TITLE: Show that the linear combination is unique?
QUESTION [6 upvotes]: I have the following question:
Let S be a subset of the vector space $\Bbb R^3$ defined by
$$
S =  \left\{  ~
\begin{bmatrix}2\\-1\\ 0\end{bmatrix},
\begin{bmatrix}1\\3\\ -2\end{bmatrix}, 
\begin{bmatrix}1\\1\\ 4\end{bmatrix} 
~\right\}
$$  

show that 
$v= \begin{bmatrix}-4\\4\\ -6\end{bmatrix}$ is in $\text{span}(S)$ by constructing it as a linear combination of the vectors in $S$.
Show that this linear combination is unique.

This is what I got after solving part 1: 
$$
c_1\begin{bmatrix}2\\-1\\ 0\end{bmatrix}
 + c_2\begin{bmatrix}1\\3\\ -2\end{bmatrix} 
+ c_3\begin{bmatrix}1\\1\\ 4\end{bmatrix} = \begin{bmatrix}-4\\4\\ -6\end{bmatrix}$$
after solving this I got  $c_1= -2$, $c_2= 1$, $c_3= -1$.
How do I show that the linear combination is unique?

REPLY [4 votes]: You said you solved the equation for $c_1, c_2, c_3$.
If, by this, you mean that you found the complete set of all solutions for these variables (as people often mean by "solve"), then you've already proven the linear combination is unique, since there was only one solution.