I have recently learned that instead of transposing the $mxn$ matrix A and finding a basis for $N(A^T)$ through direct computation, I could find the elementary row operation matrix $E$ that satisfies $EA=R$, where $R=rref(A)$. The rows of $E$ corresponding to the last $m-r$ zero rows of $R$ would then give me a basis for $N(A^T)$, where $r=rank(A)$. The detailed method is described in this answer.
I can see that these rows of $E$ are solutions to $A^Tx=0$, but I can't seem to wrap my head around why these $m-r$ rows span the left nullspace of A, and why they are guaranteed to be linearly independent. As a consequence, I also don't understand why $dim(N(A^T))=m-r$. So far all the books and resources I have consulted skimp on the explanation or proof. Could someone explain or provide a proof that the last $m-r$ rows of $E$ forms a basis for $N(A^T)$. Thank you!
