Show that $x\in{}V$ to $[x]_B\in\mathbb{R}^n$ is a linear transformation Let $V$ be a vector space, let $B=\{b_1,b_2,...,b_n\}$ be an ordered basis for $V$. Show that the coordinate mapping, which maps $x\in{}V$ to $[x]_B\in\mathbb{R}^n$ is a linear transformation.
Proving that this is a linear transformation involves $T(cu)=cT(u)$ and $T(u+v)=T(u)+T(v)$, right? But I'm not sure where to go from there...
 A: Since $\{b_i\}$ is a basis for $V$, we have, $\forall x \in V \rightarrow x=x_1b_1+\cdots x_nb_n$ and, if $V$ is a vector space over $\mathbb{R}$, $x_i\in \mathbb{R}$.
So the transformation is:
$$
T(x)= \begin{bmatrix}
x_1\\\cdot\\x_n
\end{bmatrix}
$$ 
Now, if $y=y_1b_1+\cdots +y_nb_n$, we have ( $\forall c \in \mathbb{R}$):
$$
x+cy=(x_1b_1+\cdots x_nb_n)+c(y_1b_1+\cdots +y_nb_n)
$$
and since $V$ is a vector space (addition is commutative and compatible with the multiplication by a scalar): 
$$
x+cy=(x_1+cy_1)b_1+ \cdots (x_n+cy_n)b_n
$$
so we have:
$$
T(x+cy)=
\begin{bmatrix}
(x_1+cy_1)\\ \cdot \\ (x_n+cy_n)
\end{bmatrix}
$$
and, since also $\mathbb{R}^n$ is a vector space:
$$
T(x+cy)=
\begin{bmatrix}
x_1+cy_1\\
\cdot
\\x_n+cy_n
\end{bmatrix}
=
\begin{bmatrix}
x_1\\
\cdot
\\x_n
\end{bmatrix}
+c
\begin{bmatrix}
y_1\\
\cdot\\
y_n
\end{bmatrix}=T(x)+cT(y)
$$
and this shows that $T$ is linear.
A: Let $u,v\in V$ Then 
$$T(u+v)=[u+v]_B$$ 
Now you might have proved that $[u+v]_B=[u]_B+[v]_B$ but then you wouldn't have asked the question most probably so here's a sketch of the proof. First let's express our vectors in terms of the basis$$u=c_1b_1+\dots +c_nb_n, \ v=d_1b_1+\dots d_nb_n$$
Where the $c_i,d_i$ are constants. Now evaluate $u+v$ and use the distributive property. Then use the definition of a co-ordinate vector. For example 
$$[u]_B=\begin{bmatrix}
         c_1 &   c_2 & \dots  &   c_n
        \end{bmatrix}^{T}$$
The result should follow quite simply from here. If you have problems following let me know. Proving that the transformation is closed under scalar multiplication is similar and simpler.
