# Eigenfunctions and their Eigenvalues

1. Oct 9, 2006

### g782k936

If I have two eigenfunctions of an operator with the same eigenvalue how do I construct linear combinations of my eigenfunctions so that they are orhtogonal?

My eigenfunctions are: f=e^(x) and g=e^(-x)

and the operator is (d)^2/(dx)^2

2. Oct 9, 2006

### dextercioby

Orthogonal wrt what?

You need a scalar product.

Daniel.

3. Oct 9, 2006

### g782k936

I want to have linearly independent combinations of f and g that are orthognal on the interval from (-1,1) I'm guesing that they need to be wrt f and g.

4. Oct 9, 2006

### HallsofIvy

Staff Emeritus
No, that was not the question. "Orthogonal" means that the inner product is 0 so whether or not two vectors are orthogonal depends on the inner product used.

The most common inner product for real valued functions on an interval (a, b) is $\int_a^b f(x)g(x)dx$.

Since, if two eigenvectors correspond to the same eigenvalue, any linear combination is also an eigenvector corresponding to that eigenvalue, a simple "orthogonal projection" will work.

If u and v are two vectors in an inner product space, then the "projection of v onto u" is given by
$$\frac{<u,v>}{<u,u>}\vec{u}$$
The "orthogonal projection" is v minus that:
$$\vec{v}- \frac{<u,v>}{<u,u>}\vec{u}$$

Calculate that with u= ex, v= e-x, and inner product $<u,v>= \int_{-1}^1 u(x)v(x)dx$.

5. Oct 9, 2006

### g782k936

O.k.

I think that worked. I had been trying the integral in a slightly different way using f + g instead of fg.

Thanks.