$T$ is the linear transformation of $V$ ($n$-dimensional) to $W$ ($m$-dimensional) and {$b_1,...b_n$} is the basis $B$ for $V.$ Given any x in $V$, the coordinate vector $[x]_B$ is in $R^n$ and the coordinate vector of its image, $[T(x)]_C$ is in $R^m$.
$T(x)$ =T($r_1$$b_1$ + ... + $r_nb_n$)$ = T(b_1)$ + ... + r$_nT(b_n)$
How did the above equation lead to this:
$[T(x)]_C = r_1[T(b_1]_C + .... + r_n[T(b_n]_C $ ?
Looking at $[T(x)]_C$, what will be the basis, $C$ (for $W$) based on this? I know that to get $[T(x)]_C$, you will have to find some $c_1 ... c_n$ (scalar values) that will be used to form a linear combination of $T(x)$?
Someone, please?
It seems that the function $[-]_C:W\to \mathbb{R}^m$ is an invertible linear transformation between $W$ and $\mathbb{R}^m$ (an isomorphism) identifying the elements of the basis $C$ with the standard unit vectors in $\mathbb{R}^m$. In this case the linearity condition is what is important. For any vectors $w,w'\in W$ and elements $a,b\in \mathbb{R}$ we have $$[aw+bw']_C=a[w]_C+b[w']_C$$ This is because if $$C=\{c_1,c_2,\ldots,c_m\}$$ then $$[c_1]_C=(1,0,0,\ldots,0)$$ and $$[c_2]_C=(0,1,0,\ldots,0)$$ etc., so if $a_1,\ldots,a_m$ are real numbers, then $$[a_1c_1+a_2c_2+\cdots +a_mc_m]_C=a_1[c_1]_C+\cdots +a_m[c_m]_C=(a_1,a_2,\ldots,a_m)\mathrm{.}$$
If $v=a_1c_1+a_2c_2+\cdots+a_mc_m$ and $v'=a_1'c_1+a_2'c_2+\cdots+a_m'c_m$ and $p$ and $q$ are real, then $$[pv+qv']_C=[(pa_1+qa_1')c_1+(pa_2+qa_2')c_2+\cdots+(pa_m+qa_m')c_m]_C=(pa_1+qa_1',\ldots,pa_m+qa_m')=p(a_1,a_2,\ldots,a_m)+q(a_1',a_2',\ldots,a_m')=p[v]_C+q[v']_C$$ This explains why $$[T(x)]_C=r_1[T(b_1)]_C+\cdots +r_n[T(b_n)]_C$$