# Linear Transformations proof

1. Sep 14, 2008

### kehler

1. The problem statement, all variables and given/known data
Let V be the vector space of all functions f: R->R which can be differentiated arbitrarily many times.
a)Let T:V->V be the linear transformation defined by T(f) = f'. Find the (real) eigenvalues and eigenvectors of T. More precisely, for each real eigenvalue describe the eigenspace of T corresponding to the eigenvalue. (Since the elements of the vector space are functions, some people would use the term eigenfunction instead of eigenvector
b)Now let T:V->V be the linear transformation defined by T(f) = f''. Prove that every real number eigenvalue is an eigenvalue of T

3. The attempt at a solution
I don't really know where to start :S
The eigenvectors should be functions f where
T(f) = eigenvalue x f
So, f' = eigenvalue x f
What do I do from here? I can't expand it cos I don't know the form of f(x)

2. Sep 14, 2008

### Defennder

Well I can think of a class of functions for which f(n) = $$\lambda$$f. Though I don't know if they are the only possible types of functions which satisfy this property.

3. Sep 14, 2008

### kehler

Would that be the exponential functions?

4. Sep 14, 2008

### HallsofIvy

Staff Emeritus
In other words, you need to solve the differential equations $df/dx= \lambda f$ and $d^f/dx^2= \lambda f$. That shouldn't be too hard.