In Sheldon Axler's Linear Algebra Done Right, when proving that the dimension of range T equals the column rank of $\mathcal{M}(T)$ (where $\mathcal{M}(T)$ is the matrix of the linear map T), it says
Suppose $v_1$, ... $v_n$ is a basis of $\mathbb{V}$ and $w_1$, ... $w_m$ is a basis of $\mathbb{W}$. The function that takes $w \in span(Tv_1,...,Tv_n)$ to $\mathcal{M}(w)$ is easily seen to be an isomorphism from $span(Tv_1,...,Tv_n)$ onto $span(\mathcal{M}(Tv_1)...\mathcal{M}(Tv_n))$. Thus dim $span(Tv_1,...,Tv_n)=$ dim $span(\mathcal{M}(Tv_1)...\mathcal{M}(Tv_n))$.
What I don't quite understand is what exactly is this function that takes a vector to a matrix of a vector? And what does it mean that this function is isomorphic?
Given a finite dimensional vector space $V$, over a field $F$, and a list of vectors $B = (v_1, \ldots, v_n)$ in $V$, we can define a map from $F^n$ into the vector space $V$ by $$\begin{bmatrix} a_1 \\ \vdots \\ a_n\end{bmatrix} \in F^n \mapsto a_1v_1 + \ldots + a_n v_n \in V.$$ That is, just write out a linear combination of vectors in $B$, using the entries of the vector in $F^n$ as your coefficients.
It's not hard to establish this map is linear. This map will be surjective if and only if $B$ spans $V$. It will be injective if and only if $B$ is linearly independent. (All three of these assertions make good exercises.)
That means that, when $B$ is a basis for $V$, this map is a bijection. Each vector in $F^n$ corresponds uniquely to a single element of $V$. For a given vector $v \in V$, we call the corresponding vector in $F^n$ the coordinate column vector of $v$ with respect to $B$.
The map $\mathcal{M}$ (when applied to vectors) is the inverse of this map. That is, it takes vectors in $V$, and produces the corresponding coordinate column vector in $F^n$.
It is also a linear map, and its inverse is the map above. Note that it depends on the basis chosen for $V$; with a different basis, we'll obtain a totally different map.
When we have an invertible linear map, we call this an isomorphism.
As an example, let's fix $V = P_2(\Bbb{R})$, the polynomials of degree $2$ or less over the reals, and fix the basis $B = (1, x - 1, (x - 1)^2)$ of $V$. We have that $$\mathcal{M}(x^2) = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix},$$ as $1 \cdot 1 + 2 \cdot (x - 1) + 1 \cdot (x - 1)^2 = x^2$. As $B$ is a basis, it doesn't matter which polynomial in $V$ I choose, there will always be a unique linear combination of vectors from $B$ that will produce this polynomial, and the coordinate column vector is simply these coefficients put into a column vector.