Orthonormalize the set of the vectors v1 = [ 1 0 i ] \begin{bmatrix}
1 \\
0 \\
i
\end{bmatrix} ⎣ ⎡ 1 0 i ⎦ ⎤ and v2 = [ 2 1 1 + i ] \begin{bmatrix}
2 \\
1 \\
1 + i
\end{bmatrix} ⎣ ⎡ 2 1 1 + i ⎦ ⎤
According to the Gram-Schmidt process, uk = vk - ∑ j − 1 k − 1 p r o j u j ( v k ) \sum_{j - 1}^{k - 1} proj_{uj}(v_k) ∑ j − 1 k − 1 p ro j u j ( v k ) where
p r o j u ( v ) = u ∗ v u ∗ u u proj_u(v) = \frac{u*v}{u * u}u p ro j u ( v ) = u ∗ u u ∗ v u
The normalized vector is e k = u k u k ∗ u k e_k = \frac{u_k}{\sqrt{u_k*u_k}} e k = u k ∗ u k u k
step1: u 1 = v ! = u_1 = v_! = u 1 = v ! = [ 1 0 i ] \begin{bmatrix}
1 \\
0 \\
i
\end{bmatrix} ⎣ ⎡ 1 0 i ⎦ ⎤
e 1 = u 1 u 1 ∗ u 1 = [ 2 2 0 2 i 2 ] e_1 = \frac{u_1}{\sqrt{u_1*u_1}} = \begin{bmatrix}
\frac{\sqrt{2}}{2} \\
0 \\
\frac{\sqrt{2}i}{2}
\end{bmatrix} e 1 = u 1 ∗ u 1 u 1 = ⎣ ⎡ 2 2 0 2 2 i ⎦ ⎤
step2: u 2 = v 2 − u 1 ∗ v 2 u 1 ∗ u 1 u 1 = [ 1 2 − i 2 1 3 2 − i 2 ] u_2 = v_2 - \frac{u_1*v_2}{u_1*u_1}u_1 = \begin{bmatrix}
\frac{1}{2} - \frac{i}{2} \\
1 \\
\frac{3}{2} - \frac{i}{2}
\end{bmatrix} u 2 = v 2 − u 1 ∗ u 1 u 1 ∗ v 2 u 1 = ⎣ ⎡ 2 1 − 2 i 1 2 3 − 2 i ⎦ ⎤
e 2 = u 2 u 2 ∗ u 2 = u 2 = v 2 − u 1 ∗ v 2 u 1 ∗ u 1 u 1 = [ 1 4 − i 4 1 2 3 4 − i 4 ] e_2 = \frac{u_2}{\sqrt{u_2*u_2}} = u_2 = v_2 - \frac{u_1*v_2}{u_1*u_1}u_1 = \begin{bmatrix}
\frac{1}{4} - \frac{i}{4} \\
\frac{1}{2} \\
\frac{3}{4} - \frac{i}{4}
\end{bmatrix} e 2 = u 2 ∗ u 2 u 2 = u 2 = v 2 − u 1 ∗ u 1 u 1 ∗ v 2 u 1 = ⎣ ⎡ 4 1 − 4 i 2 1 4 3 − 4 i ⎦ ⎤
Answer: the set of orthonormal vectors is
e 1 = [ 2 2 0 2 i 2 ] e_1 = \begin{bmatrix}
\frac{\sqrt{2}}{2} \\
0 \\
\frac{\sqrt{2}i}{2}
\end{bmatrix} e 1 = ⎣ ⎡ 2 2 0 2 2 i ⎦ ⎤ , e 2 = [ 1 4 − i 4 1 2 3 4 − i 4 ] e_2 = \begin{bmatrix}
\frac{1}{4} - \frac{i}{4} \\
\frac{1}{2} \\
\frac{3}{4} - \frac{i}{4}
\end{bmatrix} e 2 = ⎣ ⎡ 4 1 − 4 i 2 1 4 3 − 4 i ⎦ ⎤
Comments