Chapter 5 - Inner Product Spaces - 5.2 Orthogonal Sets of Vectors and Orthogonal Projections - Problems - Page 360: 9

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Given $v_1=(-4,0,0,1)\\ v_2=(1,2,3,4)$ Asume $u=(u_1,u_2,u_3,u_4)$ is orthogonal to both $v_1$ and $v_2$ then $=0\\ =0\\ -4.u_1+1.u_4=0\\ \rightarrow u_4=4u_1$ $=0\\ =0\\ 1.u_1+2.u_2+3.u_3+4.u_4=0\\ u_1+2u_2+3u_3+16u_1=0\\ 17u_1+2u_2+3u_3=0\\ 17u_1=-2u_2-3u_3 \rightarrow u_1=-\frac{2}{17}u_2-\frac{3}{17}u_3$ Substitute: $u_4=4(-\frac{2}{17}u_2-\frac{3}{17}u_3)\\ \rightarrow u_4=-\frac{8}{17}u_2-\frac{12}{17}u_3$ Obtain: $u=\{(-\frac{2}{17}u_2-\frac{3}{17}u_3,u_2,u_3,-\frac{8}{17}u_2-\frac{12}{17}u_3); u_2,u_3 \in R$ If we take $u_2=17,u_3=0 \rightarrow u_1=(-2,17,0,-8)$ If we take $u_2=0,u_3=17 \rightarrow u_2=(-3,0,17,-12)$ Obtain: $\begin{bmatrix} -4 & 0 & 0 & 1\\ 1& 2 & 3 & 4\\ -2 & 17 & 0 & -8\\ -3 & 0 & 17 & -12 \end{bmatrix} \approx \begin{bmatrix} 1& 2 & 3 & 4\\ -4 & 0 & 0 & 1\\ -2 & 17 & 0 & -8\\ -3 & 0 & 17 & -12 \end{bmatrix} \approx \begin{bmatrix} 1& 2 & 3 & 4\\ 0 & 8 & 12 & 17\\ 0 & 12 & 6 & 0\\ 0 & 6 & 26 & 0 \end{bmatrix} \approx \begin{bmatrix} 1& 2 & 3 & 4\\ 0 & 8 & 12 & 17\\ 0 & 7 & 2 & 0\\ 0 & 3 & 13 & 0 \end{bmatrix} \approx \begin{bmatrix} 1& 2 & 3 & 4\\ 0 & 8 & 12 & 17\\ 0 & 0 & -\frac{17}{2} & -\frac{119}{8}\\ 0 & 0 & -\frac{17}{2} & -\frac{51}{8} \end{bmatrix} \approx \begin{bmatrix} 1& 2 & 3 & 4\\ 0 & 8 & 12 & 17\\ 0 & 0 & -\frac{17}{2} & -\frac{119}{8}\\ 0 & 0 & 0 & \frac{68}{8} \end{bmatrix}$