Use of Dirac (bra-ket) notation and the inner product I have been told that, in Dirac notation, the inner product can be written as $$\langle \mathbf{v} \mid \mathbf{w} \rangle = \sum_i \langle \mathbf{v} \mid \mathbf{\epsilon}_i \rangle\langle \mathbf{\epsilon}_i \mid \mathbf{w} \rangle$$
However I don't really understand Dirac notation, so don't know how to show that this is true.
I know that I can write $$\mathbf{v} = \mid v\rangle = \sum_{i=1}^{n}v_i\mid e_i \rangle$$
and assume that this will help me in some way, but I am unsure how.
 A: If $\{|\epsilon_i\rangle\}$ is an orthonormal basis for the Hilbert space $\mathcal{H}$, then for arbitrary $|v\rangle \in \mathcal{H}$, we can write:
$$|v\rangle = \sum_i v_i |\epsilon_i\rangle \text{, where } v_i = \langle \epsilon_i| v\rangle.$$
The fact that $v_i = \langle \epsilon_i| v\rangle$ follows from orthonormality of the basis:
$$ \langle\epsilon_i |v\rangle = \sum_j v_j \langle \epsilon_i|\epsilon_j\rangle = \sum_j v_j \delta_{ij} = v_i$$
Now, from this, we can prove the following useful fact:
$$ 1 = \sum_i |\epsilon_i\rangle\langle\epsilon_i|.$$
The proof is as follows. Since $\{|\epsilon_i\rangle\}$ is an orthonormal basis, for arbitrary $|v\rangle \in \mathcal{H}$, we can write
$$ |v\rangle = \sum_i \langle \epsilon_i|v\rangle |\epsilon_i\rangle = \sum_i |\epsilon_i\rangle\langle \epsilon_i |v\rangle = \left(\sum_i |\epsilon_i\rangle\langle \epsilon_i |\right)|v\rangle,$$
meaning that $\sum_i |\epsilon_i\rangle\langle \epsilon_i|$ sends all $|v\rangle$ to $|v\rangle$.
Thus, in your expression, we are simply inserting $1$ into the inner product to write it as a sum over the basis elements:
$$\langle v|w\rangle = \langle v|1|w\rangle = \langle v|\left(\sum_i|\epsilon_i\rangle\langle\epsilon_i| \right)|w\rangle = \sum_i \langle v|\epsilon_i\rangle\langle\epsilon_i|w\rangle = \sum_i v_i^* w_i.$$
