# 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 $Ax\geq b$?

• Two possible definitions include $|Ax| >= |b|$ and element-by-element $>=$ comparisons. Is there nothing in the prior sections to give context to this relation? – abiessu Aug 29 '13 at 20:07
• In the context of linear coding I've seen it used as a coordinate-wise relation, but I don't think I've seen that anywhere else. The norm inequality @abiessu suggested seems the most likely given no prior context. – Jonathan Y. Aug 29 '13 at 20:28
• If never seen it to mean $|Ax| \geq |b|$ so the coordinate-wise inequality seems the most likely. – WimC Aug 29 '13 at 20:33
• Farkas' Lemma will definitely use a coordinate-wise relation, it's for linear programming. – Evan Aug 29 '13 at 20:51
• Rudra: In light of your responses to answers below, you should clarify what exactly you're looking for in an answer. – Cam McLeman Aug 31 '13 at 0:15

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.

• I understand that x>=0 means, that each component of vector x must be greater than or equal to zero. But can we say something from the perspective of linear maps like how is b related to the column space of A? – Rudra Murthy Aug 29 '13 at 21:35

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.

• I understand that x>=0 means, that each component of vector x must be greater than or equal to zero. But can we say something from the perspective of linear maps like how is b related to the column space of A? – Rudra Murthy Aug 29 '13 at 21:36

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.

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 $$(Ax)_i \le b_i$$