Question: Given two matrices $A\_{m\times n}$ and $B\_{n\times p}$, what is the sufficient and necessary condition for $AB$ to have full rank?

I know $r(AB)=r(B)-\dim N(A) \cap R(B)$, so is it true the above iff $\dim N(A)\cap R(B)=0$? But seems incorrect if $m\lt p$. Please help.

Answer: 
The key concept to note here is that when you differentiate a constant you get 0, this is due to the fact that the slope of the tangent line of a constant function, say $f(x) = 4$, will simply be a horizontal line spanning the x-axis with a slope of zero everywhere.

When you differentiate a particular function that has a constant at the end, say $f(x) = x^2 +2x +4$ to get $f'(x) = 2x +2$, you have no way, given only the derivative $f'(x)$, to recover the "constant information" about the original function.

This is precisely why you have to have a slope field representation of the anti-derivative of a function. When you integrate a particular function, you must add that $+C$ because it says that, the anti-derivative of the function could be one of any of the slope field lines in $\mathbb{R}^2$. The particular value for $C$ collapses it to exactly one of these slope field lines. Here is a graphical representation of a slope field and three particular values of $C$ that draw out the purple functions. The slope field anti-derivative is given by:
$$ \int \cos(x) dx = \sin(x) +C$$

![Antiderivative](https://i.stack.imgur.com/tWwMW.gif)

The three purple functions drawn out are:
$$f(x) = \sin(x) -2$$
$$f(x) = \sin(x) + 0$$
$$f(x) = \sin(x) +2 $$

It may also make more sense when you take a differential equations course, but this should be a sufficient explanation.
Sometimes you need to know all antiderivatives of a function, rather than just one antiderivative. For example, suppose you're solving the differential equation
$$
y'=-4y
$$
and there's an initial condition $y(0)=5$. You get
$$
\frac{dy}{dx}=-4y
$$
$$
\frac{dy}{y} = -4\;dx
$$
$$
\int\frac{dy}{y} = \int -4\;dx
$$
$$
\log\_e y = -4x + C.
$$
Here you've added a constant. Then
$$
y = e^{-4x+C} =e^{-4x}e^C
$$
$$
5=y(0)= e^{-4\cdot0}e^C = e^C
$$
So
$$
y=5e^{-4x}.
$$

(And sometimes you need only one antiderivative, not all of them. For example, in integration by parts, you may have $dv=\cos x\;dx$, and conclude that $v=\sin x$.)