Null space of a matrix I was referring to this lecture http://www.stanford.edu/class/ee364a/videos/video05.html (about 0:38:10) related to convex optimization and for optimization it had a certain affine function equality constraint like
$$Ax=b$$
The lecturer then obtained the equivalent optimization problem removing the equality  constraint and replacing $x$ with $Fz+x_0$
$$Ax=b \iff x=Fz+x_0\text{ for some }z.$$
According to the lecturer $x_0$ is a solution of the equality $Ax=b$ and $F$ is obtained from the null space of matrix $A$. I didn't get this null space thing and how this 
$$x = Fz+x_0$$
is derived. Can anyone please explain?

When we have $x_b$ a solution of $Ax = b$, then all other solutions of $Ax+b$ are given by $x_b + x_0$, where $x_0$ is any solution of $Ax = 0$.
Proof:
Fix $x_b$ a solution of $Ax=b$.


*

*Let $x_0$ be a solution of $Ax = 0$. Then $x_0 + x_b$ is a solution of $Ax+b$, since $A(x_0+x_b)= Ax_0 + Ax_b = 0 + b = b$.

*Now, let $y_b$ be any solution of $Ax=b$. To prove that it has the form $x_0 + x_b$ for some $x_0$ solution of $Ax = 0$, it suffices to show that $y_b - x_b$ is indeed a solution of $Ax = 0$. Since $A(y_b - x_b) = Ay_b - Ax_b = b - b = 0$, the result follows.


This is a proof of two inclusions, which together show the equality, for any $x_b$ solution of $Ax = b$: $\{ x \in \mathbb{K}^n : Ax = b \} = x_b + \{ x \in \mathbb{K}^n : Ax = 0 \} = \{ x + x_b : x \in \mathbb{K}^n, Ax = 0 \}$ (where $\mathbb{K}$ is the appropriate field and $n$ is the number of columns of $A$).
We call the set $\{ x \in \mathbb{K}^n : Ax = 0 \}$ the null space of $A$.
