linear algebra - Elementary row operations on $A$ don't change the row-space of $A$

Assume that $F$ is a field. Consider a matrix like $A \in M_{m,n}(F)$.

If we look at each row of $A$, as a member of $F^n$, the subspace produced by rows of $A$ is called the row-space of $A$.

a) Prove that elementary row operations on $A$ don't change the row-space of $A$. Then, Conclude that foreach matrix like $B$, if $A$,$B$ are row equivalent, then they have the same row-space.

b) Prove that $A \in M_{n}(F)$ is invertible iff the row-space of it is $F^n.$

Note : I'm confused with the definitions ... I have no idea how to solve this problem.

Answer

Let $v_1,\ldots,v_n$ denote the rows of $A$, and let $V$ be the row-space of $A$. In other words,
$$
V=\text{span}\,\{v_1,\ldots,v_n\}.
$$
Elementary operations could consist of a permutation of rows, which amount to permuting some $v_j$ and $v_k$ above; such operation will not change the span of $v_1,\ldots,v_n$. The same happens with multiplying by a nonzero number: $v_1,\ldots,v_n$, and $v_1,\ldots,\lambda v_k,\ldots, v_n$ span the same subspace.

The last kind of elementary operation consists of replacing $v_k$ with $v_k+\lambda v_j$. In this case, we can write

$$
\alpha_1v_1+\cdots+\alpha_nv_n=\alpha_1v_1+\cdots+\alpha_{k-1}v_{k-1}+\alpha_k(v_k+\lambda v_j)+\alpha_{k+1}v_{k+1}+\cdots (\alpha_j-\alpha_k\lambda)v_j+\cdots+\alpha_nv_n.
$$
So every linear combination of $v_1,\ldots,v_n$ is also a linear combination of $v_1,\ldots,v_n$ with $v_k$ replaced by $v_k+\lambda v_j$.

In summary, after doing any elementary operation to $v_1,\ldots,v_n$, the span doesn't change. It follows directly that if $A$ and $B$ are row equivalent, since the rows of $B$ can be obtained by elementary operations from the rows of $A$, the spans of their rows are equal.

If $A$ is invertible, then it is row equivalent to $I$, and so its row space is $F^n$. Conversely, if the row space of $A$ is $F^n$, then by writing each of $v_1,\ldots,v_n$ in terms of the canonical basis, we get a recipe to go from $I$ to $A$ by elementary operations, and so $A$ is invertible.