I have trouble in understanding the proof of 3.117 in Linear Algebra Done Right (Third Edition). The Theorem is the following:
3.117$\quad$ Dimension of range $T$ equals column rank of $\mathcal{M}(T)$
Suppose $V$ and $W$ are finite-dimensional and $T\in\mathcal{L}(V;W).$ Then $\dim\text{range } T$ equals the column rank of $\mathcal{M}(T).$
The first paragraph of the proof given is the following:
Proof
Suppose $v_1,\dots, v_n$ is a basis of $V$ and $w_1,\dots, w_m$ is a basis of $W.$ The function that takes $w\in\text{span}(Tv_1,\dots, Tv_n)$ to $\mathcal{M}(w)$ is easily seen to be an isomorphism from $\text{span}(Tv_1,\dots, Tv_n)$ onto $\text{span}\big(\mathcal{M}(Tv_1),\dots, \mathcal{M}(Tv_n)\big).$ Thus $\dim\text{span}(Tv_1,\dots, Tv_n)=\dim\text{span}\big(\mathcal{M}(Tv_1),\dots,\mathcal{M}(Tv_n)\big),$ where the last dimension equals the column rank of $\mathcal{M}(T).$
I don't understand that sentence "The function that takes $w\in\text{span}(Tv_1,\dots, Tv_n)$ to $\mathcal{M}(w)$ is easily seen to be an isomorphism from $\text{span}(Tv_1,\dots, Tv_n)$ onto $\text{span}\big(\mathcal{M}(Tv_1),\dots,\mathcal{M}(Tv_n)\big)$.".
I have checked another almost same question, But I still wondering that how does $\mathcal{M}$ map $\text{span}(Tv_1,\dots, Tv_n)$ onto $\text{span}\big(\mathcal{M}(Tv_1),\dots,\mathcal{M}(Tv_n)\big)$.
I'm new in linear algebra, any tips are appreciated.
The following is my understanding.
Suppose $T \in \mathcal{L}(V,W)$ and suppose $w=Tv$ for some $w \in \text{range }T$ and $v \in V$. $$ w = Tv = T(c_1v_1+\cdots+c_nv_n) = c_1Tv_1+\cdots+c_nTv_n $$ Thus, we can see $w \in \text{span}(Tv_1,...,Tv_n)$.
And we can have a linear map $\mathcal{M}: W \rightarrow F^{m,1}$, which is an isomorphism. $$ \mathcal{M}(w)=\mathcal{M}(c_1Tv_1+\cdots+c_nTv_n) = c_1\mathcal{M}(Tv_1)+\cdots+c_n\mathcal{M}(Tv_n) $$ Thus, $\mathcal{M}(w) \in \text{span}(\mathcal{M}(Tv_1),...,\mathcal{M}(Tv_n))$.