# Why is a positive definite quantity plotted as a negative number?

I have two boxes, $A$ and $B$ that exchange mass.

I will call the flux of mass between $A$ and $B$ $ab$; $ba$ is the reverse.

$X = ab -ba$

I am trying to understand the following statement:

$dA/dt$, $dB/dt$, $ab$, and $ba$ are positive definite since they represent a one-way process. Whereas $X$ is a vector with a sign that implies the direction of transfer.

The origin of my confusion is that when the terms $X$, $ab$, and $ba$ are plotted together over time, $ba$ is plotted with negative values. I imagine that this is justified by rewriting the equation as $X=ab + (-ba)$.

I have never seen the term "positive definite".

Here is a diagram of the problem:

-
Postive definite is a term commonly used in the context of matrix or operator algebra, and also in dynamical systems (although I am only familiar with the former 2). I am confident someone will post a competent answer, but in the meantime, have a look if you like: en.wikipedia.org/wiki/Positive_definiteness. – gnometorule Feb 7 '13 at 20:26

A real valued function $f: X \rightarrow \mathbb{R}$ on an arbitrary set $X$ is called positive-definite if $f(x)>0, \forall x \in \mathcal{X}$. The flux is in general not a scalar quantity, because it is described by the magnitude and the direction as well. Since $ab$ denotes the flux from $A$ to $B$, then the information of direction is encoded in the ordering of the characters $a$ and $b$. Hence there is no reason for a sign. Consequently, any value of $ab$ will only be positive and will denote the magnitude. The same is true for $ba$. If values of $ba$ are plotted with negative sign, that might just be an abuse of notation and its not rigorous. I believe your understanding is correct. In fact, you can see that the diagram is not very precise, i.e. instead of $ba$ they plot $-ba$, since if $ba<0$ and $ab>0$, then $X>0$, which is not the case.
When the set $X$ has an algebraic structure, e.g. when it is a monoid with respect to addition with identity element $e$, then we must have $f(x)\ge 0 \forall x \in \mathcal{X}$ and $f(x)=0 \Leftrightarrow x=e$.