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Construction of Orthogonal Set of Vectors From a Linearly Independent SetUsing Gram - Schmidt Procedure

Given a set of linearly independent vectors in a space  
\[V\]
  we can consturct an orthogonal set for  
\[V\]
  using the Gram - Scmidt procedure.
Example Find an oorthogonal set for  
\[\mathbb{R}^5\]
  using the set of vectors  
\[\left\{ \mathbf{v}_1, \mathbf{v}_2 , \mathbf{v}_3 \right\} = \left\{ \begin{pmatrix}1\\0\\0\\1\\0\end{pmatrix} , \begin{pmatrix}0\\1\\1\\-1\\0\end{pmatrix} , \begin{pmatrix}1\\1\\1\\1\\1\end{pmatrix} \right\}\]

This set of vectors is not an orthogonal set.
\[ \begin{pmatrix}1\\0\\0\\1\\0\end{pmatrix} \cdot \begin{pmatrix}0\\1\\1\\-1\\0\end{pmatrix} =0+0+0-1+0=-1 \]

\[ \begin{pmatrix}1\\0\\0\\1\\0\end{pmatrix} \cdot \begin{pmatrix}1\\1\\1\\1\\1\end{pmatrix} =1+0+0-1+0=2 \]

\[ \begin{pmatrix}0\\1\\1\\-1\\0\end{pmatrix} \cdot \begin{pmatrix}1\\1\\1\\1\\1\end{pmatrix} =0+1+1-1+0=1 \]

To construct an orthogonal set  
\[\left\{ \mathbf{u}_1, \mathbf{u}_2 , \mathbf{u}_3 \right\}\]
  first let  
\[\mathbf{u}_1= \mathbf{v}_1= \begin{pmatrix}1\\0\\0\\1\\0\end{pmatrix}\]

\[\begin{equation} \begin{aligned}\mathbf{u}_2 &= \mathbf{v}_2 - \frac{\mathbf{v}_2 \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1} \\ &= \begin{pmatrix}0\\1\\1\\-1\\0\end{pmatrix} - \frac{ \begin{pmatrix}0\\1\\1\\-1\\0\end{pmatrix} \cdot \begin{pmatrix}1\\0\\0\\1\\0\end{pmatrix}}{\begin{pmatrix}1\\0\\0\\1\\0\end{pmatrix} \cdot \begin{pmatrix}1\\0\\0\\1\\0\end{pmatrix}} \begin{pmatrix}1\\0\\0\\1\\0\end{pmatrix} \\ &= \begin{pmatrix}0\\1\\1\\-1\\0\end{pmatrix}+ \frac{1}{2} \begin{pmatrix}1\\0\\0\\1\\0\end{pmatrix} \\ &= \begin{pmatrix}1/2\\1\\1\\-1/2\\0\end{pmatrix}\end{aligned} \end{equation}\]

\[\begin{equation} \begin{aligned} \mathbf{u}_3 &= \mathbf{v}_3 - \frac{\mathbf{v}_3 \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1}- \frac{\mathbf{v}_3 \cdot \mathbf{u}_2}{\mathbf{u}_2 \cdot \mathbf{u}_2} \\ &= \begin{pmatrix}1\\1\\1\\1\\1\end{pmatrix} - \frac{ \begin{pmatrix}1\\1\\1\\1\\1\end{pmatrix} \cdot \begin{pmatrix}1\\0\\0\\1\\0\end{pmatrix}} {\begin{pmatrix}1\\0\\0\\1\\0\end{pmatrix} \cdot \begin{pmatrix}1\\0\\0\\1\\0\end{pmatrix}} \begin{pmatrix}1\\0\\0\\1\\0\end{pmatrix} - \frac{ \begin{pmatrix}1\\1\\1\\1\\1\end{pmatrix} \cdot \begin{pmatrix}1/2\\1\\1\\-1/2\\0\end{pmatrix}} {\begin{pmatrix}1/2\\1\\1\\-1/2\\0\end{pmatrix} \cdot \begin{pmatrix}1/2\\1\\1\\-1/2\\0\end{pmatrix}} \begin{pmatrix}1/2\\1\\1\\-1/2\\0\end{pmatrix} \\ &= \begin{pmatrix}1\\1\\1\\1\\1\end{pmatrix}- \begin{pmatrix}1\\0\\0\\1\\0\end{pmatrix}- \begin{pmatrix}2/5\\4/5\\4/5\\-2/5\\0\end{pmatrix}\\ &= \begin{pmatrix}-2/5\\1/5\\1/5\\2/5\\1\end{pmatrix}\end{aligned} \end{equation}\]

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