– finite dimensional real vector space
a smooth submanifold, which can be represented as a zero set of a differentiable map , whre is a real vector space and such that is surjective for each .
We want to minimize for . It won’t work to set and solving for as will not be a critical point of in general. The Lagrange multipliers are used to define another function such that solving gives extrema of the constrained extremization problem.
Theorem (Loomis-Sternberg 3.12.2) Suppose has a maximum on at . Then there is a function(al) in such that is a critical point of the function .
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 on and sets , so that . Now will be of the form and and gives
This is equations, which together with equations for give equations for unknowns . The last 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 be a real symmetric matrix. Then is diagonalizable over the real numbers.
Consider the problem of maximizing the function where is subject to the constraint . (Such an extreme point exists, say by compactness.) By the symmetry of , the gradient of is easily calculated to be , whereas the gradient of the Euclidean norm? is . At a point where a maximum is attained, we have for some Lagrange multiplier . Thus is an eigenvector of with eigenvalue . The usual arguments show that restricts to a self-adjoint operator on the hyperplane orthogonal to ; by picking an orthonormal basis of this hyperplane, we may represent this restriction of by a real symmetric matrix of size , and the argument repeats.