Let \(W\) be the subspace of \(C^3\) spanned by \(\alpha_1 = (1, 0, i)\) and \(\alpha_2 = (1 + i, 1, -1)\).
Show that \(\alpha_1\) and \(\alpha_2\) form a basis for \(W\).
Show that the vectors \(\beta_1 = (1, 1, 0)\) and \(\beta_2 = (1, i, 1 + i)\) are in \(W\) and form another basis for \(W\).
Find the coordinates of each of \(\alpha_1\) and \(\alpha_2\) in the ordered basis \(\{\beta_1, \beta_2\}\).
Solution
1. \(\alpha_1,\alpha_2\) form a basis of \(W\).
They span \(W\) by definition. They are linearly independent because if \[ a\alpha_1+b\alpha_2=0, \] the second coordinate gives \(b=0\), and then the first coordinate gives \(a=0\).
2. \(\beta_1,\beta_2\) lie in \(W\) and form another basis.
Directly, \[ \beta_1=-i\alpha_1+\alpha_2, \] and \[ \beta_2=(2-i)\alpha_1+i\alpha_2. \] Thus both are in \(W\). If \(a\beta_1+b\beta_2=0\), then using their first two coordinates gives \[ a+b=0, \qquad a+ib=0. \] Hence \((i-1)b=0\), so \(b=0\) and then \(a=0\). Thus they are independent. Since \(W\) has dimension \(2\), they form a basis.
3. The coordinates of \(\alpha_1,\alpha_2\) in the ordered basis \(\{\beta_1,\beta_2\}\) are\[ [\alpha_1]_{\beta} = \begin{bmatrix} \frac12-\frac{i}{2}\\ \frac12+\frac{i}{2} \end{bmatrix}, \]\[ [\alpha_2]_{\beta} = \begin{bmatrix} \frac32+\frac{i}{2}\\ -\frac12+\frac{i}{2} \end{bmatrix}. \]
Substitution of these coefficients into the displayed expressions for \(\beta_1,\beta_2\) gives \(\alpha_1\) and \(\alpha_2\), respectively; uniqueness follows because \(\beta_1,\beta_2\) are a basis.