Let \(\mathfrak{B} = \{\alpha_1, \alpha_2, \alpha_3\}\) be the basis for \(C^3\) defined by \[ \alpha_1 = (1, 0, -1), \quad \alpha_2 = (1, 1, 1), \quad \alpha_3 = (2, 2, 0). \] Find the dual basis of \(\mathfrak{B}\).
Solution
Let \(A\) be the matrix whose columns are the basis vectors: \[ A=\begin{bmatrix}1&1&2\\0&1&2\\-1&1&0\end{bmatrix}. \] The coordinate functionals are the rows of \(A^{-1}\). Direct computation gives \[ A^{-1}=\begin{bmatrix} 1&-1&0\\ 1&-1&1\\ -\frac12&1&-\frac12 \end{bmatrix}. \] Hence the dual basis \(\{f_1,f_2,f_3\}\) is \[ f_1(x,y,z)=x-y, \] \[ f_2(x,y,z)=x-y+z, \] \[ f_3(x,y,z)=-\frac12x+y-\frac12z. \] Indeed, the matrix identity \(A^{-1}A=I\) says exactly that \(f_i(\alpha_j)=\delta_{ij}\).