308F/G Winter 2013, Quiz 4
- The first problem was problem 13 from section 3.6 of the book. It asked to find an orthogonal basis for the span of the vectors
\[w_{1}=\left( \begin{array}{c} 0 \\ 0 \\ 1 \\ 0\end{array}\right), w_{2}=\left( \begin{array}{c} 1 \\ 1 \\ 2 \\ 1\end{array}\right), w_{3}=\left( \begin{array}{c} 1 \\ 0 \\ 1 \\ 1\end{array}\right).\]
There is more than one way to go about this, but we will use the Grahm-Schmidt process to find a basis. We first let
\[u_{1}=w_{1}=\left( \begin{array}{c} 0 \\ 0 \\ 1 \\ 0\end{array}\right),\]
then
\[u_{2}=w_{2}-\frac{u_{1}^{T}w_{2}}{||u_{1}||^{2}}u_{1}
=\left( \begin{array}{c} 1 \\ 1 \\ 2 \\ 1\end{array}\right)-\frac{0\cdot 1 +0\cdot 1 + 1 \cdot 2 + 0\cdot 1}{0^{2}+0^{2}+1^{2}+0^{2}}\left( \begin{array}{c} 0 \\ 0 \\ 1 \\ 0\end{array}\right)
=\left( \begin{array}{c} 1 \\ 1 \\ 2 \\ 1\end{array}\right)-2\left( \begin{array}{c} 0 \\ 0 \\ 1 \\ 0\end{array}\right)
=\left( \begin{array}{c} 1 \\ 1 \\ 0 \\ 1\end{array}\right)\]
and finally
\[u_{3}=w_{3}-\frac{u_{1}^{T}w_{3}}{||u_{1}||^{2}}u_{1}-\frac{u_{2}^{T}w_{3}}{||u_{2}||^{2}}u_{2}
= \left( \begin{array}{c} 1 \\ 0 \\ 1 \\ 1\end{array}\right)-\frac{0\cdot 1 + 0\cdot 0 + 1\cdot 1 + 0\cdot 1}{1}\left( \begin{array}{c} 0 \\ 0 \\ 1 \\ 0\end{array}\right)-\frac{1\cdot 1+0\cdot 1 +1\cdot 0 +1\cdot 1}{1^{2}+1^{2}+0^{2}+1^{2}}\left( \begin{array}{c} 1 \\ 1 \\ 0 \\ 1\end{array}\right)
=\left( \begin{array}{c} 1 \\ 0 \\ 1 \\ 1\end{array}\right)-\left( \begin{array}{c} 0 \\ 0 \\ 1 \\ 0\end{array}\right)-\left( \begin{array}{c}2/3 \\ 2/3 \\ 0 \\ 2/3\end{array}\right)
=\left( \begin{array}{c}1/3 \\ -2/3 \\ 0 \\ 1/3\end{array}\right).\]