Theorem: Simultaneous diagonalisation theorem.
Suppose A and B are Hermitian operators. Then $[A,B]=0$ IFF there exists an orthonormal basis such that both A and B are diagonal wrt that basis.
The forward implication:
Let $[A,B]=0$. Then, AB = BA. Let $ |a, j \rangle$ be an orthonormal basis for the eigenspace $V_{a}$ of A with eigenvalue a; $j$ labels the possible degeneracies.
Observe: AB $|a, j \rangle$ = BA $|a, j \rangle$ = $a B |a, j \rangle$ is an eigenvector equation so $|a, j \rangle$ is an element of the eigenspace $V_{a}$.
Letting $P_{a}$ be the projector onto the space $V_{a}$; $P_{a}: V_{a} \rightarrow V_{a}$.
Here, defined $B_{a} \equiv P_{a}BP_{a}$.
1) What does $B_{a} \equiv P_{a}BP_{a}$ mean geometrically?
2) What does it mean to say that $B_{a}$ restricted to the eigenspace $V_{a}$ is Hermitian on $V_{a}$?
Any help is appreciated.
1) You are only interested in the action of $B$ when restricted to $V_a$, both in domain and range. So you first project your vector $v$ to $V_a$, then take the image, and then project it again, because a part of $B \circ P_a (v)$ may not lie in $V_a$. For example, consider the matrix
$$B = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$$
and eigenspace
$$V_a = \langle (1,0) \rangle.$$
The projector with respect to this eigenspace is
$$ P_a = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}.$$
Consider the vector $v = (1,1)$. $P_a v = (1,0)$ and lies in $V_a$ as intended. However, $B P_a v = (1,1)$ and does not lie in the eigenspace anymore; project it again to get $ P_a B P_a v = (1,0)$. So $B_a (1,1) = (1,0)$; more specifically, restricted to the eigenspace $V_a$ we have $B_a (1,0) = (1,0)$ so $B_a = Id$.
2) Note that $B_a$ maps any vector of $V_a$ to a vector of $V_a$ (why?). Pick any orthonormal basis for $V_a$. The operator $B_a$ can be represented as a matrix with respect to this chosen basis; this matrix will be Hermitian. I.e. if we denote the matrix as $T$:
$$ T_{ij} = \overline{T_{ji}}.$$
This property will hold regardless of which basis you have chosen for $V_a$. The Hermitian property is important in quantum mechanics (which I assume you're dealing with?) because it implies all eigenvalues are real.