> $\{ e_i \}$ is an orthonormal basis which has the orthonormal condition as following  $$ e^T_i \cdot e_j = \delta_{ij} $$

Not quite. The reason I say this because this statement isn't the most general statement you could make about orthonormal bases for vector spaces (or at least it's not precise enough) 


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At any rate, the correct general statement is as follows. 

Let $\{ e_i\}$ be a discrete (possibly infinite) orthonormal basis for a **complex** vector space $V$ equipped with an inner product " $\cdot$ ". Then, for any $i,j$ we have that 

$$ 
e_i \cdot e_j = \delta_{ij}. \tag{a}
$$


*Wait a second. Shouldn't that be $(e_i^*)^T \cdot e_j$?*  No, and this is because the " $\cdot$" denotes an **inner product** between two vectors $v, w$---whatever that may be. That is, **the dot does not refer to matrix multiplication** necessarily, but rather refers to the inner product between two vectors $v,w\in V$.
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**Note** that the case you cited $(e_i)^T \cdot e_j = \delta_{ij}$ is a **special** case of (1) when we are representing the basis vectors in a matrix representation and (2) when it is a *real* vector space, since $e_i^\dagger := (e_i)^T$. 
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Lastly, Dirac notation is just the notational change 

\begin{align}
& e_i \to | e_i \rangle\\
& e_i^\dagger \to \langle e_i|\\
&  v \cdot w \to \langle v |w \rangle. \tag{b}
\end{align}