3 ms·
Functions satisfy the laws of a vector space (well, certain classes of functions, such as L^2 anyway -- I'm certainly no expert here). I.e., functions are vecto
by monktastic1 4y ago
Functions satisfy the laws of a vector space (well, certain classes of functions, such as L^2 anyway -- I'm certainly no expert here). I.e., functions are vectors. And the derivative is a linear operator on that space: d/dx(af + bg) = ad/dx(f) + bd/dx(g). Therefore d/dx has eigenvalues, one of which you properly identified as e^x (and more generally, e^ax, as others have pointed out).
- Tainnor 4y ago> (well, certain classes of functions, such as L^2 anyway -- I'm certainly no expert here) For any vector space V over a field F, the set of functions V -> V is a vector space over F. This is essentially because the vector space properties are carried over "point-wise". When you consider only certain classes of functions (such as L^2), the question of whether they form a vector space boils down to the question whether those classes are closed under addition and scalar multiplication. All other vector space properties are satisfied because they are in the larger vector space of all functions.