Example 2.2.1.
Obtain a set of three orthonormal vectors by the Schmidt’s method from the vectors \(u_{1}=(1,0,0)\text{,}\) \(u_{2}=(1,1,0)\text{,}\) and \(u_{3}=(1,1,1).\)
Solution.
The given vectors are linearly independent, hence set,
\begin{equation*}
v_{1}=u_{1}.
\end{equation*}
Let,
\begin{equation*}
v_{2}=u_{2}+\lambda v_{1}
\end{equation*}
We have -
\begin{equation*}
\lt v_{1}, u_{2}\gt = (1\times 1+0\times 1+0\times 0) =1
\end{equation*}
and
\begin{equation*}
\lt v_{1}, v_{1}\gt = (1\times 1+0\times 0+0\times 0) =1
\end{equation*}
\begin{equation*}
\therefore \lambda = -\frac{\lt v_{1}, u_{2}\gt}{\lt v_{1}, v_{1}\gt} = -\frac{1}{1}=-1
\end{equation*}
Hence,
\begin{equation*}
v_{2}=u_{2}-v_{1} = (1,1,0)-(1,0,0) = (1-1, 1-0, 0-0)=(0,1,0)
\end{equation*}
Now take
\begin{equation*}
v_{3}=u_{3}+\lambda_{1} v_{1} + \lambda_{2} v_{2}
\end{equation*}
where,
\begin{equation*}
\lambda_{1}= -\frac{\lt v_{1}, u_{3}\gt}{\lt v_{1}, v_{1}\gt}
\end{equation*}
and
\begin{equation*}
\lambda_{2}= -\frac{\lt v_{2}, u_{3}\gt}{\lt v_{2}, v_{2}\gt}
\end{equation*}
Here,
\begin{equation*}
\lt v_{1}, u_{3}\gt = (1\times 1+0\times 1+0\times 1) =1;
\end{equation*}
\begin{equation*}
\lt v_{1}, v_{1}\gt = (1\times 1+0\times 0+0\times 0) =1;
\end{equation*}
\begin{equation*}
\lt v_{2}, u_{3}\gt = (0\times 1+1\times 1+0\times 1)=1;
\end{equation*}
and
\begin{equation*}
\lt v_{2}, v_{2}\gt = (0\times 0+1\times 1+0\times 0) =1.
\end{equation*}
\begin{equation*}
\therefore \lambda_{1} = -\frac{1}{1} =-1, \quad \text{and}\quad \lambda_{2} = -\frac{1}{1} =-1.
\end{equation*}
Hence,
\begin{equation*}
v_{3} = u_{3}=v_{1}-v_{2} =(1-1-0, 1-0-1, 1-0-0) =(0,0,1).
\end{equation*}
The corresponding orthonormal set is
\begin{equation*}
x_{1}= \frac{v_{1}}{\parallel v_{1} \parallel} = \frac{(1,0,0)}{\sqrt{1+0+0}} = \frac{1}{\sqrt{1}}(1,0,0);
\end{equation*}
\begin{equation*}
x_{2}= \frac{v_{2}}{\parallel v_{2} \parallel} = \frac{(0,1,0)}{\sqrt{1}};
\end{equation*}
\begin{equation*}
\text{and} \quad x_{3}= \frac{(0,0,1)}{\sqrt{1}}.
\end{equation*}
