What does $Ax\geq b$ mean in Linear Algebra? I'm going through Farka's Lemma. I can understand what $Ax = b$ from a linear algebra perspective. But, I'm unable to understand what $Ax\geq b$?
 A: The meaning of that notation depends on context, but the Wikipedia linked article makes it clear: 

"Here, the notation $x \ge 0$ means that all components of the vector x are nonnegative."

so, we can guess that $A x \ge b $ is to be understood component wise, or equivalently $A x - b \ge 0 $ with the above interpretation.
A: For Farkas Lemma, $Ax \geq b$ means you multiply $Ax$ to get a vector $v$, and then $v_i \geq b_i$ for each index $i$ for the vectors.  Similarly using the same interpretation, in linear programming problems, people often use $Ax \leq b$ where $x$ is the vector of variables for the linear programming problem and $A,b$ are numeric, to define the constraints for the linear programming problem. 
A: Oh I understand your question now. 
Here is something:
Consider the one-row case, $a^T x \geq b$.
This is geometrically saying that $x$ lies on one side of the hyperplane $a^T x=b$ (pictures available for the two dimensional case, you should draw). The hyperplane is perpendicular to $a$.
Taking all rows together, each row gives a hyperplane cut, and tells you that $x$ is one one of side of the cut. Then these define some convex region for $x$ (not necessarily bounded though)
In terms of the column space, I can't think of anything... but I don't think it leads anywhere that will lend you intuition for understanding Farkas' lemma.
A: This is a short hand notation used in linear programming. For $A \in \mathbb{K}^{m \times n}$, $x \in \mathbb{K}^n$, $b \in \mathbb{K}^m$ it means
$$
A x \le b \iff
\forall i \in \{1,\ldots, m\}: a_i^\top x \le b_i
$$
where $a_i$ is the $i$-th row vector of $A$. 
This can be interpreted as $m$ linear inequality constraints of the type $\le$ on the $n$ unknowns $x_j$.
$$
\sum_{j=1}^n a_{ij} x_j \le b_i
$$ 
Agreeing with leonbloy, it can be interpreted component-wise like this
$$
(Ax)_i \le b_i
$$
