The row space of a matrix $A$ is the span $\langle a_1, a_2, \dots, a_n \rangle$ of the $n$ rows of $A$. Does anyone know a reference, or quick proof, for the claim that there exist a subset of these vectors, $(a_{r_1}, \dots, a_{r_k})$, where $k = \operatorname{rank}(A)$, that spans the row space of $A$?
I am able to prove this by explicitly generating such a subset using row reduction. I.e., the rows containing pivots (and undoing any row transpositions) after $A$ is put into row echelon form gives one subset of these vectors. However, this seems like it should be an easily accessible 'fact'.
Here is a sketch of a quick proof using induction on the rank of a matrix. The claim is clearly true when the rank is 1. We will now show that the claim is true for matrices $A$ with rank $k+1$, assuming that the claim holds for matrices of rank $k$. Let $S = \{a_{r_1}, a_{r_2}, \dots, a_{r_l}\}$ denote the subset of rows of $A$ with minimal cardinality that spans the row space of $A$, and suppose for a contradiction that $l > k + 1$. Since this set has minimal cardinality, it must be possible to remove a row $a_{s_{k+1}}$ such that $S \setminus \{a_{s_{k+1}}\}$ has rank $k$. By the induction hypothesis there exist a subset of $S \setminus \{a_{s_{k+1}}\}$, comprising of the rows $\{a_{s_1}, a_{s_2}, \dots, a_{s_k}\}$, of cardinality $k$, that has rank $k$. But $\{a_{s_1}, a_{s_2}, \dots, a_{s_{k+1}}\}$ has rank $k + 1$, a contradiction to the minimality of the cardinality of $S$.