There is another way to understand what $\mathcal{M}(T)$ is. We have shown that any two vector spaces with the same dimension are isomorphic, meaning $V\cong\mathbb{R}^n$ and $W\cong\mathbb{R}^m$. Set $\{e_1,...,e_n\}$ to be the standard basis for $\mathbb{R}^n$ and $\{f_1,...,f_m\}$ to be the standard basis of $\mathbb{R}^m$. There is an isomorphism $\upphi_V:\mathbb{R}^n\to V$ such that $\upphi_V(e_i)=v_i$ for $i=1,...,n$ and another isomorphism $\upphi_W:\mathbb{R}^m\to V$ such that $\upphi_W(f_i)=w_i$. The composition $\upphi^{-1}_W\circ T\circ\upphi_V:\mathbb{R}^n\to\mathbb{R}^m$ is a linear map, and can be represented by a matrix. That matrix is $\mathcal{M}(T)$. $$\begin{CD} \mathbb{R}^n @>{\mathscr{M}(T)}>> \mathbb{R}^m; \\ @V{{S_V}}VV @AA{S_W^{-1}}A \\ V @>>{T}> W\\ \end{CD}$$ ## Remark This essentially proves that $\mathcal{M}$ is a linear map.