Linear Independence of Tensor Product. So I am trying to prove (or disprove) linear independence of following normalized vectors given in $n-$tensor product :
$$\{ V_{\theta_i} = \frac{1}{\sqrt{2}} \left(|0\rangle+ e^{\iota\theta_i}|1\rangle \right) \otimes \cdots \otimes \frac{1}{\sqrt{2}} \left(|0\rangle+e^{\iota\theta_i}|1\rangle \right) \mid i=1 \dots M \}$$
I know that $[\Bbb C^2]^{\otimes k} \equiv \Bbb C^{2^k}$. So, presenting it in a different way, what I am trying to do is to find our if the $M$ vectors generated by above tensor products are actually linearly independent or not.
For example when $n=3$ then we have
$$V_{\theta_i}=\frac{1}{\sqrt{2}} \left(|0\rangle+ e^{\iota\theta_i}|1\rangle \right) \otimes \frac{1}{\sqrt{2}} \left(|0\rangle+ e^{\iota\theta_i}|1\rangle \right) \otimes \frac{1}{\sqrt{2}} \left(|0\rangle+ e^{\iota\theta_i}|1\rangle \right)$$
Which is equivalent to
$$
V_i = 
\begin{pmatrix}
1 & e^{\iota\theta_i} & e^{\iota\theta_i} & e^{\iota2\theta_i} & e^{\iota\theta_i} & e^{\iota2\theta_i} & e^{\iota2\theta_i} & e^{\iota3\theta_i}
\end{pmatrix}
$$
Now my question is : Are the vectors say $V_1,V_2,V_3,V_4$ and $V_5$ in the above for linearly independent for all $\theta_i$ different?
 A: For your question as stated, the answer is no. Let $A$ denote the matrix
$$
A = 
\pmatrix{
1 & e^{\iota\theta_1} & e^{\iota\theta_1} & e^{\iota2\theta_1} & e^{\iota\theta_1} & e^{\iota2\theta_1} & e^{\iota2\theta_1} & e^{\iota3\theta_1}\\
&&&&\vdots \\
1 & e^{\iota\theta_M} & e^{\iota\theta_M} & e^{\iota2\theta_M} & e^{\iota\theta_M} & e^{\iota2\theta_M} & e^{\iota2\theta_M} & e^{\iota3\theta_M}
}
$$
so that $A$ has $M$ rows. Let's say that $1 \leq M \leq 2^3$. This matrix has linearly independent rows if and only if there exists an invertible $M \times M$ matrix. However, it is clear that if $M \geq 5$, then any $M \times M$ submatrix has a repeated column and therefore fails to be invertible.
On the other hand, for any $M \leq 4$, the matrix will have linearly independent rows. To see this, it suffices to note that the $(k+1) \times (k+1)$ matrix
$$
\pmatrix{1 & e^{\iota \theta_1} & \cdots & e^{\iota k\theta_1}\\
&&\vdots \\
1 & e^{\iota \theta_{k+1}} & \cdots & e^{\iota k \theta_{k+1}}}
$$
is a Vandermonde matrix with distinct rows and is therefore invertible.
By extending this logic, we can see that the set
$$
\{ V_{\theta_i} =\overbrace{ \frac{1}{\sqrt{2}} \left(|0\rangle+ e^{\iota\theta_i}|1\rangle \right) \otimes \cdots \otimes \frac{1}{\sqrt{2}} \left(|0\rangle+e^{\iota\theta_i}|1\rangle \right)}^n \mid i=1 \dots M \}
$$
for distinct $\theta_i$ will be linearly independent if and only if $M \leq n+1$.
