$v=(-1,2,3,3)$ $S=\{(1,1,1,1),(0,1,1,1),(0,0,1,1)\}$
Normally what I would do is row reduce the augmented matrix with columns of the vectors in S.
$$A=\left[ {\begin{array}{rrr|r} 1&0&0&-1\\ 1&1&0&2\\ 1&1&1&3\\ 1&1&1&3\\ \end{array} } \right]$$
Which I find to be
$$A=\left[ {\begin{array}{rrr|r} 1&0&0&-1\\ 0&1&0&3\\ 0&0&1&1\\ 0&0&0&0\\ \end{array} } \right]$$
But the point of this is supposed to be finding an x such that $Ax=v$ and that the existence of such an x allows me to conclude that the matrix is in the spawn. However I can't multiply a 4x3 matrix with a 4x1 matrix so I don't understand why this works.
I have noticed however that these numbers, (-,1,3,1) correspond to a linear combination of vectors -1(1,1,1,1)+3(0,1,1,1)+1(0,0,1,1)=(-1,2,3,3). So this process does show me it's in the spawn but not in the way I thought it was supposed to. My thought is that since the space of up to 3rd degree polynomials is dimension 4 and I only have 3 vectors in this set, that is why this is happening but I'm not sure why.
Could someone explain to me why this works?
By the first system we are looking for $x, y, z$ such that
$$v=xv_1+yv_2+zv_3 \iff \begin{bmatrix} 1&0&0\\ 1&1&0\\ 1&1&1\\ 1&1&1\\ \end{bmatrix} \begin{bmatrix} x\\ y\\ z\\ \end{bmatrix} =\begin{bmatrix} -1\\ 2\\ 3\\ 3\\ \end{bmatrix}$$
and by RREF we have found the solution $(x,y,z)=(-1,3,1)$ and therefore
$$v=-v_1+3v_2+v_3$$