Question

Use the Gram-Schmidt process to find an orthonormal basis for the subspace of R4 spanned by the vectors (1, 0, 1, 1), (1, 0, 1, 0), (0, 0, 1, 1).

Answer 1

Answer:

$e_1 = \begin{pmatrix}\frac{\sqrt{\textbf{3}}}{\textbf{2}}\\\\\textbf{0} \\\\\frac{\sqrt{\textbf{3}}}{\textbf{3}}\\\\\frac{\sqrt{\textbf{3}}}{\textbf{3}}\end{pmatrix}$         $e_2 = \begin{pmatrix}\frac{\sqrt{\textbf{6}}}{\textbf{6}}\\\\\textbf{0} \\\\\frac{\sqrt{\textbf{6}}}{\textbf{6}}\\\\\frac{\sqrt{\textbf{-6}}}{\textbf{3}}\end{pmatrix}$        $e_3 = \begin{pmatrix}\frac{\sqrt{\textbf{-2}}}{\textbf{2}}\\\\\textbf{0} \\\\\frac{\sqrt{\textbf{2}}}{\textbf{2}}\\\\0\end{pmatrix}$

Step-by-step explanation:

we have to orthonormalize the vectors:

$v_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \\ 1 \end{pmatrix}$    $v_2 = \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix}$      $v_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \end{pmatrix}$

According to Gram - Schmidt process, we have:

$u_k = v_k - \sum_{j = 1} ^ {k - 1} proj_{uj} (v_k)$ where, $proj_u (v) = \frac{u . v}{u . u}u$

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

Now, the first step.

$v_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \\ 1 \end{pmatrix}$ = u₁

Therefore, e₁ = $\frac{u_1}{\sqrt{u_1.u_1}}$

$= \begin{pmatrix}\frac{\sqrt{\textbf{3}}}{\textbf{2}}\\\\\textbf{0} \\\\\frac{\sqrt{\textbf{3}}}{\textbf{3}}\\\\\frac{\sqrt{\textbf{3}}}{\textbf{3}}\end{pmatrix}$

Now, we find e₂.

$u_2 = v_2 - \frac{u_1.v_2}{u_1.u_1}u_1$

$= \begin{pmatrix} \frac{1}{3}\\ \\ 0 \\\\ \frac{1}{3}\\\\ \frac{-2}{3} \end{pmatrix}$

Therefore, $e_2 = \frac{u_2}{\sqrt{u_2.u_2}}$

$e_2 = \begin{pmatrix}\frac{\sqrt{\textbf{6}}}{\textbf{6}}\\\\\textbf{0} \\\\\frac{\sqrt{\textbf{6}}}{\textbf{6}}\\\\\frac{\sqrt{\textbf{-6}}}{\textbf{3}}\end{pmatrix}$

To find e₃:

$u_3 = v_3 - \frac{u_1. v_3}{u_1.u_1}u_1 - \frac{u_2. v_3}{u_2.u_2} u_2$

$= \begin{pmatrix} \frac{-1}{2} \\\\ 0\\ \\ \frac{1}{2} \\\\ 0 \\\end{pmatrix}$

$e_3 = \frac{u_3}{\sqrt{u_3.u_3}}$

$e_3 = \begin{pmatrix}\frac{\sqrt{\textbf{-2}}}{\textbf{2}}\\\\\textbf{0} \\\\\frac{\sqrt{\textbf{2}}}{\textbf{2}}\\\\0\end{pmatrix}$

So, we have the orthonormalized vectors $e_1, e_2, e_3$.

Hence, the answer.