# 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.

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 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 $x4,x5,x6,x7$
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 $N(S)$.