I'm currently working through the proof that the column and row rank of a matrix $A$ over a field $F$ are equal, using the concept of linear combinations (more specifically I'm following the proof sketch given on wikipedia). There seems to be several posts on this forum handling this proof, but none of them seem to answer a confusion I have at a certain step in the proof:
Assuming we have an $m \times n$ matrix $A$ over a field $F$, the beginning of the proof shows that we can find an $m \times r$ matrix $C$ and an $r \times n$ matrix $R$, such that $A=CR$ (where $r$ is the column rank of $A$). I assume you are familiar with the proof, because I won't explain the details here. The proof then proceeds to show that the rows of $R$ span the row space of $A$, which is the step I'm confused at:
so if $R_1,\ldots R_r$ are the rows of $R$, it is shown that Span$(R_1,\ldots , R_r)= $rowsp$(A)$. At this point I'm only able to show that each element of rowsp$(A$) is a linear combination of the rows of $R$, and thus Span$(R_1,\ldots , R_r)\supseteq$ rowsp$(A)$. But I'm having difficulty understanding why the other inclusion holds. At least I'm assuming that it should hold, because that is the meaning of the rows of R span the row space of A.
I hope someone could shed some light on this.
Best Answer
We'll denote by $r_c, r_r$ the column rank and row rank of an $m\times n$ matrix $A$ (respectively), and we decompose $A=CR$ with $C$ an $m\times r_c$ matrix and $R$ an $r_c\times n$ matrix.
You've shown that $\operatorname{rowspan}(A)\subseteq\operatorname{rowspan}(R)$, therefore $\dim(\operatorname{rowspan}(A))\leq\dim(\operatorname{rowspan}(R))$.
Notice that $R$ has $r_c$ rows, therefore $\dim(\operatorname{rowspan}(R))\leq r_c$, so in total we get $$\dim(\operatorname{rowspan}(A))\leq\dim(\operatorname{rowspan}(R))\leq r_c \tag{$*$}$$ but from the definition of the row-rank of a matrix, $r_r=\dim(\operatorname{rowspan}(A))$, therefore $r_r\leq r_c$.
Applying this same argument to $A^T$ gives us that $r_c\leq r_r$, so in total we get $r_c=r_r$ as we wanted.
Furthermore, once we have this equality, we get that the inequalities in $(*)$ are equalities, that is we have $\dim(\operatorname{rowspan}(A))=\dim(\operatorname{rowspan}(R))$, combining this with $\operatorname{rowspan}(A)\subseteq\operatorname{rowspan}(R)$ we get $\operatorname{rowspan}(A)=\operatorname{rowspan}(R)$.
So after we've shown $r_r=r_c$ we can say that indeed the rows of $R$ span the row-space of $A$ (before that we could only say they span a vector space which has the row-space of $A$ as a subspace)