Answer to Question #151362 in Linear Algebra for Ashweta Padhan

Question

Consider C^n, with the standard inner product. Find an orthonormal basis for the subspace spanned by v1=(1, 0, i) and v2=(2, 1, 1+i)

Answer 1

Orthonormalize the set of the vectors v₁ = $\begin{bmatrix} 1 \\ 0 \\ i \end{bmatrix}$ and v₂ = $\begin{bmatrix} 2 \\ 1 \\ 1 + i \end{bmatrix}$

According to the Gram-Schmidt process, u_k = v_k - $\sum_{j - 1}^{k - 1} proj_{uj}(v_k)$ where

$proj_u(v) = \frac{u*v}{u * u}u$

The normalized vector is $e_k = \frac{u_k}{\sqrt{u_k*u_k}}$

step1: $u_1 = v_! =$ $\begin{bmatrix} 1 \\ 0 \\ i \end{bmatrix}$

$e_1 = \frac{u_1}{\sqrt{u_1*u_1}} = \begin{bmatrix} \frac{\sqrt{2}}{2} \\ 0 \\ \frac{\sqrt{2}i}{2} \end{bmatrix}$

step2: $u_2 = v_2 - \frac{u_1*v_2}{u_1*u_1}u_1 = \begin{bmatrix} \frac{1}{2} - \frac{i}{2} \\ 1 \\ \frac{3}{2} - \frac{i}{2} \end{bmatrix}$

$e_2 = \frac{u_2}{\sqrt{u_2*u_2}} = u_2 = v_2 - \frac{u_1*v_2}{u_1*u_1}u_1 = \begin{bmatrix} \frac{1}{4} - \frac{i}{4} \\ \frac{1}{2} \\ \frac{3}{4} - \frac{i}{4} \end{bmatrix}$

Answer: the set of orthonormal vectors is

$e_1 = \begin{bmatrix} \frac{\sqrt{2}}{2} \\ 0 \\ \frac{\sqrt{2}i}{2} \end{bmatrix}$, $e_2 = \begin{bmatrix} \frac{1}{4} - \frac{i}{4} \\ \frac{1}{2} \\ \frac{3}{4} - \frac{i}{4} \end{bmatrix}$