Answer
See below
Work Step by Step
Determine basic for rowspace $(A)$.
$A=\begin{bmatrix}
1 & 3 & 5\\
-1 & -3 & 1\\
0 & 2 & 3\\
1& 5 & 2\\
1 & 5 & 8
\end{bmatrix} \approx \begin{bmatrix}
1 & 3 & 5\\
0 & 0 & 6\\
0 & 2 & 3\\
0& 2 & -3\\
0 & 2 & 3
\end{bmatrix}\approx \begin{bmatrix}
1 & 3 & 5\\
0 & 0& 1\\
0 & 2 & 3\\
0& 4 & 0\\
0 & 0 & 0
\end{bmatrix}\approx \begin{bmatrix}
1 & 0 & 0\\
0 & 0 & 1\\
0 & 0 & 0\\
0& 4 & 0\\
0 & 0 & 0
\end{bmatrix} \approx \begin{bmatrix}
1 & 0 & 0\\
0 & 0 & 1\\
0 & 0 & 0\\
0& 1 & 0\\
0 & 0 & 0
\end{bmatrix}\approx \begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
0& 0 & 0\\
0 & 0 & 0
\end{bmatrix}$
Hence, basic for rowspace $(A):\{(1,0,0),(0,1,0),(0,0,1)\}$
Determine basic for colspace $(A)$.
$A=\begin{bmatrix}
1 & 3 & 5\\
-1 & -3 & 1\\
0 & 2 & 3\\
1& 5 & 2\\
1 & 5 & 8
\end{bmatrix} \approx \begin{bmatrix}
1 & 0 & 0\\
-1 & 0 & 6\\
0 & 2 & 3\\
1& 2 & -3\\
1 & 2 & 3
\end{bmatrix}\approx \begin{bmatrix}
1 & 0 & 0\\
-1 & 0& 2\\
0 & 1 & 1\\
1 & 1 & -1\\
1 & 1 & 1
\end{bmatrix}\approx \begin{bmatrix}
1 & 0 & 0\\
-1 & 0 & 2\\
-1 & 1 & 0\\
0& 1 & -2\\
0 & 1 & 0
\end{bmatrix} \approx \begin{bmatrix}
1 & 0 & 0\\
-1 & 0 & 1\\
-1 & 1 & 0\\
0& 1 & -1\\
0 & 1 & 0
\end{bmatrix}$
Hence, basic for rowspace $(A):\{(1,-1,-1,0,0),(0,0,1,1,1),(0,1,0,-1,0)\}$
Assume $x_1=(1,-1,-1,0,0)\\
x_2=(0,1,2)$
Apply Gram-Schmidt:
$v_1=x_1=(1,-1,-1,0,0)\\
v_2=x_2-\frac{}{||v_1||^2}\\
=(0,0,1,1,1)-\frac{}{||(1,-1,-1,0,0)||^2}(1,-1,-1,0,0)\\
=(0,0,1,1,1)-\frac{0.1+0.(-1)+1.(-1)+1.0+1.0}{||1^2+(-1)^2+(-1)^2+0^2+0^2}(1,-1,-1,0,0)\\
=(0,0,1,1,1)+\frac{1}{3}(1,-1,-1,0,0)\\
=\frac{1}{3}(1,-1,2,3,3)\\
w_3=y_3-\frac{}{||w_1||^2}-\frac{}{||w_2||^2}\\
=(0,1,0,-1,0)-\frac{}{||(1,-1,-1,0,0)||^2}(1,-1,-1,0,0)-\frac{}{||\frac{1}{3}(1,-1,2,3,3)||^2}\frac{1}{3}(1,-1,2,3,3)\\
=(0,1,0,-1,0)-\frac{0.1+1.(-1)+0(-1).0+0.0}{1^2+(-1)^2+(-1)^2+0^2+0^2}(1,-1,-1,0,0)-\frac{\frac{1}{3}(0.1+1.(-1)+0.2+(-1).3+0.3)}{\frac{1}{3}.\frac{1}{3}(1^2+(-1)^2+2^2+3^2+3^2)}\frac{1}{3}(1,-1,2,3,3)\\
=(0,1,0,-1,0)+\frac{1}{3}(1,-1,-1,0,0)+\frac{1}{6}(1,-1,2,3,3)\\
=\frac{1}{2}(1,1,0,-1,1)$
To determine an orthogonal set, we obtain:
$\frac{w_1}{||w_1||}=\frac{(1,-1,-1,0,0)}{\sqrt 1^2+(-1)^2+(-1)^2+0^2+0^2}=\frac{1}{\sqrt 3}(1,-1,-1,0,0)\\
\frac{w_2}{||w_2||}=\frac{\frac{1}{3}(1,-1,2,3,3)}{\sqrt \frac{1}{3}.\frac{1}{3}.1^2+(-1)^2+2^2+3^2+3^2}=\frac{1}{2\sqrt 6}(1,-1,2,3,3)\\
\frac{w_3}{||w_3||}=\frac{\frac{1}{2}(1,1,0,-1,1)}{\sqrt \frac{1}{2}.\frac{1}{2}.1^2+1^2+0^2+(-1)^2+1^2}=\frac{1}{2}(1,1,0,-1,1)$
The orthogonal basis for colspace $(A)$ is:
$\{\frac{1}{\sqrt 3}(1,-1,-1,0,0),\frac{1}{2\sqrt 6}(1,-1,2,3,3),\frac{1}{2}(1,1,0,-1,1)\}$
