Question

Use Gram-Schmidt orthogonalisation process to find an orthonormal basis for the subspace of C^4 generated by the vectors (1,i,0,−i), (−i,0,1,2) and (0,−i,1,1).

Answer 1

Answer on Question #42862 – Math - Linear Algebra

Problem. Use Gram-Schmidt orthogonalisation process to find an orthonormal basis for the subspace of $\mathbb{C}^4$ generated by the vectors $(1,i,0,-i)$, $(-i,0,1,2)$ and $(0,-i,1,1)$.

Solution.

$$\begin{array}{l} a_1(1,i,0,-i), a_2(-i,0,1,2), a_3(0,-i,1,1). \\ b_1 = a_1 = (1,i,0,-i). \\ b_2 = a_2 - \frac{(a_2,b_1)}{(b_1,b_1)}b_1 = (-i,0,1,2) - (1,i,0,-i) \cdot \frac{(-i) \cdot 1 + 0 \cdot (-i) + 1 \cdot 0 + 2 \cdot i}{1 \cdot 1 + i \cdot (-i) + 0 \cdot 0 + (-i) \cdot i} = \\ = (-i,0,1,2) - (1,i,0,-i) \cdot \frac{i}{3} = (-i,0,1,2) - \left(\frac{i}{3}, \frac{-1}{3}, 0, \frac{1}{3}\right) = \frac{1}{3} (-4i, 1,3,5). \\ b_3 = a_3 - \frac{(a_3,b_2)}{(b_2,b_2)}b_2 - \frac{(a_3,b_1)}{(b_1,b_1)}b_1 = \\ = (0,-i,1,1) - \frac{1}{3} (-4i, 1,3,5) \cdot \frac{\frac{1}{3} (0 \cdot 4i + (-i) \cdot 1 + 1 \cdot 3 + 1 \cdot 5)}{1/3 ((-4i) \cdot 4i + 1 \cdot 1 + 3 \cdot 3 + 5 \cdot 5)} - (1,i,0,-i) \cdot \frac{0 \cdot 1 + (-i) \cdot (-i) + 1 \cdot 0 + 1 \cdot i}{1 \cdot 1 + i \cdot (-i) + 0 \cdot 0 + (-i) \cdot i} = \\ = (0,-i,1,1) - (-4i,1,3,5) \cdot \frac{8 - i}{51} - (1,i,0,-i) \cdot \frac{i - 1}{3} = \\ = (0,-i,1,1) - \frac{1}{51} (-32i - 4, 8 - i, 24 - 3i, 40 - 5i) - \frac{1}{3} (i - 1, -i - 1, 0, i + 1) = \\ = \frac{1}{51} (0 + 32i + 4 - 17i + 17, -51i - 8 + i + 17i + 17, 51 - 24 + 3i + 0, 51 - 40 + 5i - 17i - 17) = \\ = \frac{1}{51} (15i + 21, -33i + 9, 27 + 3i, -6 - 12i) = \frac{1}{17} (5i + 7, -11i + 3, 9 + i, -2 - 4i). \\ b_1(1,i,0,-i), b_2\left(-\frac{4i}{3}, \frac{1}{3}, 1, \frac{5}{3}\right), b_3\left(\frac{5i + 7}{17}, \frac{-11i + 3}{17}, \frac{9 + i}{17}, \frac{-2 - 4i}{17}\right). \\ |b_1|^2 = 1 \cdot 1 + i \cdot (-i) + 0 \cdot 0 + (-i) \cdot i = 3; |b_1| = \sqrt{3}. \\ |b_2|^2 = \left(-\frac{4i}{3}\right) \cdot \frac{4i}{3} + \frac{1}{3} \cdot \frac{1}{3} + 1 \cdot 1 + \frac{5}{3} \cdot \frac{5}{3} = \frac{51}{3^2}; |b_2| = \frac{\sqrt{51}}{3}. \\ |b_3|^2 = \left(\frac{7 + 5i}{17}\right) \cdot \left(\frac{7 - 5i}{17}\right) + \left(\frac{3 - 11i}{17}\right) \cdot \left(\frac{3 + 11i}{17}\right) + \left(\frac{9 + i}{17}\right) \cdot \left(\frac{9 - i}{17}\right) + \left(\frac{-2 - 4i}{17}\right) \left(\frac{-2 + 4i}{17}\right) = \frac{306}{17^2}; |b_3| = \frac{\sqrt{306}}{17}. \\ \end{array}$$

Vectors $b_1, b_2, b_3$ form an orthogonal basis. We will normalize vectors $b_1, b_2, b_3$.

$$c_1 = \frac{b_1}{|b_1|} = \frac{1}{\sqrt{3}} (1,i,0,-i).$$ $$c_2 = \frac{b_2}{|b_2|} = \frac{1}{\sqrt{51}} (-4i, 1,3,5).$$ $$c_3 = \frac{b_3}{|b_3|} = \frac{1}{\sqrt{306}} (5i + 7, -11i + 3,9 + i, -2 - 4i).$$

Answer: $\frac{1}{\sqrt{3}} (1,i,0,-i), \frac{1}{\sqrt{51}} (-4i,1,3,5), \frac{1}{\sqrt{306}} (5i + 7, -11i + 3,9 + i, -2 - 4i)$.

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