Let the vectors be [ v_1 = (1, i, 0, 1), \quad v_2 = (1, 0, i, 0), \quad v_3 = (-i, 0, 1, -1) ] We want an orthonormal basis using the Gram-Schmidt process in (\mathbb{C}^4). The inner product in (\mathbb{C}^4) is (\langle x, y \rangle = \sum_{k=1}^4 x_k \overline{y_k}). --- Step 1: Set (u_1 = v_1) [ u_1 = v_1 = (1, i, 0, 1) ] Compute its norm: [ |u_1| = \sqrt{\langle u_1, u_1 \rangle} = \sqrt{1\cdot 1 + i\cdot(-i) + 0 + 1\cdot 1} = \sqrt{1 + 1 + 0 + 1} = \sqrt{3} ] So the first orthonormal vector is [ e_1 = \frac{u_1}{|u_1|} = \frac{1}{\sqrt{3}}(1, i, 0, 1) ] --- Step 2: Orthogonalize (v_2) against (u_1) [ u_2 = v_2 - \frac{\langle v_2, u_1 \rangle}{\langle u_1, u_1 \rangle} u_1 ] Compute (\langle v_2, u_1 \rangle): [ \langle v_2, u_1 \rangle = 1\cdot 1 + 0\cdot (-i) + i\cdot 0 + 0\cdot 1 = 1 ] (\langle u_1, u_1 \rangle = 3), so [ u_2 = v_2 - \frac{1}{3} u_1 = (1,0,i,0) - \frac{1}{3}(1,i,0,1) = \left(1-\frac{1}{3}, 0-\frac{i}{3}, i-0, 0-\frac{1}{3}\right) ] [ u_2 = \left(\frac{2}{3}, -\frac{i}{3}, i, -\frac{1}{3}\right) ] Compute (|u_2|^2 = \langle u_2, u_2 \rangle): [ \langle u_2, u_2 \rangle = \left(\frac{2}{3}\right)^2 + \left(-\frac{i}{3} \cdot \frac{i}{3}\right) + (i \cdot -i) + \left(-\frac{1}{3}\right)^2 ] Step by step: * ((2/3)^2 = 4/9) * ((-i/3) \cdot (i/3) = (-i)(-i)/9 = (-i)(-i)/9 = -i(-i)/9 = -(-i^2)/9 = -(1)/9 = -1/9) Wait carefully: Actually, ((-i/3) \cdot \overline{-i/3} = (-i/3)(i/3) = (-i \cdot i)/9 = (-i^2)/9 = -(-1)/9 = 1/9). ✅ * (i \cdot \overline{i} = i \cdot -i = -i^2 = 1) * ((-1/3)^2 = 1/9) Sum: (4/9 + 1/9 + 1 + 1/9 = 6/9 + 1 = 2/3 + 1 = 5/3) (|u_2| = \sqrt{5/3} = \frac{\sqrt{15}}{3}) So [ e_2 = \frac{u_2}{|u_2|} = \frac{3}{\sqrt{15}} \left(\frac{2}{3}, -\frac{i}{3}, i, -\frac{1}{3}\right) = \frac{1}{\sqrt{15}}(2, -i, 3i, -1) ] --- Step 3: Orthogonalize (v_3) against (u_1) and (u_2) [ u_3 =