if $A$ and $B$ are row equivalent and rows of $A$ are linearly independent then each row of $B$ expressible as a linear combination of $A$?

Question

Here's the question:

If $A$ and $B$ are row equivalent and rows of $A$ are linearly independent then is each row of $B$ expressible as a linear combination of rows of $A$?

Seems legit. However, I'm not sure how to prove it. I know that say that a linear combination is by definition obtained by a sequence of elementary row operations. Also, I know that elementary row operations do not affect row rank. Thus it must be the case that the rows of $B$ must also be linearly independent. Can I get some hints?

Answer 1

Here is a short conceptual argument for "column equivalent" matrices, that can be adapted by taking transposes, or just dualizing:

Column operations correspond to multiplication by invertible elementary matrices.

In particular, given column-equivalent linear maps $T,S:V \to W$, there is an invertible matrix $F:W \to W$ so that $F \circ T=S$. In particular, if we take the column space of $T$ as a spanning set for its image, then surely every column in $S$ can be written as $F(v)$, for $v \in \mathrm{Im}(T)$, and in particular, it is a linear combination.