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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 $[c_1,...,c_k]^T$.

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Suppose that $r_j = \sum_{p=1}^k b_{pj} r_p'$ for scalars $b_{pj}$. We can then write $$ \sum_{j=1}^k c_j r_j = \sum_{j=1}^k c_j \sum_{p=1}^k b_{pj} r_p' = \sum_{p=1}^k \left(\sum_{j=1}^k b_{pj} c_j\right)r_p'. $$

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  • $\begingroup$ sorry a bit confused about the second line...can you add a bit of word explanation? $\endgroup$ Nov 7 '20 at 1:12
  • $\begingroup$ are you saying we let rj equal to r'p and cj equal to the sum of $b_pjc_j$=$c_j'$? $\endgroup$ Nov 7 '20 at 1:26
  • $\begingroup$ @jamesblack Yes ${}{}{}{}{}$ $\endgroup$ Nov 8 '20 at 1:42

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