[proofplan]
The proof uses the defining property of a basis: every vector has a unique expansion in the ordered basis $\mathcal{B}$. First we record that this makes the coordinate map well-defined. Then we compare the unique coordinate expansion of $\alpha v+\beta w$ with the coordinate expansions of $v$ and $w$ to prove linearity. Finally, injectivity follows from uniqueness of the zero coordinate expansion, and surjectivity follows by building a vector with prescribed coordinates.
[/proofplan]
[step:Use the ordered basis to define coordinates uniquely]
For each $x \in V$, since $\mathcal{B}=(v_1,\ldots,v_n)$ is a basis of $V$, there exist unique scalars $a_1,\ldots,a_n \in k$ such that
\begin{align*}
x=\sum_{i=1}^{n} a_i v_i.
\end{align*}
Define $[x]_{\mathcal{B}} \in k^n$ by
\begin{align*}
[x]_{\mathcal{B}}=(a_1,\ldots,a_n).
\end{align*}
The uniqueness of the scalars $a_1,\ldots,a_n$ shows that $[x]_{\mathcal{B}}$ is well-defined, and therefore $\Phi_{\mathcal{B}}:V \to k^n$ is a well-defined map.
[/step]
[step:Compare coordinate expansions to prove linearity]
Let $v,w \in V$ and let $\alpha,\beta \in k$. Write
\begin{align*}
[v]_{\mathcal{B}}=(a_1,\ldots,a_n)
\end{align*}
and
\begin{align*}
[w]_{\mathcal{B}}=(b_1,\ldots,b_n),
\end{align*}
where $a_1,\ldots,a_n,b_1,\ldots,b_n \in k$. By the definition of coordinates,
\begin{align*}
v=\sum_{i=1}^{n} a_i v_i
\end{align*}
and
\begin{align*}
w=\sum_{i=1}^{n} b_i v_i.
\end{align*}
Using the [vector space](/page/Vector%20Space) operations in $V$, we obtain
\begin{align*}
\alpha v+\beta w=\sum_{i=1}^{n}(\alpha a_i+\beta b_i)v_i.
\end{align*}
By uniqueness of coordinates in the basis $\mathcal{B}$,
\begin{align*}
[\alpha v+\beta w]_{\mathcal{B}}=(\alpha a_1+\beta b_1,\ldots,\alpha a_n+\beta b_n).
\end{align*}
The vector space operations in $k^n$ are componentwise, so
\begin{align*}
(\alpha a_1+\beta b_1,\ldots,\alpha a_n+\beta b_n)=\alpha(a_1,\ldots,a_n)+\beta(b_1,\ldots,b_n).
\end{align*}
Therefore
\begin{align*}
\Phi_{\mathcal{B}}(\alpha v+\beta w)=\alpha\Phi_{\mathcal{B}}(v)+\beta\Phi_{\mathcal{B}}(w).
\end{align*}
Thus $\Phi_{\mathcal{B}}$ is linear.
[guided]
We prove linearity by reducing the statement to uniqueness of coordinates. Let $v,w \in V$ and let $\alpha,\beta \in k$. Since $\mathcal{B}$ is a basis, there are unique scalars $a_1,\ldots,a_n \in k$ and $b_1,\ldots,b_n \in k$ such that
\begin{align*}
v=\sum_{i=1}^{n} a_i v_i
\end{align*}
and
\begin{align*}
w=\sum_{i=1}^{n} b_i v_i.
\end{align*}
Equivalently,
\begin{align*}
[v]_{\mathcal{B}}=(a_1,\ldots,a_n)
\end{align*}
and
\begin{align*}
[w]_{\mathcal{B}}=(b_1,\ldots,b_n).
\end{align*}
Now form the linear combination $\alpha v+\beta w$ inside $V$. Substituting the basis expansions of $v$ and $w$ gives
\begin{align*}
\alpha v+\beta w=\alpha\sum_{i=1}^{n}a_i v_i+\beta\sum_{i=1}^{n}b_i v_i.
