Let $\varphi\_k\in\mathbb{C}$ be a primitive $k$-th root of unity, and define the sets
$$S\_\ell:=\left\{\left[\begin{matrix}x\\x\varphi\_k^\ell\end{matrix}\right]\in\mathbb{C}^{2n}\;\middle|\;x\in\mathbb{R}^n\right\}$$
with $\ell=1,\ldots,k\leq 2n$. Taking randomly (independent and identically random with some continuous distribution) one element $y\_\ell\in S\_\ell$ from each set, can I conclude that $y\_1,\ldots,y\_k$ are linear independent with high probability?


**Example** $n=2, k=3$:
$$y\_1\in\left\{\left[\begin{matrix}x\\xe^{i\frac{2}{3}\pi}\end{matrix}\right]\;\middle|\;x\in\mathbb{R}^2\right\},\;
y\_2\in\left\{\left[\begin{matrix}x\\xe^{i\frac{4}{3}\pi}\end{matrix}\right]\;\middle|\;x\in\mathbb{R}^2\right\},\;
y\_3\in\left\{\left[\begin{matrix}x\\x\end{matrix}\right]\;\middle|\;x\in\mathbb{R}^2\right\}$$



Are $\left[\begin{matrix}x_\ell \\ x_\ell\varphi_k^\ell\end{matrix}\right]$ linearly independent?
**Edit:** In response to asker comment, edited significatively the answer as it failed to take into account the case $n<k$




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Here is my argument why the probability of linear independence is 1:


Case $k\leq n$: 


Notice $y\_1,...,y\_k$ linearly dependent implies $x\_1,...,x\_k$ are linearly dependent. So the probability that $y\_1,...,y\_k$ are linearly dependent is less or equal than the probability that $x\_1,...,x\_k$ are linearly dependent. In this case, the probability of $x\_1,...,x\_k$ being linearly dependent is 0. 


The proof is similar to one answer to [this question](https://math.stackexchange.com/questions/474628/the-probability-that-two-vectors-are-linearly-independent). The idea is that, for each $l\leq k$, the probability that $x\_l$ is in the same subspace than $x\_1,...,x\_{l-1}$ is 0. 


Case $n<k\leq 2n$: 


Let $x\_{1},...,x\_{k}$ s.t. $y\_{1},...,y\_{k}$ are linearly dependent. Since the probability that $x\_{1},...,x\_{n}$ are linearly independent is 1, we can assume w.l.o.g $x\_{1},...,x\_{n}$ will be linearly independent. Then $x\_{1},...,x\_{n}$ are a basis of $\mathbb{R}^{n}$. In particular, for $l\in\{n+1,...,k\}$, there are unique $w\_{1l},...,w\_{nl}$ s.t. 
$x\_{l}=\sum\_{i=1}^{n}w\_{il}x\_{i}$.


Now, since $y\_{1},...,y\_{k}$ are linearly dependent, there is $
\lambda\in\mathbb{C}^{k}\setminus\{0\}$ s.t. 
$$
0=\sum\_{i=1}^{k}\lambda\_{i}y\_{i}=\sum\_{i=1}^{k}\lambda\_{i}\left(\begin{array}{c}
x\_{i}\\
\varphi\_{k}^{i}x\_{i}
\end{array}\right)
$$


Expanding $$
0=\sum\_{i=1}^{k}\lambda\_{i}\left(\begin{array}{c}
x\_{i}\\
\varphi\_{k}^{i}x\_{i}
\end{array}\right)
$$
we get 
\begin{eqnarray}
0 & = & \sum\_{i=1}^{n}\lambda\_{i}\left(\begin{array}{c}
x\_{i}\\
\varphi\_{k}^{i}x\_{i}
\end{array}\right)+\sum\_{l=n+1}^{k}\lambda\_{l}\left(\begin{array}{c}
x\_{l}\\
\varphi\_{k}^{l}x\_{l}
\end{array}\right)
\\
& = & \sum\_{i=1}^{n}\lambda\_{i}\left(\begin{array}{c}
x\_{i}\\
\varphi\_{k}^{i}x\_{i}
\end{array}\right)+\sum\_{l=n+1}^{k}\lambda\_{l}\left(\begin{array}{c}
\sum\_{i=1}^n w\_{il}x\_i\\
\varphi\_{k}^{l}\sum\_{i=1}^n w\_{il}x\_i
\end{array}\right)\\
& = & \sum\_{i=1}^{n}\left(\begin{array}{c}
x\_{i}\left(\lambda\_i+\sum\_{l=n+1}^k\lambda\_lw\_{il}\right)\\
x\_{i}\left(\varphi\_{k}^i\lambda\_i+\sum\_{l=n+1}^k\lambda\_l\varphi\_k^lw\_{il}\right)
\end{array}\right)
\end{eqnarray}
Using $x\_1,...,x\_n$ are linearly independent, we deduce, for all $i\in\{1,...,n\}$,
\begin{eqnarray}
\lambda\_{i}+\sum\_{l=n+1}^{k}\lambda\_{l}w\_{il} & = & 0\\
\lambda\_{i}+\sum\_{l=n+1}^{k}\lambda\_{l}\varphi\_{k}^{l-i}w\_{il} & = & 0
\end{eqnarray}
which implies $\sum\_{l=n+1}^{k}w\_{il}\lambda\_{l}(1-\varphi\_{k}^{l-i})=0$
and $(\lambda\_{n+1},...,\lambda\_{k})\neq0$. In other words, for all $i$, $(w\_{in+1},...,w\_{ik})$ are linearly dependent, which happens with probability 0.


