Answer
See below
Work Step by Step
Determine basic for rowspace $(A)$.
$A=\begin{bmatrix}
1 & 2 &6\\
2 & 1 & 6\\
0 & 1 & 2\\
1& 0 & 2
\end{bmatrix} \approx \begin{bmatrix}
1 & 2 &6\\
0 & -3 & -6\\
0 & 1 & 2\\
0 & -2 & -4
\end{bmatrix} \approx \begin{bmatrix}
1 & 2 &6\\
0 & 0 & 0\\
0 & 1 & 2\\
0& 0 & 0
\end{bmatrix}\approx \begin{bmatrix}
1 & 2 &6\\
0 & 1 & 2\\
0 & 0 & 0\\
0& 0 & 0
\end{bmatrix}$
Hence, basic for rowspace $(A):\{(1,2,6),(0,1,2)\}$
Assume $x_1=(1,2,6)\\
x_2=(0,1,2)$
Apply Gram-Schmidt:
$v_1=x_1=(1,2,6)\\
v_2=x_2-\frac{}{||v_1||^2}\\
=(0,1,2)-\frac{}{||(1,2,6||^2}(1,2,6)\\
=(0,1,2)-\frac{0.1+1.2+2.6}{||1^2+2^2+6^2}(1,2,6)\\
=(0,1,2)-\frac{14}{41}(1,2,6)\\
=\frac{1}{41}(-14,13,-2)$
To determine an orthogonal set, we obtain:
$\frac{v_1}{||v_1||}=\frac{(1,2,6)}{\sqrt 1^2+2^2+6^2}=\frac{1}{\sqrt 41}(1,2,6)\\
\frac{v_2}{||v_2||}=\frac{\frac{1}{41}(-14,13,-2)}{\sqrt \frac{1}{41}.\frac{1}{41}.(-14)^2+13^2+(-2)^2}=\frac{1}{3\sqrt 41}(-14,13,-2)$
Determine basic for colspace $(A)$.
$A=\begin{bmatrix}
1 & 2 &6\\
2 & 1 & 6\\
0 & 1 & 2\\
1& 0 & 2
\end{bmatrix} \approx \begin{bmatrix}
1 & 2 &3\\
2 & 1 & 3\\
0 & 1 & 1\\
1 & 0 & 1
\end{bmatrix} \approx \begin{bmatrix}
1 & 0 & 0\\
2 & -3 & -3\\
0 & 1 & 1\\
1& -2 & -2
\end{bmatrix}\approx \begin{bmatrix}
1 & 0 &0\\
2 & -3 & 0\\
0 & 1 & 0\\
1& -2 & 0
\end{bmatrix}$
The basic for colspace $(A):\{(1,2,0,1),(0,-3,1,-2)\}$
Assume $x_1=(1,2,0,1)\\
x_2=(0,-3,1,-2)$
Apply Gram-Schmidt:
$v_1=x_1=(1,2,0,1)\\
v_2=x_2-\frac{}{||v_1||^2}\\
=(0,-3,1,-2)-\frac{}{||(1,2,0,1)||^2}(1,2,0,1)\\
=(0,-3,1,-2)-\frac{0.1+(-3).2+1.0+(-2).1}{||1^2+2^2+0^2+1^2}(1,2,0,1)\\
=(0,-3,1,-2)+\frac{4}{3}(1,2,0,1)\\
=\frac{1}{3}(4,-1,3,-2)$
To determine an orthogonal set, we obtain:
$\frac{w_1}{||w_1||}=\frac{(1,2,0,1)}{\sqrt 1^2+2^2+0^2+1^2}=\frac{1}{\sqrt 6}(1,2,0,1)\\
\frac{w_2}{||w_2||}=\frac{\frac{1}{3}(4,-1,3,-2)}{\sqrt \frac{1}{3}.\frac{1}{3}.4^2+(-1)^2+3^2+(-2)^2}=\frac{1}{\sqrt 30}(4,-1,3,-2)$
