$(v,w) \geq 0$, where (a,b) denotes an inner product
This is linear in the right argument, and anti-linear in the first argument (ie. take complex conjugates)
(v,v) = 0 implies v = 0
$v = \Sigma_{i} e_{i} (e_{i}, v)$ always holds in Hilbert spaces (ie. can always decompose into an eigenbasis, where $(e_{i}, e_{j}) = \delta_{ij}$)
This can be extended to non-orthonormal bases (ie. $(e_{i}, e_{j}) = G_{ij}$), which implies that $(v,w) = \Sigma_{ij} (v_{i}e_{i})G_{ij}^{-1} (e_{j}, w)$
You can add Hilbert spaces of different dimensions together
The new Hilbert space has the equivalence relationship of $<v_{1}+v_{2},w_{1}+w_{2}> = <v_{1},w_{1}> + <v_{2},w_{2}>$
The total dimensionality of the new Hilbert space is just the sum of the previous two
You can also multiply two Hilbert spaces together
The equivalence relation which must hold for the product is that $\lambda <v,w> = <\lambda v,w> = <v,\lambda w>$
You can decompose this product Hilbert space as a sum of tensor products of the eigenbases of each prior Hilbert space (ie. $H_{prod} = \Sigma_{ia} \Phi_{ia} e_{i} \otimes e_{a}$)
We also define $\Pi* \Pi = 1$, which implies $\Pi = \Pi^{-1}$ and that the eigenstates are $\pm 1$
Suppose that $\phi$ is an eigenstate of $\Pi$. Then $(\phi x \phi) = 0$ (Apply parity definition, transfer parity operator to states, apply eigen equation $\Pi \psi = \epsilon \psi$), and then note that you get $-q =q $, which implies q=0
As a consequence of this, if you Hamiltonian commutes with parity, then you know that you can write you state as a sum of even and odd states