Proof of Theorem BRS

From Theorem REMRS we know that $\rsp{A}=\rsp{B}$. If $B$ has any zero rows, these are columns of $\transpose{B}$ that are the zero vector. We can safely toss out the zero vector in the span construction, since it can be recreated from the nonzero vectors by a linear combination where all the scalars are zero. So $\rsp{A}=\spn{S}$.

Suppose $B$ has $r$ nonzero rows and let $D=\set{d_1,\,d_2,\,d_3,\,\ldots,\,d_r}$ denote the indices of the pivot columns of $B$. Denote the $r$ column vectors of $\transpose{B}$, the vectors in $S$, as $\vectorlist{B}{r}$. To show that $S$ is linearly independent, start with a relation of linear dependence \begin{equation*} \lincombo{\alpha}{B}{r}=\zerovector \end{equation*}

Now consider this vector equality in location $d_i$. Since $B$ is in reduced row-echelon form, the entries of column $d_i$ of $B$ are all zero, except for a leading 1 in row $i$. Thus, in $\transpose{B}$, row $d_i$ is all zeros, excepting a 1 in column $i$. So, for $1\leq i\leq r$, \begin{align*} 0 &=\vectorentry{\zerovector}{d_i}&&\knowl{./knowls/definition.ZCV.knowl}{\text{Definition ZCV}}\\ &=\vectorentry{\lincombo{\alpha}{B}{r}}{d_i} &&\knowl{./knowls/definition.RLDCV.knowl}{\text{Definition RLDCV}}\\ &= \vectorentry{\alpha_1\vect{B}_1}{d_i}+ \vectorentry{\alpha_2\vect{B}_2}{d_i}+ \vectorentry{\alpha_3\vect{B}_3}{d_i}+ \cdots+ \vectorentry{\alpha_r\vect{B}_r}{d_i} &&\knowl{./knowls/definition.MA.knowl}{\text{Definition MA}}\\ &= \alpha_1\vectorentry{\vect{B}_1}{d_i}+ \alpha_2\vectorentry{\vect{B}_2}{d_i}+ \alpha_3\vectorentry{\vect{B}_3}{d_i}+ \cdots+ \alpha_r\vectorentry{\vect{B}_r}{d_i} &&\knowl{./knowls/definition.MSM.knowl}{\text{Definition MSM}}\\ &= \alpha_1(0)+ \alpha_2(0)+ \alpha_3(0)+ \cdots+ \alpha_i(1)+ \cdots+ \alpha_r(0) &&\knowl{./knowls/definition.RREF.knowl}{\text{Definition RREF}}\\ &=\alpha_i \end{align*}

So we conclude that $\alpha_i=0$ for all $1\leq i\leq r$, establishing the linear independence of $S$ (Definition LICV).