[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.[/step]

[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: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]