nLab
Lagrange multiplier

Idea following Loomis-Sternberg

$X$ – finite dimensional real vector space

$U\subset X$ open

$F:U\to R$ 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 $(x_1,\dots ,x_n)$ on $U$ and sets $Y=R^m$, so that $G=(g^1,\dots ,g^n)$. Now $l:Y\to R$ will be of the form $l(y_1,\dots ,y_m)={\sum }_{i=1}^m{\lambda }_i{y}_i$ and $F-l\circ G=F-{\sum }_{i=1}^m{\lambda }_i{g}^i$ and $d(F-l\circ G)=0$ gives

$$\frac{\partial F}{\partial x_j}-\sum _{i=1}^m{\lambda }_i\frac{\partial g^i}{\partial x_j}=0,j=1,\dots ,n.$$\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 $G=(g^1,\dots ,g^n)=0$ for $S$ give $m+n$ equations for $m+n$ unknowns $x_1,\dots ,x_n,{\lambda }_1,\dots ,{\lambda }_m$. The last $m$ variables here are the Lagrange multipliers.

Applications

To spectral theory

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.

Proposition

Let $A$ be a real symmetric $n\times n$ matrix. Then $A$ is diagonalizable over the real numbers.

Proof

Consider the problem of maximizing the function $f(x)=⟨x\mid A\mid x⟩$ where $x\in {ℝ}^n$ is subject to the constraint $⟨x\mid x⟩=1$. (Such an extreme point exists, say by compactness.) By the symmetry of $A$, the gradient of $f$ is easily calculated to be $\nabla f(x)=2Ax$, whereas the gradient of the Euclidean norm? $⟨x\mid x⟩$ is $2x$. At a point $x$ where a maximum is attained, we have $\nabla f(x)=2Ax=\lambda (2x)$ 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.

Literature

• 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