Why can an inner product of an eigenvector also be used as an eigenvector? In quote box below, there is an inner product of an angular momentum eigenvector. Why can you use this inner product as a new eigenvector for the next part of the work?
And why do they "of course" satisfy the same relations?


In coordinate representation, the eigenvectors of angular momentum are known as spherical harmonics
  $$ \langle \theta, \phi | lm_l \rangle = Y_{lm_l}(\theta, \phi)$$
  They of course satisfy the same relations:
  $$\hat{\vec{L}}^2 Y_{l m_l}(\theta, \phi) = \hbar^2 l (l + 1)Y_{l m_l}(\theta, \phi)
\qquad
\hat{L}_z Y_{l m_l} (\theta, \phi) = \hbar m_l Y_{l m_l}(\theta, \phi)$$


 A: Recall that in linear algebra there are two types of objects you usually deal with: vectors $|\psi\rangle$ and linear maps $\hat{L}:|\psi\rangle \mapsto |\phi\rangle$. The vectors geometrically represent arrows or quantum states whilst the linear operators represent rotations, scalings, shears or various transformations on quantum states. However the abstract approach is hard to actually do calculations with, so we put everything in terms of concrete numbers by projecting onto a basis $\{|x\rangle\}$.
$$\psi(x) = \langle x | \psi \rangle$$
Likewise the matrix elements of an operator become:
$$L(x,x')=\langle x' | \hat{L} | x \rangle$$
So there are two types of thing: the abstract geometric objects, and their components. The rules for multiplying the components (ie matrix multiplication) are chosen to match up with the rules for manipulating the geometric objects. So if you have an operator $\hat{L}$ with coordinate description $L(x,x')$ and an eigenvector $|\Lambda\rangle$ with coordinate description $\Lambda(x)$ then the fact that
$$\hat{L}|\Lambda\rangle = \lambda |\Lambda \rangle$$
must mean that when you 'multiply' the coordinate representations of the things on the left, you get the coordinate representation of the things on the right, ie:
$$L(x,x')[\Lambda(x')] = \lambda \Lambda(x)$$
if you didn't, then the whole scheme of converting things to coordinates, multiplying and then converting back to abstract quantities wouldn't work.
A: For angular momentum you have the equations
\begin{align}
\hat{L}_z|l,m\rangle &= m\hbar|l,m\rangle,\\
\hat{L}^2|l,m\rangle &= l(l+1)\hbar^2|l,m\rangle.
\end{align}
What you have in your equations is simply the representation of the states $|l,m\rangle$ in the real space, i.e.
\begin{align}
\langle \theta,\phi|\hat{L}_z|l,m\rangle &= m\hbar\langle \theta,\phi|l,m\rangle,\\
\langle \theta,\phi|\hat{L}^2|l,m\rangle &= l(l+1)\hbar^2\langle \theta,\phi|l,m\rangle.
\end{align}
If you want to know how $\hat{L}_z$  and $\hat{L}^2$ are represented in real space, you can check this, Eqs. (3.76.26) and (3.6.28).
