Linear Algebra matrix notation My question is referring to the following $4 \times 6$ matrix:
$$\begin{bmatrix}
0 & 1 & 2 & 0 & 0 & 2 \\ 
0 & 0 & 0 & 1 & 0 & 3 \\ 
0 & 0 & 0 & 0 & 1 & 4 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix}$$
Then the book states "It is useful to denote the column numbers in which the leading entries $1$ occur, by  $c_1, c_2, ...c_r$, with remaining columns being $c_{r+1} ... c_n$ where $r$ is the number of non-zero rows. For this matrix is gives the example of $r=3$, $c_1 = 2$, $c_2=4$, $c_3=5$, $c_4=1$, $c_5=3$, $c_6=6$.
I don't understand this, what does it mean? How did they get all those numbers for each $c$?
Thanks everyone for your time.
 A: I haven't seen this notation before, and I don't see the purpose of it, which is probably why it is so hard to understand.  This seems to be some sort of intermediate (and complicated...) way of describing reduced row echelon forms.
The definition, stated a little less haphazardly, is:


*

*(1) $r$ is the number of nonzero rows in the matrix

*(2) $c_j$, for $j \le r$ : look at the $j^{th}$ nonzero row, $c_j$ is the column of the first nonzero value

*(3) $c_j$, for $j > r$ : look at the $j^\text{th}$ column that doesn't contain a pivot.  Suppose the $j^{th}$ column without a pivot is the $n^\text{th}$ column in the matrix.  $c_j = n$.


Example:
Here is your example matrix, with boxes around the pivots (the first non zero elements per row):
$$\begin{bmatrix}
0 & \boxed{1} & 2 & 0 & 0 & 2 \\ 
0 & 0 & 0 & \boxed{1} & 0 & 3 \\ 
0 & 0 & 0 & 0 & \boxed{1} & 4 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix}$$
This matrix has $3$ nonzero rows, so $r=3$:
$${
\boxed{\begin{array} {c c c c c c}
0 & 1 & 2 & 0 & 0 & 2 \\ 
0 & 0 & 0 & 1 & 0 & 3 \\ 
0 & 0 & 0 & 0 & 1 & 4 \\
\end{array}}\\
\begin{array} {c c c c c c}
0 & 0 & 0 & 0 & 0 & 0 \\
\end{array}
}$$
So now we want to find $c_1, c_2, c_3$.  To find $c_1$, look at the first nonzero row, and find out what column the pivot is in:
$$\begin{bmatrix}
0 & \boxed{1} & 2 & 0 & 0 & 2 \\ 
0 & 0 & 0 & 1 & 0 & 3 \\ 
0 & 0 & 0 & 0 & 1 & 4 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix} \text{First nonzero row, pivot in column 2}$$
So $c_1 = 2$.  For $c_2$:
$$\begin{bmatrix}
0 & 1 & 2 & 0 & 0 & 2 \\ 
0 & 0 & 0 & \boxed{1} & 0 & 3 \\ 
0 & 0 & 0 & 0 & 1 & 4 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix} \text{Second nonzero row, pivot in column 4}$$
So $c_2 = 4$.  For $c_3$:
$$\begin{bmatrix}
0 & 1 & 2 & 0 & 0 & 2 \\ 
0 & 0 & 0 & 1 & 0 & 3 \\ 
0 & 0 & 0 & 0 & \boxed{1} & 4 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix} \text{Third nonzero row, pivot in column 5}$$
So $c_3 = 5$.  Now we want to find the rest of the $c$ values, look at the columns without pivots:
$$
\boxed{\begin{array} {c} 0 \\ 0 \\ 0 \\ 0 \end{array}}
\begin{array} {c} 1 \\ 0 \\ 0 \\ 0 \end{array}
\boxed{\begin{array} {c} 2 \\ 0 \\ 0 \\ 0 \end{array}}
\begin{array} {c} 0 \\ 1 \\ 0 \\ 0 \end{array}
\begin{array} {c} 0 \\ 0 \\ 1 \\ 0 \end{array}
\boxed{\begin{array} {c} 2 \\ 3 \\ 4 \\ 0 \end{array}}
$$
That's columns $1, 3$, and $6$.  So $c_4 = 1$, $c_5 = 3$, and $c_6 = 6$.
