Creating orthogonal states of tensor products from a set of given (possibly linearly independent) states I have less experience with tensor products, so I am having some difficulty in dealing with it.
I want to understand how to orthogonalize a system of 2-dimensional complex states.
Suppose I have $n$ sets of states like
$$\{ |1_i\rangle\otimes|2_i\rangle\otimes|3_i\rangle\otimes \cdots\otimes|k_i\rangle \mid i=1 \dots n \}$$ where $i=1, \dots, n$ is number of states in the sets with $k$ tensor products of elements from $\mathbb{C}^2$.
Now I am interested in orthogonalizing these set of $n$ states. Especially I want to use the Gram-Schmidt orthogonalization procedure here.
I can apply Gram-Scmidt decomposition by writing these vectors in terms of $2^k \times 1$ matrix (because it is tensor products of $\mathbb{C}^2$)
Now I don't know how to write the obtained vectors back in terms of the tensor product.
 A: The essence of your question seems to be the following: given a vector $x \in \Bbb C^{2^k}$, how can you find the corresponding vector in $[\Bbb C^2]^{\otimes k} := \underbrace{\Bbb C^2 \otimes \cdots \otimes \Bbb C^2}_k$?
To that end, we need to understand the standard interpretation of the representation of a state $|\phi \rangle$ in a tensor product as a vector $x = (x_0,\dots,x_{2^{k-1}}) \in \Bbb C^{2^k}$. Based on your original question, these states represent qubits, which means that each copy of $\Bbb C^2$ has the standard basis $\{ |0\rangle,|1\rangle\}$.  The basis $\{|0\rangle,|1\rangle\}$ of $\Bbb C^2$ induces the following basis of $[\Bbb C^2]^{\otimes k}$:
$$
\mathcal B = \{|a_0 \rangle |a_1 \rangle \cdots |a_{k-1}\rangle : a_i \in \{0,1\} \text{ for } i = 0,\dots,k-1\}.
$$
Importantly, this basis is taken in lexicographical order. If $|\phi_i\rangle$ denotes the $i$th element of $\mathcal B$, then we say that $x$ is the vector corresponding to $|\phi\rangle$ if
$$
|\phi\rangle = x_0 |\phi_0 \rangle + \cdots + x_{k-1} |\phi_{k-1} \rangle.
$$
Now, an interesting consequence of lexicographical ordering is that there is a nice correspondence between the $i$th element $|\phi_i\rangle$ and the binary representation of $i$: if the $k$-digit binary representation of $i$ has digits $a_0,a_1,\dots,a_{k-1}$, then
$$
|\phi_i \rangle = |a_0\rangle |a_1\rangle \cdots |a_{k-1}\rangle.
$$
For example: with $k = 5$, the $5$-digit binary representation of $11$ is $01011_2$, which means that $|\phi_{11}\rangle$ (the $12$th element of the induced basis on the tensor product) is
$$
|\phi_{11} \rangle = |0 \rangle |1 \rangle |0 \rangle|1 \rangle |1 \rangle.
$$
Note: it is often the case in the literature that the state $|\phi_i \rangle \in [\Bbb C^{2}]^{\otimes k}$ is more simply denoted as $|i \rangle$, but this seems like it might conflict with your notation.

For a quick example, consider the vector
$$
\frac 12 (0,1,0,i,0,-1,-i,0) \in \Bbb C^{2^3}.
$$
The corresponding element of the tensor product will be
$$
|\phi\rangle = \frac 12 |\phi_1 \rangle + \frac i2 |\phi_3\rangle - \frac 12 |\phi_5\rangle - \frac i2 |\phi_6\rangle
\\ = \frac 12 |0\rangle|0\rangle|1\rangle + \frac i2 |0\rangle|1\rangle|1\rangle 
- \frac 12 |1\rangle|0\rangle|1\rangle
- \frac i2 |1\rangle|1\rangle|0\rangle.
$$
