Why do I mistakenly think there's an error in this proof by Hoffman and Kunze?

Question

Theorem 4 (Chapter 1) in Hoffman and Kunze's Linear Algebra, 2nd edition (HK) says

Theorem 4: Every $m \times n$ matrix over the field $F$ is row-equivalent to a row-reduced matrix.

Before supplying the proof, I'll note their definition of a row-reduced matrix:

Definition. An $m \times n$ matrix $R$ is called row-reduced if: (a) the first non-zero entry in each non-zero row of R is equal to 1; (b) each column of $R$ which contains the leading non-zero entry of some row has all its other entries 0.

Their proof of Theorem 4 is now given as below:

Proof. Let A be an $m \times n$ matrix over $F$. If every entry in the first row of $A$ is 0, then condition (a) is satisfied insofar as row 1 is concerned. If row 1 has a non-zero entry, let $k$ be the smallest positive integer $j$ for which $A_{1j} \neq 0$. Multiply row 1 by $A_{1k}^{-1}$, and then condition (a) is satisfied with regard to row 1. Now for each $i \geq 2$, add ($-A_{ik}$) times row 1 to row i. Now the leading non-zero entry of row 1 occurs in column $k$, that entry is 1, and every other entry in column $k$ is 0. Now consider the matrix which has resulted from above. If every entry in row 2 is 0, we do nothing to row 2. If some entry in row 2 is different from 0, we multiply row 2 by a scalar so that the leading non-zero entry is 1. In the event that row 1 had a leading non-zero entry in column $k$, this leading non-zero entry of row 2 cannot occur in column $k$; say it occurs in column $k' \neq k$. By adding suitable multiples of row 2 to the various rows, we can arrange that all entries in column $k'$ are 0, except the 1 in row 2. The important thing to notice is this: In carrying out these last operations, we will not change the entries of row 1 in columns $1, . . . , k$; nor will we change any entry of column $k$. Of course, if row 1 was identically 0, the operations with row 2 will not affect row 1. Working with one row at a time in the above manner, it is clear that in a finite number of steps we will arrive at a row-reduced matrix.

My concern is with the bolded parts, and in particular the claim that "these last operations, we will not change the entries of row 1 in columns $1, . . . , k$". Surely if $k>k'$ this need not be true in general, for then any entry $A_{1m}$ with $k' \leq m <k$ will be changed since it is permitted that $A_{2m} \neq 0$ for such $m$? Is it possible that HK make an error here and instead meant to say "these last operations, we will not change the entries of row 1 in columns $1, . . . , k'$"?

Or, perhaps, if we consult this proof from ProofWiki, we see that the first step is to consider the first non-zero column $j$, and then take any row which contains a non-zero entry in said column. HK do no such thing, and just plug on ahead. If they did do this, I think their proof would be rescued since we would always have $k<k'$, right?

This did not appear as an erratum in this thread, so I am concerned that I am misunderstanding the proof.

Answer 1

The proof is okay as stated.

If $k' < k$, then entries $A_{1,1} \ldots A_{1,k-1}$ are zero by the time we look at row 2. In particular, $A_{1,k'} = 0$, so no change is needed to row 1. Or more exactly, the algorithm says to add $-A_{1,k'}$ times row 2 to row 1, but that's just adding zeros and leaves all of row 1 unchanged, not just its first $k$ entries.

(And as you did realize, if $k' > k$, then since $A_{2,1} \ldots A_{1,k'-1}$ are all zero, the entries $A_{1,1} \ldots A_{1,k}$ are not changed by this step.)