V0ODO0CH1LD said:
In which sense are the solution spaces to ordinary linear differential equations and partial linear differential equations, respectively, finite and infinite? Are you treating function spaces as vector spaces? In which case what are some of its basis sets?
Also, is the theory of linear operators "equivalent" to the theory of vector spaces? In other words, is it necessarily the case that if a map between two mathematical object is linear those mathematical objects are vectors in some vector space? Or can linear operators be maps between other things that aren't vectors?
There is no concept of "linear" except for maps between vector spaces, in the same way that you can't talk about functions between arbitrary sets being "homomorphisms" or "continuous"; those notions only make sense for functions between groups and between topological spaces.
Given any set X, the set F(X) of functions with domain X and codomain \mathbb{R} can be made into a vector space over \mathbb{R} by defining vector addition and scalar multiplication pointwise:<br />
f + g : x \mapsto f(x) + g(x), \\<br />
af : x \mapsto af(x). If X = \{x_1, \dots, x_n\} is finite then we can describe any f \in F(X) as a linear combination of the functions \phi_i : X \to \mathbb{R} where <br />
\phi_i(x) = \begin{cases} 1, & x = x_i, \\<br />
0, & x \neq x_i,\end{cases} so that <br />
f = \sum_{i=1}^n f(x_i)\phi_i.<br /> These functions \phi_i are then a
basis for F(X). Since there are a finite number of them we say that F(X) is finite-dimensional.
But if X is not finite then we need at least a countable number of functions \phi_y : X \to \mathbb{R}, with \phi_y(x) = 1 if x = y and \phi_y(x) = 0 otherwise, to describe an arbitrary f \in F(X) as a linear combination in the same say, and F(X) is then infinite-dimensional.
Letting X = \mathbb{R}, we see that \mathbb{R} is not finite, so the space F(\mathbb{R}) is infinite-dimensional, and the subspace C^{\infty}(\mathbb{R}) of smooth functions is also infinite-dimensional.
We can define a linear operator (or function, or map, the terms are equivalent) D : C^{\infty}(\mathbb{R}) \to C^{\infty}(\mathbb{R}) : f \mapsto f' which takes each function to its derivative. If P is a polynomial of order n in D with coefficients in \mathbb{R} then we can consider <br />
L : C^{\infty}(\mathbb{R}) \to C^{\infty}(\mathbb{R}) : f \mapsto P(D)f<br /> which is a linear operator, and <br />
L(f) = 0<br /> is then a homogenous linear differential equation of order n.
The set \{f \in C^{\infty}(\mathbb{R}) : L(f) = 0\} is a subspace called the
kernel of L, and it turns out that this kernel is finite-dimensional. The basis depends on the roots of P. In general we can factorise P over \mathbb{R} to obtain <br />
P(\lambda) = A\prod_{i = 1}^{r} (\lambda - \lambda_i)^{n_i} \prod_{i=1}^{c} ((\lambda - p_i)^2 + q_i^2)^{m_i}<br /> where A \neq 0 is a constant and r and c are respectively the number of distinct real roots \lambda_i and distinct pairs of complex conjugate roots p_i \pm iq_i; \sum_{i=1}^r n_i + 2\sum_{i=1}^c m_i = n, the degree of P.
For each distinct \lambda_i we acquire the basis functions <br />
e^{\lambda_i x}, xe^{\lambda_i x}, \dots, x^{n_i - 1}e^{\lambda_i x}<br /> and for each distinct (p_i, q_i) we acquire the basis functions <br />
e^{p_i x}\cos(q_i x), xe^{p_i x}\cos(q_i x), \dots, x^{m_i - 1}e^{p_i x}\cos(q_i x), \\<br />
e^{p_i x}\sin(q_i x), xe^{p_i x}\sin(q_i x), \dots, x^{m_i - 1}e^{p_i x}\sin(q_i x).
