Let $$L : D(L) \to L^2([0, \pi] ; \mathbb{C}$$ be a linear differential operator defined by $$(Lu)(x)=u'(x)$$ where $x \in [0, \pi]$
The domain $D(L)$ of $L$ is given by $$D(L)=\{ u \in L^2([0,\pi]): Lu \in L:^2([0, \pi])\}$$
$u$ is continuously differentiable on $[0,\pi]$ and $u(0)=u(\pi)$
I am trying to determine the eigenvalues of $L$ and the corresponding eigenvectors.
In the solutions I am told that the associated boundary value problem is
$$u'(x)=\lambda u(x)$$ where $u(0)=u(\pi)$
Why has $L$ been replaced by $\lambda$? Is this always the case for the associated boundary value problem?
The general solution is $$u(x)=Ae^{\lambda x}$$
Applying the boundary condition gives us $$e^{\lambda \pi} =1$$
So $$\lambda=2ik$$ where $k \in \mathbb{Z}$
How does $\lambda$ equal $2ik$ this and is this the eigenvalue?
How do you find the corresponding eigenvectors?
By definition, if $V$ is a complex vector space and $T: V_1 \to V_2$ a linear map between subspaces of $V$, a complex number $\lambda \in \mathbb C$ is an eigenvalue for $T$ if there exists $v \neq 0$ in $V$ such that $Tv = \lambda v$. The vector $v$ is then called an eigenvector of $T$ associated to $\lambda$. So, in your situtation, $V_1 = D(L)$, $v = u$ would be a function in $D(L)$, and $Tu = u'$. So the equation $Tv = \lambda v$ becomes $u' = \lambda u$, or $u'(x) = \lambda u(x)$ for almost all $x \in [0,\pi]$ (hence all $x$ by continuity).
Accordingly, after writing out the definitions, you are trying to find all $\lambda \in \mathbb{C}$ for which there exists a continuously differentiable function $u = u_\lambda: [0,\pi] \to \mathbb{C}$ such that $u \neq 0$ (we want a nonzero vector, which in this situation means a function not identically zero)
$$ u'(x) = \lambda u(x),\text{ for all $x \in [0,\pi]$ (since $Lu = \lambda u$)} $$
and $u(0) = u(\pi)$ (so that $u \in D(L)$).
Now onto your second set of questions, we do, indeed, know that the general solution to the ODE $u' = \lambda u$ is of the form $u(x) = Ae^{\lambda x}$. Hence substituting $u(\pi) = u(0)$, we get $Ae^{\lambda \pi} = Ae^{\lambda \cdot 0} = A$. Since $u$ is not identically zero, we know $A \neq 0$, and so we may cancel the $A$'s on both sides to get $e^{\lambda \pi} = 1$. Writing $\lambda$ as $x + iy$ with $x,y \in \mathbb{R}$, we see that $e^{x\pi} = |e^{x\pi}e^{iy\pi}| = |1| = 1$, and so $e^{x\pi} = 1$. This is only possible if $x = 0$ and $\lambda = iy$. Finally, apply Euler's formula to conclude that $\lambda\pi = 2\pi i k$ and $\lambda = 2 i k$ as claimed. The corresponding eigenvector is $u = Ae^{2\pi i k}$ for any choice of $A \neq 0$.