- #1

- 1,266

- 11

## Homework Statement

http://img130.imageshack.us/img130/2217/49541511.gif [Broken]

## The Attempt at a Solution

I'm not sure how I need to start but I think I need to use the Gram-Schmidt orthogonalization process to first to transform this into an orthogonal basis, and then normalize them to obtain an orthonormal basis. So we have:

[tex]w_1=(1,1,1,0)^T[/tex]

[tex]w_2=(0,1,1,1)^T[/tex]

[tex]v_1=w_1=(1,1,1,0)^T[/tex]

[tex]v_2=w_2 - proj_{w_1}w_2 = (1,1,1,0)^T - \frac{w_2 . v_1}{\| v_1 \|}v_1[/tex]

[tex]= \left(\begin{array}{ccc}0\\1\\1\\1\end{array}\right) - \frac{\left(\begin{array}{ccc}0\\1\\1\\1\end{array}\right).\left(\begin{array}{ccc}1\\1\\1\\0\end{array}\right)}{\left\| \left(\begin{array}{ccc}1\\1\\1\\0\end{array}\right) \right\|} \left(\begin{array}{ccc}1\\1\\1\\0\end{array}\right)[/tex]

[tex]= \left(\begin{array}{ccc}0\\1\\1\\1\end{array}\right)- \frac{2}{3} \left(\begin{array}{ccc}1\\1\\1\\0\end{array}\right) = \left(\begin{array}{ccc}-2/3\\1/3\\1/3\\1/3\end{array}\right)[/tex]

Now I normalize the the vectors:

[tex]\left\| v_1 \right\| = \sqrt{3}[/tex]

[tex]\left\| v_2 \right\| = \sqrt{\left(\frac{-2}{3}\right)^2+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^2} = \sqrt{\frac{7}{9}}[/tex]

[tex]q_1 = \frac{v_1}{\|v_1 \|} = \left(\begin{array}{ccc}1/\sqrt{3}\\1/\sqrt{3}\\1/\sqrt{3}\\0\end{array}\right)[/tex]

[tex]q_2 = \frac{v_2}{\|v_2 \|} = \left(\begin{array}{ccc}-2/3 \div \sqrt{7/9}\\1/3 \div \sqrt{7/9}\\1/3 \div \sqrt{7/9}\\1/3\div \sqrt{7/9}\end{array}\right)[/tex]

I don't know if my method and my results are correct. Is there anything wrong with my working?

Last edited by a moderator: