Silliness: $\exists~X~\text{s.t.}~AX=B \iff B\in R(L_A)$ So, I am asked to prove that the system of linear equations $AX=B$ has $\color{black}{a~solution}$ if and only if $B\in R(L_A)$. $R$ denotes the "range of" and $L_A$ is left multiplication by $A$. If the question says "a solution" does that mean that I can rely on the fact that it's invertible?
Friedberg's 1989 edition of Linear Algebra says the above are equivalent, namely that $\exists~X~\text{s.t.}~AX=B \iff B\in R(L_A)$:

How am I to "prove" this?
 A: I think it's best if we state the definition of range first.
Definition: Let $A$ be an $m\times n$ matrix and consider the linear mapping $L_A:\mathbb{F}^n\rightarrow \mathbb{F}^m$ induced by $A$ as $L_A(\mathbf{x}) = A\mathbf{x}$. We denote the range or image of $L_A$ as the set of all vectors $\mathbf{b}\in\mathbb{F}^m$ such that $\mathbf{b}=L_A(\mathbf{x})$ for some $\mathbf{x}\in \mathbb{F}^n$. We denote the range as 
$$\mathrm{R}(L_A) = \{\mathbf{b}\in\mathbb{F}^m\mid \exists \mathbf{x}\in\mathbb{F}^n\ L_A(\mathbf{x})=\mathbf{b}\}$$
$$$$
Now, if there exists $\mathbf{x}$ such that $A\mathbf{x} = \mathbf{b}$ then by definition of $L_A$ we have $A\mathbf{x} = L_A(\mathbf{x}) = \mathbf{b}$. This means that $\mathbf{b}$ is a member of the range since there exists some $\mathbf{x}$ such that $L_A(\mathbf{x})=\mathbf{b}$.
Conversely, suppose that $\mathbf{b}$ is in the range of $L_A$. By definition of range, there exists some $\mathbf{x}$ such that $L_A(\mathbf{x}) = \mathbf{b}$. By definition of $L_A$ we have $L_A(\mathbf{x}) = A\mathbf{x} = \mathbf{b}$ so $\mathbf{x}$ satisfies the system.
A: $A$ is of the shape
\begin{eqnarray}
\overset{m}{\boxed{\huge A}}\overset{\overset{\overset{\overset{\overset{\overset{\overset{\Large n}{}}{}}{}}{}}{}}{}}{},
\end{eqnarray}
and $X$ is of the shape
\begin{eqnarray}
\overset{n}{\boxed{\huge X}}\overset{\overset{\overset{\overset{\overset{\overset{\overset{\Large 1}{}}{}}{}}{}}{}}{}}{},
\end{eqnarray}
so $AX$ has the shape
\begin{eqnarray}
\overset{m}{\boxed{\huge A}}\overset{\overset{\overset{\overset{\overset{\overset{\overset{\Large n}{}}{}}{}}{}}{}}{}}{}~\overset{\overset{\overset{\overset{\overset{\overset{\overset{\Large \times}{}}{}}{}}{}}{}}{}}{}~~\overset{n}{\boxed{\huge X}}\overset{\overset{\overset{\overset{\overset{\overset{\overset{\Large 1}{}}{}}{}}{}}{}}{}}{} ~\overset{\overset{\overset{\overset{\overset{\overset{\overset{\Large =}{}}{}}{}}{}}{}}{}}{} ~~\overset{m}{\boxed{\huge AX}}\overset{\overset{\overset{\overset{\overset{\overset{\overset{\Large 1}{}}{}}{}}{}}{}}{}}{}.
\end{eqnarray}
$AX$—that is, $B$—is an $m$-tuple for which it is thus a element of the range of $L_A$, for if $A$ is an $m\times n$ matrix with entries from a field $F$, then we denote by $L_A$ the mapping $L_A:F^n\rightarrow F^m$ defined by $L_A(X)=AX$ (the matrix product of $A$ and $X$) for each column vector $x\in F^n$.