\end{align*}
Distributivity of scalar multiplication over finite sums gives
\begin{align*}
\alpha v+\beta w=\sum_{i=1}^{n}\alpha a_i v_i+\sum_{i=1}^{n}\beta b_i v_i.
\end{align*}
Combining terms with the same basis vector $v_i$ gives
\begin{align*}
\alpha v+\beta w=\sum_{i=1}^{n}(\alpha a_i+\beta b_i)v_i.
\end{align*}
Because coordinates in a basis are unique, the coordinate tuple of $\alpha v+\beta w$ must be the tuple of coefficients appearing in this expansion:
\begin{align*}
[\alpha v+\beta w]_{\mathcal{B}}=(\alpha a_1+\beta b_1,\ldots,\alpha a_n+\beta b_n).
\end{align*}
On the other hand, addition and scalar multiplication in $k^n$ are defined componentwise, so
\begin{align*}
\alpha[v]_{\mathcal{B}}+\beta[w]_{\mathcal{B}}=\alpha(a_1,\ldots,a_n)+\beta(b_1,\ldots,b_n).
\end{align*}
The right-hand side is
\begin{align*}
(\alpha a_1+\beta b_1,\ldots,\alpha a_n+\beta b_n).
\end{align*}
Thus
\begin{align*}
[\alpha v+\beta w]_{\mathcal{B}}=\alpha[v]_{\mathcal{B}}+\beta[w]_{\mathcal{B}}.
\end{align*}
Since $\Phi_{\mathcal{B}}(x)=[x]_{\mathcal{B}}$ for every $x \in V$, this is exactly
\begin{align*}
\Phi_{\mathcal{B}}(\alpha v+\beta w)=\alpha\Phi_{\mathcal{B}}(v)+\beta\Phi_{\mathcal{B}}(w).
\end{align*}
Therefore $\Phi_{\mathcal{B}}$ is linear.
[/guided]
[/step]
[step:Use the zero coordinate tuple to prove injectivity]
Let $0_V$ denote the zero vector of $V$, and let $0_{k^n}=(0,\ldots,0)$ denote the zero vector of $k^n$. Suppose $v \in V$ satisfies $\Phi_{\mathcal{B}}(v)=0_{k^n}$. Then
\begin{align*}
[v]_{\mathcal{B}}=(0,\ldots,0).
\end{align*}
By the definition of coordinates,
\begin{align*}
v=\sum_{i=1}^{n}0v_i=0_V.
\end{align*}
Thus $\ker \Phi_{\mathcal{B}}=\{0_V\}$. Since $\Phi_{\mathcal{B}}$ is linear, this proves that $\Phi_{\mathcal{B}}$ is injective.
[/step]
[step:Build a vector with prescribed coordinates to prove surjectivity]
Let $y \in k^n$. Then there exist scalars $a_1,\ldots,a_n \in k$ such that
\begin{align*}
y=(a_1,\ldots,a_n).
\end{align*}
Define $v \in V$ by
\begin{align*}
v=\sum_{i=1}^{n}a_i v_i.
\end{align*}
By the definition of the coordinate tuple with respect to $\mathcal{B}$,
\begin{align*}
[v]_{\mathcal{B}}=(a_1,\ldots,a_n)=y.
\end{align*}
Therefore $\Phi_{\mathcal{B}}(v)=y$. Since $y \in k^n$ was arbitrary, $\Phi_{\mathcal{B}}$ is surjective.
[/step]
[step:Conclude that the coordinate map is a linear isomorphism]
We have shown that $\Phi_{\mathcal{B}}:V \to k^n$ is linear, injective, and surjective. Hence $\Phi_{\mathcal{B}}$ is a bijective [linear map](/page/Linear%20Map). Therefore $\Phi_{\mathcal{B}}$ is a linear isomorphism of vector spaces over $k$.
[/step]
