When writing solution set in vector form, are the vectors necessarily independent? If I solve a system of linear equations, and its REF has k free variables $\{x_1,...,x_k \}$, are the vectors $\mathbf{v}_i$ in the vector form solution set
$\mathbf{x}=x_1\mathbf{v}_1+...+x_k\mathbf{v}_k$ necessarily linearly independent? 
Up until now I've been working under assumption that they are, unfortunately I cannot think of a simple way to prove it. Is there a particular step that I do when writing vector form of the solution based the REF of augmented matrix that forces them to be linearly independent?
 A: The vector $\mathbf x$ can be written as $\begin{bmatrix} \hat x_1 & \hat x_2 & \cdots & \hat x_m & x_1 & x_2 & \cdots & x_k\end{bmatrix}^{\mathsf T}$, where $x_1, x_2, \dots, x_k$ are the free variables and $\hat x_1, \hat x_2, \dots, \hat x_m$ are the pivot variables; after row reduction, we obtain expressions for $\hat x_1, \hat x_2, \dots, \hat x_m$ in terms of the free variables.
The key property we need is this: if $\mathbf x = \mathbf 0$, then $x_1 = x_2 = \dots = x_k = 0$. In this representation of $\mathbf x$, it holds because $x_1, x_2, \dots, x_k$ are entries of $\mathbf x$. So if all entries of $\mathbf x$ are $0$, then in particular these entries are $0$.
When you convert this representation of $\mathbf x$ to a linear combination
$$
    \mathbf x = x_1 \mathbf v_1 + x_2 \mathbf v_2 + \dots + x_k \mathbf v_k,
$$
the meaning of the variables $x_1, x_2, \dots, x_k$ stays the same and so this key property is retained. But now it turns into the definition of the linear independence of $\mathbf v_1, \mathbf v_2, \dots, \mathbf v_k$:
$$
    \text{if }x_1 \mathbf v_1 + x_2 \mathbf v_2 + \dots + x_k \mathbf v_k=\mathbf 0, \text{ then } x_1 = x_2 = \dots = x_k = 0.
$$

A less intuitive but more concrete way to see this: in each vector $\mathbf v_i$, the entry in $x_i$'s position is $1$, and each entry in $x_j$'s position for $j \ne i$ is $0$. So if a linear combination is $0$, each coefficient must be $0$, because every vector $\mathbf v_i$ has a coordinate no other vectors control.
A: Take the following linear system :
(1) $\{\hat x_1 + x_3 + 2*x_4 + x_5 = 3 $
(2) $\{\hat x_2 + 2*x_4 + x_5 = 7 $
It is in Row Echelon Form (and in Reduced Row Echelon Form, too); $\hat x_1$ and $\hat x_2$ are the pivot variables and $x_3$, $x_4$ and $x_5$ are the free variables.  
Here, it is obvious that the subsystem that you are mentioning:
(3) $\{x_3 + 2*x_4 + x_5 = 0 $
(4) $\{2*x_4 + x_5 = 0 $
can admit solutions where $x_3=0$, $x_4=-1$ and $x_5=2$. 
So, (3, 7, 0, -1, 2) is a solution to the system (1)&(2),
(0, -1, 2) is a solution to the system (3)&(4),
but not all free variables are equal to zero. To say the same thing in another form : the set of vectors {(1,0),(2,2),(1,1)} are linearly dependent.
--  
Further reading
For solving linear systems, using the coefficient matrix
$$
    \begin{bmatrix}
    1 & 0 & 1 & 2 & 1 \\
    0 & 1 & 0 & 2 & 1 \\
    \end{bmatrix}
$$
and the augmented coefficient matrix
$$
    \begin{bmatrix}
    1 & 0 & 1 & 2 & 1 & 3\\
    0 & 1 & 0 & 2 & 1 & 7\\
    \end{bmatrix}
$$
may be helpful, as finding the rank of these matrices will help you a lot to know if the system has a solution (and in which form), or no solution. 
Read more on wikipaedia
I hope this helps!
