Proof that free variables yield basis for Nullspace Let $A$ be a matrix with rank $r$ and number of columns $m$ where $r<m$. We can write the reduced eschelon form $R$ as:
$\begin{bmatrix} I & F \\ 0 & 0 \\ \end{bmatrix}$ or 
$\begin{bmatrix} I & F \\ \end{bmatrix}$
We have that each row where we don't have a pivor correspond to a free variable.
It is stated on many courses that a ways to obtain a basis for the nullspace of that matrix is to:
1) Write $x$ as a function of the free variables.
2) Generate $m-r$ vectors by setting the free variable $i$ to $1$ and all others to $0$.
How to prove that this always holds?
 A: We can write the associated system of equations as
$$
Rx = 0 \implies \pmatrix{I & F\\0&0} \pmatrix{\mathbf x^{(p)} \\ \mathbf x^{(f)}} = 0
$$
where $\mathbf x^{(p)}$ is the column-vector of pivot variables and $\mathbf x^{(f)}$ is the column-vector of free variables.  Now, either using block-matrix multiplication or by verifying directly, we see that we can rewrite this equation as
$$
I\mathbf x^{(p)} + F\mathbf x^{(f)} = 0 \implies \mathbf x^{(p)} = -F\mathbf x^{(f)}
$$
Writing this out, we have
$$
\pmatrix{x^{(p)}_1 \\ x^{(p)}_2 \\ \vdots  \\ x^{(p)}_r} = 
-F \pmatrix{x^{(f)}_1 \\ x^{(f)}_2 \\ \vdots \\ x^{(f)}_{n-r}}
$$
Thus, we can characterize the solution space by writing each pivot variable as a function of the free variables.  Moreover, we can rewrite this as
$$
\pmatrix{x^{(p)}_1 \\ x^{(p)}_2 \\ \vdots  \\ x^{(p)}_r} = 
-x^{(f)}_1F \pmatrix{1\\0\\ \vdots \\ 0}
-x^{(f)}_2F \pmatrix{0\\1\\ \vdots \\ 0} - \cdots 
-x^{(f)}_{n-r}F \pmatrix{0\\0\\ \vdots \\ 1}
$$
Now, looking at the second term for example, we see that 
$$
\pmatrix{x^{(p)}_1 \\ x^{(p)}_2 \\ \vdots  \\ x^{(p)}_r} = 
-F \pmatrix{0\\1\\ \vdots \\ 0}
$$
is exactly what we get by setting $x^{(f)}_2 = 1$ and all other free variables equal to $0$.
