I have trouble in understanding the first paragraph of proof of 3.177 in Axler's book 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 here:
Proof $\quad $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 have no trouble in understanding the remaining part of the proof, except the 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 thought for a long time in order to prove that such a map $w\in\text{span}(Tv_1,\dots, Tv_n) \mapsto \mathcal{M}(w)$ is indeed an isomorphism, but failed at last. What I have done is blow.
Let $A$ denote the matrix of the map $T$ with respect to the given bases $v_1,\dots, v_n$ in $V$ and $w_1,\dots, w_m$ in $W.$ Since $A$ is not necessarily invertible, and $Tv_1,\dots, Tv_n$ is not necessarily linearly independent, for every $w\in\text{span}(Tv_1,\dots, Tv_n),$ the coefficients $a_1,\dots, a_n$ such that $$w=aTv_1+\cdots+a_nTv_n$$ are not necessarily unique. Thus, how could I show that $w\mapsto \mathcal{M}(w)$ is injective?
Actually, I have shown that if $w\in\text{span}(Tv_1,\dots, Tv_n),$ with a representation as $$w=a_1Tv_1+\cdots+a_nTv_n,$$ then, if let $a$ denote the column vector consisting of $a_1,\dots, a_n,$ then $Aa$ is in $\text{span}\big(\mathcal{M}(Tv_1),\dots, \mathcal{M}(Tv_n)\big).$ If $T$ is injective, then $Aa=\lambda,$ if $w=\lambda_1w_1+\cdots+\lambda_mw_m,$ where the column vector $\lambda$ consisting of $\lambda_1,\dots, \lambda_m.$ Then all things left are easy. But if $T$ is not injective, I do not how to continue. Can anyone help me?
Fix a basis $w_1, \dots, w_m$ of $W$. The map $\mathcal{M}$ that takes $w \in W$ to $\mathcal{M}(w)$ is an isomorphism of $W$ onto the $m$-by-$1$ column vectors with entries in the scalar field (think about the definition of basis to see why this is true).
Now suppose $v_1, \dots, v_n$ is a basis of $V$ and $T$ is a linear map from $V$ to $W$. The isomorphism $\mathcal{M}$ defined above is injective when restricted to any subspace, and in particular when restricted to $\text{span}(Tv_1, \dots, Tv_n)$. Thus there is no need to think about whether a vector in this span has a unique representation as a linear combination of $Tv_1, \dots, Tv_n$.
I think you can fill in the rest of the details. I hope that this helps.