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: 
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 $v=\sin x$.)
The indefinite integral $\int f(x)\,dx$ is defined to be the general class of *functions* whose derivatives are $f(x)$. 

Now, given any function $F$ with $F'=f$, it follows that $F+C$ is also an antiderivative of $f$:
$$
(F+C)'=F'+C'=F'+0=F'=f.
$$
Conversely if $G$ is an antiderivative of $f$, then $G$ has the form $F+C$ for some constant $C$.

That's why we write, for example:
$$
\tag{1}
\int x^2\,dx={x^3\over 3}+C.
$$
The derivative of ${x^3\over 3}+C$ is $x^2$. Moreover if $F'(x)=x^2$, then $F$ must have the form $F(x)={x^3\over3}+C$.

So, the general antiderivative of $f(x)=x^2$ has the form $F(x)={x^3\over3}+C$, and equation (1) is just stating this fact.

The differential equation you are considering has more than one solution. The *general solution* is as you have, with the arbitrary parameter $C$. It does belong there. 
Any function of that form will be a solution to the equation.

If you were given an initial value problem where, say, you needed $y(1)=2$, then that constant $C$ could be determined.