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: 
It is simply a special case of the ubiquitous *linearity* principle that the *general* solution of a *nonhomogeneous* linear equation is given by adding any *particular* solution to the general solution of the associated *homogeneous* equation. Namely, if $\rm\:D\:$ is a linear map then one easily proves

**Lemma** $\ \ $ If $\rm\ D\:f\_1\ =\ g\ $ then $\rm\ D\:f\_2\ =\ g\ \iff\ 0\ =\ D\:f\_1 - D\:f\_2\ =\ D\:(f\_1-f\_2)$ 

Therefore $\rm\ D^{-1}(g)\ =\ f\_1\ +\ ker\ D\ =\:\: $ particular + homogeneous solution, as in linear algebra.

In particular: $\rm\ \int g\ =\ f\_1\: +\ c,\ $ for $\rm\ c\in ker\ \dfrac{d}{d\:x}=\: $ constants w.r.t. the derivation $\rm\ D = \dfrac{d}{d\:x}\:.$

Compare this to $\rm \ x\ =\ 3 + 5\ \mathbb Z,\:$ the solution of $\rm\ 2x \equiv 6\pmod{\!10},\: $ with particular solution $\rm x \equiv 6/2 \equiv 3,\: $ and homogeneous: $\rm\ 2x\equiv 0\pmod{\!10}\iff 10\mid 2x\iff 5\mid x\iff x\in 5 \mathbb Z$
Say every month you earned 10 dollars on a principal of 1000 dollars and put it aside. What have you got after 10 months ?

Not always 100 dollars or 180 dollars or 130. That is so, only if the box contained nothing not at start.

If it contained 35 dollars at start, that is the constant added to the accumulated (integrated) amount $A:$

$ A = m.10 + c = m.10 + 35 $ dollars or 135 or 235 dollars..

The same applies to more complicated integrals. It is also the *initial value* in integration of a variable function with initial value prescribed to evaluate boundary value differential equations.