Express vector as linear combination of vectors let's say I have two vectors:
$u = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ and $v = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$
I can easily express   $\; w = \begin{pmatrix} 3 \\ 2 \\ 2 \end{pmatrix}$ as a linear combination of u and v.
My problem is: how can I express a vector v as a linear combination of n vectors?
I searched all around and I just found about expressing a $R^n$ vector as linear combination of exactly $n$ vectors from $R^n$.

*

*$[u_1 \; u_2 \; u_3 \; | \; v]$

*apply gaussian elimination;

But I didn't discover how to express it with k vectors ($ k \ne n$), for example, if I apply this method with u and v to find w, this is what happens:
$ \left[
    \begin{array}{cc|c}
      1&2&3\\
      1&1&2\\
      1&1&2
    \end{array}
\right] $   applying elimination: $ \left[
    \begin{array}{cc|c}
      1&2&3\\
      0&-1&-1\\
      0&0&0
    \end{array}
\right] $
And I know the answer for $\;au + bv = w\;$ is $\;(a = 1, \;b = 1)$ but I couldn't extract this information from the matrix.
Formalizing: Given a $R^n$ vector, how can I express it as a linear combination of $k$ vectors, where $k \ne n$ ?
 A: You wish to determine how to solve $\lambda_1 u+\lambda_2 v=w$ where
\begin{align*}
u &= \langle 1,1,1\rangle &
v &= \langle 2,1,1\rangle &
w &= \langle a,b,c\rangle
\end{align*}
To do so, form the augmented system
$$
M = 
\left[\begin{array}{rr|r}
1 & 2 & a \\
1 & 1 & b \\
1 & 1 & c
\end{array}\right]
$$
The system $M$ can be row-reduced with the following steps


*

*add $-1$ times row 1 to row 2

*add $-1$ times row 1 to row 3

*scale row 2 by $-1$

*add $-2$ times row 2 to row 1

*add $1$ times row 2 to row 3


These row-reductions gives
$$
\left[\begin{array}{rr|r}
1 & 0 & -a + 2 \, b \\
0 & 1 & a - b \\
0 & 0 & -b + c
\end{array}\right]
$$
This shows that $\lambda_1 u+\lambda_2 v=w$ has a solution if and only if $b=c$. If $b=c$ then $\lambda_1 u+\lambda_2 v=w$ is solved by
\begin{align*}
\lambda_1 &=-a+2\,b &
\lambda_2 &= a-b
\end{align*}
In the comments you ask how to express $v=(6,7,6)$ in terms of 
\begin{align*}
u_1 &= (1,1,1) &
u_2 &= (1,1,2) &
u_3 &= (1,2,1) &
u_4 &= (2,1,1)
\end{align*}
This is equivalent to solving the system
$$
\left[\begin{array}{rrrr|r}
1 & 1 & 1 & 2 & 6 \\
1 & 1 & 2 & 1 & 7 \\
1 & 2 & 1 & 1 & 6
\end{array}\right]
$$
Row-reducing gives
$$
\left[\begin{array}{rrrr|r}
1 & 0 & 0 & 4 & 5 \\
0 & 1 & 0 & -1 & 0 \\
0 & 0 & 1 & -1 & 1
\end{array}\right]
$$
This shows that 
$$
(5-4\,\lambda) u_1 +
\lambda u_2 +
(1+\lambda)u_3 +
\lambda u_4=v
$$
for any $\lambda\in\Bbb R$.
