# Complete Set

1. May 10, 2005

### jwsiii

I am studying Sturm-Liouville eigenvalue problems and their eigenfunctions form a "complete set". Can someone explain to me what this means?

2. May 11, 2005

### HallsofIvy

Staff Emeritus
A "complete" set of eigenvectors is a basis for the vector space consisting entirely of eigenvectors for a given linear transformation. If, for example, in a finite dimensional vector space, you can find a complete set of eigenvectors for a linear transformation, using those eigenvectors as the basis, you can write the linear transformation as a diagonal matrix, simplifying the problem greatly. It can be shown that "self-adjoint" linear operators always have a complete set of eigenvectors.

In working with linear differential equations, the underlying vector space is the space of infinitely differentiable functions which is infinite dimensional but using the "eigenfunctions" as a basis will still simplify the problem. The Sturm-Liouville differential operators are precisely the self-adjoint operators in that space. The simplest example is the differential operator $$\frac{d^2}{dx^2}$$ with x between 0 and $$\pi$$. It is easy to show that the eigenfunctions are cos(nx), sin(nx) and using those as a basis gives the Fourier series for a function.