Find a basis of a nullspace of a matrix 
Find a basis of nullspace of S.
I have already found a parameterized solution of $S*x=0$. But I dont Know how to get a basis out of that.
 A: Note that 
$$
S\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\\0\end{bmatrix}
$$
if and only if
\begin{align*}
x_1 &= 8x_4+2x_5-x_6+x_7\\
x_2 &= x_4-x_5+2x_6+x_7\\
x_3 &= -x_4+2x_5+3x_6-x_7
\end{align*}
It follows that the vectors in $\DeclareMathOperator{Null}{Null}\Null(S)$ are exactly the vectors of the form
$$
\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7\end{bmatrix}
=\begin{bmatrix} 8x_4+2x_5-x_6+x_7\\x_4-x_5+2x_6+x_7\\-x_4+2x_5+3x_6-x_7\\x_4\\x_5\\x_6\\x_7\end{bmatrix}=
x_4
\begin{bmatrix}
8\\1\\-1\\1\\0\\0\\0
\end{bmatrix}
+
x_5
\begin{bmatrix}
2\\-1\\2\\0\\1\\0\\0
\end{bmatrix}
+
x_6
\begin{bmatrix}
-1\\2\\3\\0\\0\\1\\0
\end{bmatrix}
+
x_7
\begin{bmatrix}
1\\1\\-1\\0\\0\\0\\1
\end{bmatrix}
$$
Hence
$$
\Null(S)=
\DeclareMathOperator{Span}{Span}\Span\left\{
\begin{bmatrix}
8\\1\\-1\\1\\0\\0\\0
\end{bmatrix},
\begin{bmatrix}
2\\-1\\2\\0\\1\\0\\0
\end{bmatrix},\begin{bmatrix}
-1\\2\\3\\0\\0\\1\\0
\end{bmatrix},\begin{bmatrix}
1\\1\\-1\\0\\0\\0\\1
\end{bmatrix}
\right\}
$$
So, to provide a basis for $\Null(S)$ we need only show that these four vectors are linearly independent. Can you show this?
A: Hint:
A basis for a subspace $H$ of $R^n$ is a linearly independent set in H that spans $H$.
Find the Null space by finding the set of solutions of $S*x=0$.
In your case there are $4$ free variables.
Represent your column vector $x$ in terms of these four variables $x4,x5,x6,x7$
A: Put $S$ in row-reduced echelon form (this is already the case) and then write the solution to $S\vec{x}=\vec{0}$ in parametric form.  Then the null-space becomes clear as any linear combination of $4$ vectors (corresponding to the four free variables in your parametric form), which are exactly the basis vectors of the null-space $N(S)$.
