l\circ G) = 0$ gives
This is $n$ equations, which together with $m$ equations $G = (g^1,\ldots, g^n) = 0$ for $S$ give $m+n$ equations for $m+n$ unknowns $x_1,\ldots, x_n, \lambda_1,\ldots,\lambda_m$. The last $m$ variables here are the Lagrange multipliers.
The method of Lagrange multipliers affords an elementary proof of the spectral theorem for finite-dimensional real vector spaces, one which does not involve passage to the complex numbers and the fundamental theorem of algebra.
Let $A$ be a real symmetric $n \times n$ matrix. Then $A$ is diagonalizable over the real numbers.
Consider the problem of maximizing the function $f(x) = \langle x \vert A \vert x \rangle$ where $x \in \mathbb{R}^n$ is subject to the constraint $\langle x \vert x \rangle = 1$. (Such an extreme point exists, say by compactness.) By the symmetry of $A$, the gradient of $f$ is easily calculated to be $\nabla f (x) = 2 A x$, whereas the gradient of the Euclidean norm $\langle x \vert x \rangle$ is $2 x$. At a point $x$ where a maximum is attained, we have $\nabla f(x) = 2 A x = \lambda (2 x)$ for some Lagrange multiplier $\lambda$. Thus $x$ is an eigenvector of $A$ with eigenvalue $\lambda$. The usual arguments show that $A$ restricts to a self-adjoint operator on the hyperplane orthogonal to $x$; by picking an orthonormal basis of this hyperplane, we may represent this restriction of $A$ by a real symmetric matrix of size $(n-1) \times (n-1)$, and the argument repeats.
Named after Joseph-Louis Lagrange.
wikipedia Lagrange multiplier
Springer eom: Lagrange multipliers, Pontrjagin maximum principle
Lynn H. Loomis, S. Sternberg, Advanced calculus, section 3.12
“Submanifolds and Lagrange multipliers” section 13.2 of Loomis Sternberg Advanced Calculus
Last revised on December 15, 2021 at 19:09:00. See the history of this page for a list of all contributions to it.