Why does putting linearly dependent vectors in a matrix and row reducing yield their coordinates? Say we have the vectors $a = (1,2,3,4),$ $b = (2,-1,1,1),$ $c = (-1,8,7,10)$ and $d = (-5, 5, 0, 1).$ They are linearly dependent and span a plane. We can prove this by putting them as columns in a matrix $$(a \:\:\: b \:\:\: c \:\:\: d) = \begin{pmatrix}1&2&-1&-5\\2&-1&8&5\\3&1&7&0\\4&1&10&1
\end{pmatrix},$$ and row reducing to find its rank. Row reducing yields $$ \begin{pmatrix} 
1 & 0 & 3 & 1\\
0 & 1 & -2& -3 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0\end{pmatrix},$$ we see the rank is two and that $a$ and $b$ span the plane.
My question is the following: if we let $a$ and $b$ be a basis for the plane that these four vectors span, we see that the first two rows tell us the coordinates with respect to this (ordered) basis, for example $c$ with respect to the basis $a$ and $b$, has coordinates $(3,-2)$ since $c = 3a -2b$. Why is it the case that this shows up in this row reduced matrix? I've been thinking about it for a while and I can't seem to prove it or even explain it with intuition.
 A: Let's begin with what I suspect you already know about row reduction: an $m \times n$ matrix $A$ and its row-reduced version $R$ have the same nullspace. In other words,
$$
Ax = 0 \iff Rx = 0, \quad x \in \Bbb R^n.
$$
Now, let $a_1,\dots,a_n$ denote the columns of $A$. Suppose that $a_1,a_2$ are linearly independent, but $a_3 = 3a_1 - 2a_2$. What does this tell us about the columns of $R$?
Notice that this equation actually gives us a solution to the equation $Ax = 0$. In particular,
$$
a_3 = 3a_1 - 2a_2 \implies 3a_1 - 2a_2 - a_3 = 0 \implies\\
\pmatrix{a_1 & a_2 & a_3 & a_4& \cdots & a_n} \pmatrix{3\\-2\\-1\\0\\ \vdots \\ 0} = 0.
$$
That is, $x = (3,-2,-1,0,\dots,0)$ is an element of the nullspace of $A$. The same must hold for $R$.
On the other hand, note that because $a_1,a_2$ are linearly independent, each of these becomes a pivot column in the RREF matrix $R$ (more generally, $a_k$ becomes a pivot column iff $a_k$ is outside of the span of $a_1,\dots,a_{k-1}$). With that, we know that $R$ has the form
$$
R = \pmatrix{r_1 & r_2 & r_3 & r_4 & \cdots & r_n} = \pmatrix{1 & 0 & ? & \cdots\\
0 & 1 & ? & \cdots\\
0 & 0 & ? & \cdots \\
\vdots & \vdots & \vdots & \cdots}.
$$
From the fact that $x = (3,-2,-1,0,\dots,0)$ is in the nullspace of $R$, we have
$$
Rx = 0 \implies 3r_1 - 2r_2 - r_3 = 0 \implies r_3 = 3r_1 - 2r_2.
$$
So, we can conclude that $r_3 = (3,-2,0,\dots,0)$.
A: Starting with the matrix $A$ as follows
$A = [ v_1 , v_2, ..., v_m ] $
Reduce it to reduced-row echelon form $\widetilde{A} $, this is equivalent to premultiplying $A$ with the matrix $E$
$ \widetilde{A} = E A = [E v_1, Ev_2, ..., E v_m] $
where $E v_k$ is either an elementary unit vector $e_i$ or a linear combination of the other $e_i$'s.  Thus if
$ v_k = \displaystyle \sum_{i=1, i \ne k}^m \alpha_i v_i $
Then
$ E v_k = \displaystyle \sum_{i=1, i \ne k}^m \alpha_i E v_i $
Thus we can identify the vectors that are linearly independent, by identifying those columns that reduce to different $e_i$'s, while the others are linearly dependent on them.  And the coefficients of this dependency are the entries in the reduced row echelon form for that column.
