Let $V$ be an $n$-dimensional vector space, let $W$ be an $m$-dimensional vector space, and let $T$ be any linear transformation from $V$ to $W$ . To associate a matrix with $T$, choose (ordered) bases $B$ and $C$ for $V$ and $W$, respectively.
Given any $x$ in $V$ , the coordinate vector $[x]_B$ is in $\mathbb R^n$, and the coordinate vector of its image, $[T(x)]_C$ , is in $\mathbb R^m$, as shown in Figure 1:
The connection between $[x]_B$ and $[T(x)]_C$ is easy to find.
Let $\{b_1,\cdots,b_n\}=B$ be the basis for $V$.
If $x=r_1b_1+\cdots+r_nb_n$, then $$[x]_B=\begin{bmatrix}r_1\\ \vdots \\r_n\end{bmatrix}$$ and $$T(x)=T(r_1b_1+\cdots+r_nb_n)=r_1T(b_1)+\cdots+r_nT(b_n)$$ because $T$ is linear.
The coordinate mapping from $W$ to $\mathbb R^m$ is linear, which leads to $$[T(x)]_C=r_1[T(b_1)]_C+\cdots+r_n[T(b_n)]_C$$
Since base-$C$ coordinate vectors are in $\mathbb R^m$, the vector equation above can be written as a matrix equation, namely $[T(x)]_C = M[x]_B$ where $$M=\begin{bmatrix} [T(b_1)]_C & [T(b_2)]_C & \cdots & [T(b_n)]_C \end{bmatrix}$$
My question: I'm confused on the jump from $T(x)=r_1T(b_1)+\cdots+r_nT(b_n)$ to $[T(x)]_C=r_1[T(b_1)]_C+\cdots+r_n[T(b_n)]_C$.
Since the basis $B=\{b_1,\cdots,b_n\}$, I see how when $T$ maps it to $W$, the basis $C=\{T(b_1),\cdots,T(b_n)\}$. Then from $T(x)=r_1T(b_1)+\cdots+r_nT(b_n)$, the $r$s can be arranged as $[x]_C=\begin{bmatrix}r_1\\ \vdots \\r_n\end{bmatrix}$ and $M$ as $M=\begin{bmatrix} [T(b_1)] & [T(b_2)] & \cdots & [T(b_n)] \end{bmatrix}$. To get, $x=M[x]_C$, but I don't see how $M$ is in $\mathbb{R}^m$.