Hệ vecto đã cho độc lập tuyến tính
Đặt \(\left\{{}\begin{matrix}x_1=\left(1;0;1\right)\\x_2=\left(1;-1;1\right)\\x_3=\left(1;2;0\right)\end{matrix}\right.\)
Chọn \(y_1=x_1\)
Chọn \(y_2=x_2+tx_1\) với \(t=-\frac{< x_2;y_1>}{< y_1;y_1>}=-\frac{1.1+0.\left(-1\right)+1.1}{1^2+0^2+1^2}=-1\)
\(\Rightarrow y_2=\left(1;-1;1\right)+\left(-1;0;-1\right)=\left(0;-1;0\right)\)
Chọn \(y_3=x_3+t_1y_1+t_2y_2\) với:
\(t_1=-\frac{< x_3;y_1>}{< y_1;y_1>}=-\frac{1.1+0.2+1.0}{1^2+0^2+1^2}=-\frac{1}{2}\)
\(t_2=-\frac{< x_3;y_2>}{< y_2;y_2>}=-\frac{1.0+2.\left(-1\right)+0.0}{0^2+\left(-1\right)^2+0^2}=-\frac{-2}{1}=2\)
\(\Rightarrow y_3=\left(1;2;0\right)+\left(-\frac{1}{2};0;-\frac{1}{2}\right)+\left(0;-2;0\right)=\left(\frac{1}{2};0;-\frac{1}{2}\right)\)
Vậy ta có hệ trực giao: \(\left\{{}\begin{matrix}y_1=\left(1;0;1\right)\\y_2=\left(0;-1;0\right)\\y_3=\left(\frac{1}{2};0;-\frac{1}{2}\right)\end{matrix}\right.\)
