I have several vectors $y_1, y_2, y_3 \cdots y_n \in R^n$ that I need to linearly map to a single vector $x_0 \in C^n$. The same matrix should also map $z_1, z_2, z_3, \cdots z_n \in R^n$ to a single vector $x_1 \in C^n$ .
I need to construct this matrix. So, the ys are distant from the zs in the $R^n$ space. But within the y's, the vectors are close to each other, for example, $y_2$ is a noisy version of $y_1$, and so on...So, all the y's need to be mapped to $x_0$ and all the z's to $x_1$.
I thought about using least squares. For example, doing
V = lsqminnorm( X', D' )' in matlab, where D is a matrix of repeated entries of $x_0$ and $x_1$ and X are the stacking of $y_1 y_2 y_3... z_1 z_2, ...$
But I don't know if this is valid since this matrix has a null space and, therefore, is noninvertible. In this case, would the least norms work?
Another possibility is creating D with some arbitrarily small noise so that the columns for the repeated $x_0$ terms and $x_1$ terms are not identical.
In practice, is this mapping possible?
In short, identifying $\mathbb C^n$ with $\mathbb R^{2n}$, you want to find a linear map from $\mathbb R^n$ to $\mathbb R^{2n}$ mapping each $y_i$ to $x_0$ and each $z_i$ to $x_1$. That is a non-homogeneous linear system of equations. In the generality of your question, it may not have solutions (for instance, if $y_2=2y_1$ and $x_0 \neq 0$).
In fact, since the family of vectors $(y_1, \ldots, y_n, z_1, \ldots, z_n)$ is a family of $2n$ vectors in $\mathbb R^n$, it is never linearly independent and so unless you have some very specific situation you should not expect that this is possible.
In any case, if there are solutions, you can find them numerically like any such linear system, with e.g. the Gauss algorithm.