Consider \(C^3\), with the standard inner product. Find an orthonormal basis for the subspace spanned by \(\beta_1 = (1, 0, i)\) and \(\beta_2 = (2, 1, 1 + i)\).
Solution
1. Apply Gram–Schmidt. Let \(u_1 = \beta_1 = (1, 0, i)\).
first step of Gram–Schmidt.
2. \(\|u_1\|^2 = |1|^2 + |0|^2 + |i|^2 = 1 + 0 + 1 = 2\), so \(e_1 = \frac{u_1}{\|u_1\|} = \frac{1}{\sqrt 2}(1, 0, i)\).
normalize.
3. \(u_2 = \beta_2 - \langle \beta_2, e_1 \rangle e_1\).
Gram–Schmidt.
4. \(\langle \beta_2, e_1 \rangle = \langle (2, 1, 1+i), \frac{1}{\sqrt 2}(1, 0, i) \rangle = \frac{1}{\sqrt 2}(2 \cdot 1 + 1 \cdot 0 + (1+i)\overline{i}) = \frac{1}{\sqrt 2}(2 + (1+i)(-i)) = \frac{1}{\sqrt 2}(2 - i + 1) = \frac{3 - i}{\sqrt 2}\).
compute the inner product (using \(\overline{i} = -i\)).
5. Hence \(u_2 = (2, 1, 1+i) - \frac{3-i}{\sqrt 2} \cdot \frac{1}{\sqrt 2}(1, 0, i) = (2, 1, 1+i) - \frac{3-i}{2}(1, 0, i)\).
3 and 4.
6. \(u_2 = (2 - \frac{3-i}{2}, 1, 1+i - \frac{(3-i)i}{2}) = (\frac{1+i}{2}, 1, 1+i - \frac{3i+1}{2}) = (\frac{1+i}{2}, 1, \frac{1+i}{2})\).
5, simplifying.
7. \(\|u_2\|^2 = |\frac{1+i}{2}|^2 + |1|^2 + |\frac{1+i}{2}|^2 = \frac{1}{2} + 1 + \frac{1}{2} = 2\), so \(e_2 = \frac{u_2}{\sqrt 2} = \frac{1}{\sqrt 2}(\frac{1+i}{2}, 1, \frac{1+i}{2})\).
normalize.
8. Hence an orthonormal basis is \(\{e_1, e_2\}\) with \(e_1 = \frac{1}{\sqrt 2}(1, 0, i)\) and \(e_2 = \frac{1}{\sqrt 2}(\frac{1+i}{2}, 1, \frac{1+i}{2})\).
2 and 7.
9. Q.E.D.
8.