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I am reading a paragraph from an introductory Linear Algebra book, and I am not exactly grasping what the author means. The excerpt is as follows:

Let $F^{m,n}$ denote the set of all $m$-by-$n$ matrices with entries in some generic field $F$. Each $m$-by-$n$ matrix $A$ induces a linear map from $F^{n,1}$ to $F^{m,1}$, namely the matrix multiplication function that takes $x\in F^{n,1}$ to $Ax\in F^{m,1}$. The result above can be used to think of every linear map (from one finite-dimensional vector space to another FDVS) as a matrix multiplication map after suitable relabeling via the isomorphism given by $\mathcal{M}$. Specifically, if $T\in\mathcal{L}(V,W)$ and we identify $v\in V$ and $\mathcal{M}(v)\in F^{n,1}$, then the result above says that we can identify $Tv$ with $\mathcal{M}(T)\mathcal{M}(v)$.

The notation for $\mathcal{M},\mathcal{L}(\cdot,\cdot)$ can be found in the below reference's page 84-85.

My Question: What does the bold statement mean? Can someone explain more clearly?

Reference: Axler, Sheldon J. Linear Algebra Done Right, New York: Springer, 2015.

Martin Sleziak
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Frank Swanton
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    Every matrix $A \in F^{m,n}$ induces by matric multiplication $x \in F^{n,1} \mapsto Ax \in F^{m,1}$ a linear map. And the bold statement essentially tells you, that you get all linear maps $F^{n,1} \to F^{m,1}$ from this. Now I assume on page 84-85, the author has shown that every finite dimensional vector space is isomorphic to a vector space of the form $F^{n,1}$ for some $n$. Hence he says that every linear map on finite dimensional vector spaces can be thought of as matrix multiplication. –  Jun 24 '19 at 07:04
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    Check out the second half of my answer here: https://math.stackexchange.com/a/3177904/248286 – Theo Bendit Jun 24 '19 at 07:49
  • To both thanks. @TheoBendit I have yet to read more carefully, but the graph really looks cool and I hope it sheds more light on my thought process! – Frank Swanton Jun 24 '19 at 08:04

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