Answer
See below
Work Step by Step
Given $A=\begin{bmatrix}
1 & -4 & 7\\
-2 & 6 & -8\\
-1 & 0 & 5
\end{bmatrix}$
Determine basis for rowspace $(A)$:
$\begin{bmatrix}
1 & -4 & 7\\
-2 & 6 & -8\\
-1 & 0 & 5
\end{bmatrix} \approx \begin{bmatrix}
1 & -4 & 7\\
0 & -2 & 6\\
0 & -4 & 12
\end{bmatrix} \approx \begin{bmatrix}
1 & -4 & 7\\
0 & -1 & 3\\
0 & -1 & 3
\end{bmatrix} \approx \begin{bmatrix}
1 & -4 & 7\\
0 & -1 & 3\\
0 & 0 & 0
\end{bmatrix}$
Basic for rowspace$(A)$ is $\{(1,-4,7),(0,-1,3)\}$
According to Gram-Schmidt process, we have:
$v_1=x_1=(1,-4,7)\\
v_2=x_2-\frac{}{||v_1||^2}v_1\\
=(0,-1,3)-\frac{)}{||(1,-4,7)||^2}(1,-4,7)\\
=(0,-1,3)-\frac{0.1+(-1).(-4)+3.7}{1^2+(-4)^2+7^2}(1,-4,7)\\
=(0,-1,3)-\frac{25}{66}(1,-4,7)\\
=\frac{1}{66}(-25,34,23)$
Hence, an orthogonal basis for rowspace $(A)$ is:
$\{(0,-4,7),\frac{1}{66}(-25,34,23)\}$
Determine basis for colspace $(A)$:
$\begin{bmatrix}
1 & -4 & 7\\
-2 & 6 & -8\\
-1 & 0 & 5
\end{bmatrix} \approx \begin{bmatrix}
1 & 0 & 0\\
-2 & -2 & 6\\
-1 & -4& 12
\end{bmatrix} \approx \begin{bmatrix}
1 & 0 & 0\\
-2 & 1 & 1\\
-1 & 2 & 2
\end{bmatrix} \approx \begin{bmatrix}
1 & 0 & 0\\
-2 & 1 & 0\\
-1 & 2 & 0
\end{bmatrix}$
Basis for colspace $(A):\{(1,-2,5);(0,1,-2)\}$
Assume $y_1=(1,-2,-1)\\
y_2=(0,1,2)$
Apply Gram- Schmidt
$v_1=y_1=(1,-2,-1)\\
v_2=y_2-\frac{}{||v_1||^2}v_1\\
=(0,1,2)-\frac{}{||(1,-2,-1||^2}(1,-2,-1)\\
=(0,1,2)-\frac{0.1+1.(-2)+2.(-1)}{1^2+(-2)^2+(-1)^2}(1,-2,-1)\\
=(0,1,2)+\frac{2}{3}(1,-2,-1)\\
=\frac{1}{3}(2,-1,4)$
Consequently, an orthogonal basis for colspace $(A)$ is
$\{(1,-2,-1),\frac{1}{3}(2,-1,4)\}$
