Consider the function

\[L\from \RNrSpc{n}\longrightarrow V,\qquad L(x_1,\dots ,x_n) \DefEq x_1\Vect{b}_1+ \cdots +x_n\Vect{b}_n\]

Then \(L\) is linear as it is the unique linear map associated to the function of basis vectors:

\[\mathcal{B}\longrightarrow V,\qquad L(\StdBssVec{1})\DefEq \Vect{b}_1,\ \dots\ , L(\StdBssVec{n})\DefEq \Vect{b}_n\]

This linear map \(L\) satisfies: \(L(x_1,\dots, x_n) = \Vect{0}\) only if \((x_1,\dots ,x_n) = (0,\dots ,0)\), because the vectors \(\Vect{b}_1,\dots ,\Vect{b}_n\) are linearly independent. So, \(L\) is an isomorphism.. But the inverse of \(L\) is the coordinate vector operation \(M\). After all,

\[M(\Vect{x}) = \CoordVect{\Vect{x}}{B} = (x_1,\dots ,x_n)\quad \text{if and only if}\quad \Vect{x} = x_1\Vect{b}_1+\cdots + x_n\Vect{b}_n = L(x_1,\dots ,x_n)\]

So, \(M\) is linear and is an isomorphism.