Answer
See answers below
Work Step by Step
a) Assume that $A=\begin{bmatrix}
1 & 4 & 1 & 3 \\
2 & 8& 3 & 5\\
1 & 4 & 0 & 4 \\
2 & 8 & 2 & 6
\end{bmatrix}$
then we reduce $A$: $\begin{bmatrix}
1 & 4 & 1 & 3 \\
2 & 8& 3 & 5\\
1 & 4 & 0 & 4 \\
2 & 8 & 2 & 6
\end{bmatrix} \approx \begin{bmatrix}
1 & 4 & 1 & 3 \\
0 & 0& 1 & -1\\
0 & 0 & -1 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix} \approx \begin{bmatrix}
1 & 4 & 1 & 3 \\
0 & 0& 1 & -1\\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix} $
The basic for rowspace $A$ is $\{(1,4,1,3);(0,0,1,-1)\}$
Hence $S$ is basic of given subspace.
b) Assume that column of $A$ formed by given vectors,
then $\begin{bmatrix}
1 & 2 & 1 & 2 \\
4 & 8 & 4 & 8\\
1 & 3 & 0 & 2\\
3 & 5 & 4 & 6
\end{bmatrix} \approx \begin{bmatrix}
1 & 2 & 1 & 2 \\
0 & 0 & 0 & 0\\
0 & 1 & -1 & 0\\
0 & -1 & 1 & 0
\end{bmatrix} \approx \begin{bmatrix}
1 & 2 & 1 & 2 \\
0 & 0 & 0 & 0\\
0 & 1 & -1 & 0\\
0 & 0 & 0 & 0
\end{bmatrix} $
We notice that the first and second bolumns are independent. The basic for colspace $A$ is: $\{(1,4,1,3);(2,8,3,5)\}$
Hence $S$ is a basic of given subspace.