# Linear combination of Matrix Row Vectors

For $$A \in M_{m×n}$$ we define $$r(A)$$ to be the smallest integer k ≥ 1 such that $$A =\sum^k_{j=1} c_j r_j$$ for some $$c_j$$ column vectors and $$r_j$$ row vectors.

If k = row-rank(A), then letting $$r_1' , . . . , r_k'$$ be a subset of the rows of A forming a basis of the row-space, show $$A = \sum^k_{j=1} c_j' r_j'$$ for some column vectors $$c_j'$$.

I know that A is basically the sum of a series of $$c_j' r_j'$$ and every row should equal to a linear combination of $$r_1' , . . . , r_k'$$, i.e. $$c_1r_1' + . . . +c_k r_k'$$. But I am trying to find the connection between how we are guaranteed with a set of column vectors such that $$A=\sum c_j' r_j'$$, i.e. how sum of $$c_j$$'s equal to $$[c_1,...,c_k]^T$$.

Suppose that $$r_j = \sum_{p=1}^k b_{pj} r_p'$$ for scalars $$b_{pj}$$. We can then write $$\sum_{j=1}^k c_j r_j = \sum_{j=1}^k c_j \sum_{p=1}^k b_{pj} r_p' = \sum_{p=1}^k \left(\sum_{j=1}^k b_{pj} c_j\right)r_p'.$$
• are you saying we let rj equal to r'p and cj equal to the sum of $b_pjc_j$=$c_j'$? Nov 7 '20 at 1:26
• @jamesblack Yes ${}{}{}{}{}$ Nov 8 '20 at 1:42