Chapter 5 - Inner Product Spaces - 5.3 The Gram-Schmidt Process - Problems - Page 365: 13

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Given $A=\begin{bmatrix} 3 & 1 & 4\\ 1 & -2 & 1\\ 1 & 5 & 2 \end{bmatrix}$ Determine basis for rowspace $(A)$: $\begin{bmatrix} 3 & 1 & 4\\ 1 & -2 & 1\\ 1 & 5 & 2 \end{bmatrix} \approx \begin{bmatrix} 1 & -2 & 1\\ 3 & 1 & 4\\ 1 & 5 & 2 \end{bmatrix} \approx \begin{bmatrix} 1 & -2 & 1\\ 0 & 7 & 1\\ 0 & 7 & 1 \end{bmatrix} \approx \begin{bmatrix} 1 & -2 & 1\\ 0 & 7 & 1\\ 0 & 0 & 0 \end{bmatrix} $ Basic for rowspace$(A)$ is $\{(1,-2,1),(0,7,1)\}$ According to Gram-Schmidt process, we have: $v_1=x_1=(1,-2,1)\\ v_2=x_2-\frac{}{||v_1||^2}v_1\\ =(0,7,1)-\frac{)}{||(1,-2,1)||^2}(1,-2,1)\\ =(0,7,1)-\frac{0.1+7.(-2)+1.1}{1^2+(-2)^2+1^2}(1,-2,1)\\ =(0,7,1)-\frac{13}{6}(1,-2,1)\\ =\frac{1}{6}(13,16,19)$ Hence, an orthogonal basis for rowspace $(A)$ is: $\{(0,7,1),\frac{1}{6}(13,16,19)\}$ Determine basis for colspace $(A)$: $\begin{bmatrix} 3 & 1 & 4\\ 1 & -2 & 1\\ 1 & 5 & 2 \end{bmatrix} \approx \begin{bmatrix} 1 & -2 & 1\\ -2 & 1 & 1\\ 5 & 1 & 2 \end{bmatrix} \approx \begin{bmatrix} 1 & -2 & 1\\ -2 & 7 & 9\\ 5 & -14 & -18 \end{bmatrix} \approx \begin{bmatrix} 1 & 0 & 0\\ -2 & 1 & 1\\ 5& -2 & -2 \end{bmatrix} \approx \begin{bmatrix} 1 & 0 & 0\\ -2 & 1 & 0\\ 5& -2 & 0 \end{bmatrix}$ Basis for colspace $(A):\{(1,-2,5);(0,1,-2)\}$ Assume $y_1=(1,-2,5)\\ y_2=(0,1,-2)$ Apply Gram- Schmidt $v_1=y_1=(1,-2,5)\\ v_2=y_2-\frac{}{||v_1||^2}v_1\\ =(0,1,-2)-\frac{}{||(1,-2,5)(-2)^2+5^2}(1,-2,5)\\ =(0,1,-2)-\frac{0-2-10}{1+4+25}(1,-2,5)\\ =(0,1,-2)+\frac{2}{5}(1,-2,5)\\ =\frac{1}{5}(2,1,0)$ Consequently, an orthogonal basis for colspace $(A)$ is $\{(1,-2,5),\frac{1}{5}(2,1,0)\}$