Lagrange multiplier

$X$ – 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

$\frac{\partial F}{\partial x_j} - \sum_{i = 1}^m\lambda_i\frac{\partial g^i}{\partial x_j} = 0,\,\,\,\,\,\,\,\,j = 1,\ldots, n.$

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.

- 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 - Robert Hermann,
*Some differential-geometric aspects of the Lagrange variational problem*, Illinois J. Math.**6**, 1962, 634–673, MR145457,euclid - Juan Carlos Marrero, David Martín de Diego, Ari Stern,
*Lagrangian submanifolds and discrete constrained mechanics*, slides, pdf - Manuel de León, David Martín Diego,
*Solving non-holonomic Lagrangian dynamics in terms of almost product structures*, Extracta Math. 11 (1996), no. 2, 325–347, pdf MR97m:58079

Last revised on October 12, 2017 at 18:15:55. See the history of this page for a list of all contributions to it.