How to prove that x $\longrightarrow$ x$_B$ gives a bijective linear map between V and $\mathbb{R}^n$? How to prove that x $\longrightarrow$ x$_B$ gives a bijective linear map between V and $\mathbb{R}^n$ ?
Given:
Let V and W be vector space and $ A: V \longrightarrow W $ be a linear map.
x$_B $ is the column consisting of the coordinates of x with respect to B, where x $\in$ V can be uniquely written as x = $ x_1 $$ $b$ _1 $ + ... + $ x_n $$ $b$ _n $ and where B is an ordered basis, B = {b _1, ..., b _n }.
I got this assignment for school and I think I should start proving injectivity and bijectivity, and I know that for a linear map that is bijective that ker(A) = {0} and A(V) = W. However, I really don't know how to start with the proof and how to work with x $\longrightarrow$ x$_B$.
Thanks in advance for your help!
 A: Let $V$ is a vector space with a ordered basis $B=\{v_{1},v_{2},\ldots,v_{n}\}$, also let
\begin{eqnarray*}
[\cdot]_{B}: (V,F,+,\cdot) &\to & \mathbb{R}^{n}\\
v\mapsto [v]_{B}&=&(\alpha_{1},\alpha_{2},\ldots,\alpha_{n})
\end{eqnarray*}
where $v=\alpha_{1}v_{1}+\alpha_{2}v_{2}+\cdots \alpha_{n}v_{n}$, so $[\cdot]_{B}$ is a bijection from $V$ in $\mathbb{R}^{n}$.
Statement: $[\cdot]_{B}$ is a linear transformation. Indeed, you can prove that

*

*$[\alpha v]_{B}=\alpha[v]_{B}$

*$[v+w]_{B}=[v]_{B}+[w]_{B}$
Now, you know that $$[\cdot]_{B} \quad \text{is bijection} \quad \iff [\cdot]_{B} \quad \text{injective} \quad \wedge \quad [\cdot]_{B} \quad \text{surjective}$$
Now, note that
\begin{eqnarray*}
\mathbf{ker}([v]_{B})=\left\{v \in V: [v]_{B}=\vec{0}_{\mathbb{R}^{n}} \right\} \overbrace{\implies}^{solving} \mathbf{ker}([v]_{B})=\vec{0}_{V} \implies [v]_{B} \quad \text{is injective}.
\end{eqnarray*}
Also, you can see that
\begin{eqnarray*}
\mathbf{im}([v]_{B})=\left\{w \in \mathbb{R}^{n}:\exists v \in V: [v]_{B}=w \right\} \overbrace{\implies}^{solving} \mathbf{im}([v]_{B})=\mathbb{R}^{n} \implies [v]_{B} \quad \text{is surjective}.
\end{eqnarray*}
So, $[v]_{B}$ is a bijection from $V$ to $\mathbb{R}^{n}$.
