Why linear maps act like matrix multiplication? In Linear Algebra Done Right, it said

Suppose $T \in \mathcal{L}(V,W)$ and $v \in V$. Suppose $v_1,...,v_n$ is a basis of $V$ and $w_1,...,w_m$ is a basis of $W$. Then $$M(Tv) = M(T)M(v)$$

$M(T)$ is the m-by-n matrix whose entries $A_{j,k}$ are defined by $Tv_k = A_{1,k}w_1 + ... + A_{m,k}w_m$ suppose $T \in \mathcal{L}(V,W)$ and $v_1,...,v_n$ is a basis of $V$ and $w_1,...,w_m$ is a basis of $W$.
$M(v)$ is the matrix of vector $v$.
I generally follow the following proof:
Suppose $v = c_1v_1 + ... + c_nv_n$, where $c_1,...,c_n \in \mathbb{F}$. Thus 
$$Tv = c_1Tv_1 +...+c_nTv_n$$
Hence
\begin{equation}
\begin{split}
M(Tv) &= c_1M(Tv_1) + ...+ c_nM(Tv_n)\\
& = c_1M(T)_{.,1} +...+c_nM(T)_{.,n} \\
& = M(T)M(v)
\end{split}
\end{equation}
But I have questions on the meaning of the proof. The book said it means each m-by-n matrix $A$ induces a linear map from $\mathbb{F}^{n,1}$ to $\mathbb{F}^{m,1}$. The result can be used to think of every linear map as a matrix multiplication map after suitable relabeling via the isomorphisms given by $M$.


*

*Is the shape of $M(Tv)$ m by 1, $M(T)$ m by n, and $M(v)$ n by 1?

*What is meant by suitable relabeling via the isomorphisms given by $M$? Does it just mean $M(T)$ is a isomorphism linear map between $M(v)$ and $M(Tv)$?

 A: *

*Yes, those would be the shapes of those vectors when represented as matrices. Given that we're multiply by vectors on the right.

*There is a theorem that if $V$ is an $n-$dimensional vector space over a field $F,$ then $V$ is isomorphic to $F^n.$ Here the isomorphic mappings assign coordinates to our vectors and our linear transformation. It doesn't mean that $M$ is an isomorphism between $M(v)$ and $M(Tv).$ These are particular vectors. The map $M$ actually induces an isomorphism from $V\to F^n$, isomorphism from $T\to F^{n\times m}$, and an isomorphism from $W\to F^m.$ 
I actually like the way that this is done. The Author is telling you that you're representation of $T$ by a matrix depends on your choice of basis in $F^n$. A fact that is important to remember.
A: In answer to your first question, yes to all three: $v$ is an element of the $n$-dimensional space $V$, so the coordinate vector with respect to the basis will be an $n \times 1$ column vector. Similarly, $Tv \in W$, which is an $n$-dimensional space, so $M(Tv)$ will be an $m \times 1$ column vector. Finally, $M(T)$ is built from transforming the $n$ basis vectors of the domain, forming each an $m \times 1$ coordinate column vector, which are put into an $m \times n$ matrix.
In answer to your second question, consult the following commutative diagram (made in Paint :( ):

The process of applying $T$ to a vector $v \in V$ is the top row of the diagram. However, there's a parallel process happening between $\Bbb{F}^n$ and $\Bbb{F}^m$, mirroring the same process.
The isomorphism being referred to are the double arrows, taking us between $V$ and $\Bbb{F}^n$ and $W$ and $\Bbb{F}^m$, by way of coordinate vectors. The coordinate vector map on $V$ is a linear map between $V$ and $\Bbb{F}^n$ that is invertible, making it an isomorphism (and similarly for $W$). That is, the two spaces are structurally identical, and anything we can do with one space, we can view it in the other.
In $V$, we have some abstract vectors, and an abstract linear transformation $T$ that maps vectors in $V$ to vectors in $W$. However, using this isomorphism, we can view $V$ slightly differently as $\Bbb{F}^n$, and similarly for $W$, which means $T$ boils down to a linear map from $\Bbb{F}^n$ to $\Bbb{F}^m$, which can be characterised as matrix multiplication. The matrix, in particular, is $M(T)$.
