Writing a vector has a linear combination of two vectors: is the solution unique? Problem:
Consider the vectors $u = (1,2,3)$ and $v = (2,3,1)$ in $R^3$. Write
$w = (1,3,8)$ as a linear combination of $u$ and $v$.
Answer:
$$ c_1( 1,2,3) + c_2( 2,3,1 ) = (1,3,8) $$
Now to find $c_1$ and $c_2$ we setup a matrix.
$$
\begin{bmatrix}
1 & 2 & 1 \\
2 & 3 & 3 \\
3 & 1 & 8
\end{bmatrix}
\sim
\begin{bmatrix}
1 & 2 & 1 \\
0 & -1 & 1 \\
0 & -5 & 5
\end{bmatrix}
$$
This gives us $c_2 = -1$.
\begin{align*} 
c_1 + 2c_2 &= 1 \\
c_1 + 2(-1) &= 1 \\
c_1 &= 3 \\
\end{align*}
Hence the solution is $w =  3u - v$. This answer matches the back of the book; however it seems to me that there should be more than one solution. Is this solution unique?
 A: If $\{ v_1, v_2 \}$ is linearly independent then the linear combinations are unique. This is also true in the case $\{v_1, \ldots, v_n \}.$
For see this suppose we have $w=c_1 v_1 + c_2 v_2 = d_1v_1 + d_2v_2$. Then
$$ (c_1 - d_1) v_1 + (c_2-d_2)v_2=0. $$
As $\{v_1, v_2\}$ is linearly independent we know that $c_1-d_1=0$ and $c_2 - d_2=0$, i.e. $c_1=d_1$ and $c_2=d_2$.
A: To check for uniqueness, the basic approach is to assume there exists two different solutions, and see what happens. If the problem is truly unique, then you will run into a contradiction.
Suppose there were two combinations, $c_1, c_2$ and $d_1, d_2$. Then,
$$(1,3,8) = c_1(1,2,3) + c_2(2,3,1) = d_1(1,2,3) + d_2(2,3,1)$$
This gives,
\begin{align}
c_1 + 2c_2 &= d_1 + 2d_2 \\
2c_1 + 3c_2 &= 2d_1 + 3d_2 \\
3c_1 + c_2 &= 3d_1 + d_2 \\
\end{align}
Rearranging, we get
$$ c_1 - d_1 = 2(d_2 - c_2) = \frac{3}{2} (d_2 - c_2) = \frac{1}{3} (d_2 - c_2) $$
If the two combinations are different (i.e. $c_1 \neq d_1, c_2 \neq d_2$), then the above yields a contradiction that $2 = \frac{3}{2} = \frac{1}{3}$. To avoid this contradiction, the two combinations must be the same. Therefore, the problem has at most one unique solution.
