Now denote the location of the columns containing the leading entries of the successive rows of a k\times n matrix A in reduced row echelon form (the pivots) as (L_1, \dots, L_j), with : 0 where j \le k is the dimension of the
row space of the matrix. The data (k, n, L_1, \ldots, L_j) will be called the
shape of A, which has leading non-zero entries \{ A_{i, L_i} =1\}_{i=1, \dots, j}, the entries in the column L_i above and below it vanish, and so do all those to the left of it within the same row, as well as all entries in the ith row for i>j: :\begin{align} A_{i, L_i} =1\qquad &\text{for } i=1, \dots, j,\\ A_{l, L_i}=0\qquad &\text{for } l\ne i,\\ A_{i, l} =0\qquad &\text{for } lj \end{align}. Since all other entries are arbitrary elements of the base field K, the set A(k, n, L_1, \ldots, L_j) of all reduced echelon form matrices with shape (k, n, L_1, \ldots, L_j) is a -affine space of dimension :\text{dim}(A(k,n, L_1, \dots, L_j))=nj -\frac{1}{2}j(j-1)- \sum_{i=1}^j L_i. To see this, note that, of the nj possible matrix entries within the first j rows, j^2 are determined as 0's and 1's because they are in the columns (L_1, \dots, L_j) containing the pivots. A further \sum_{i=1}^j(L_i-1) are also required to be 0, because they are to the left of the pivots, but of these, ::\sum_{i=0}^{j-1}i = \frac{1}{2}j(j-1) are also in the columns (L_1, \dots , L_j). Therefore, the total number of entries that are not fixed to be equal to 0 or 1 is :: nj - j^2 +\frac{1}{2}j(j-1) - \sum_{i=1}^j L_i +j = nj - \frac{1}{2}j(j-1)- \sum_{i=1}^j L_i. ==Notes==