Dgdrxrt

Subtract $2n$ from both sides and then multiply by $-1$; this gives the equivalent inequality $defrkoperatornamerank(n-rk A)+(n-rk B)geq n-rk(AB)$. In other words (by the rank-nullity theorem) we must show that the sums of the dimensions of the kernels of $A$ and of $B$ gives at least the dimension of the kernel of $AB$. Intuititively vectors that get killed by $AB$ either get killed by $B$ or (later) by $A$, and the dimension of $ker(AB)$ therefore cannot exceed $dimker A+dimker B$. Here is how to make that precise.

One has $ker Bsubseteq ker(AB)$, so if one restricts (the linear map with matrix) $B$ to $ker(AB)$, giving a map $tilde B:ker(AB)to K^n$, one sees that $ker(tilde B)=ker(B)$ (both inclusions are obvious). Since the image of $tilde B$ is contained in $ker A$ by the definition of $ker(AB)$, one has $rktilde Bleqdimker A$. Now rank-nullity applied to $tilde B$ gives $$dimker(AB)=dimkertilde B+rktilde Bleqdimker B+dimker A,$$ as desired.

As left or right multiplications by invertible matrices (and in particular, elementary matrices) don't change the rank of a matrix, we may assume WLOG that $A=pmatrixI_r&0\ 0&0$. Therefore,
beginalign*
operatornamerank(A)+operatornamerank(B)
&=operatornamerank(A)+operatornamerank(AB+(I-A)B)\
&leoperatornamerank(A)+operatornamerank(AB)+operatornamerank((I-A)B)\
&le r+operatornamerank(AB)+(n-r)\
&=operatornamerank(AB)+n.
endalign*
The last but one line is due to the fact that the first $r$ rows of $(I-A)B$ are zero.

$$n+r(AB)=r(M)=r beginpmatrix
I_n & 0 \
0 & AB
endpmatrix$$
Using generalized elementary transformation to $M$:
$$M=beginpmatrix
I_n & 0 \
0 & AB
endpmatrix
to beginpmatrix
I_n & 0 \
A & AB
endpmatrix
to beginpmatrix
I_n & -B \
A & 0
endpmatrix
to beginpmatrix
B & I_n \
0 & A
endpmatrix,$$
hence
$$n+r(AB)=r(M)=
rbeginpmatrix
B & I_n \
0 & A
endpmatrixgeq r(A)+r(B):$$

This solution is from here.

Denote by $r(A)$ the rank of $A$ and by $n(A)$ the nullity of $A$, i.e. the dimension of the nullspace of $A$. The rank plus nullity theorem says that
$r(A)+n(A)=n$ and $r(B)+n(B)=n$ and so $2n= r(A)+r(B)+n(A)+n(B)$.

Let $N(A)$ be the nullspace of $A$ and $R(B)$ the range-space of $B$.
Then $r(AB)=n - n(B)- dim(N(A) cap R(B))$ (try proving this, hint: what is the nullspace of $AB$?). This gives $n = r(AB) +n(B)+ n(N(A) cap R(B))$ and so $n + r(AB) = r(A)+r(B) +n(A) - n(N(A) cap R(B)) ge
r(A)+r(B)$.

Let $T:Vto W, S:Wto V$ be linear transformations which correspond to $A,B$. Equivalently we want prove that $r(T)+r(S)leq r(TS)+n$ also similarly show that $$dim ker(T) + dim ker(S) gt dim ker(TS).$$ I know that $ker (S) subset ker (TS)$. Let $a_1,a_2,ldots,a_r$ be a base for $ker (S)$; expand this base for $ker(ST)$ to $a_1,a_2,ldots,a_r,a_r+1,a_r+2,ldots,a_s$ and then show that $S(a_r+1), S(a_r+2),ldots,S(a_s)$ are a set of independent elements s.t. $S(a_i)in ker (T) , forall i:r+1,ldots,s$ and indicate $ker (T) ge s-r,$ so $dim ker T+ dim ker Sge s-r+r=dim ker TS$. $$ 1.dim ker T+ dim ker Sge dim ker TS$$ $$2.dim ker T+r(T)=n$$ $$3.dim ker S T+r(S)=n$$ $$4.dim ker TS+r(TS)=n.$$ By 1, 2, 3, 4 easily conclude $r(T)+r(S)leq r(TS)+n$.

In response to Mr. Albanese's correct critique. I submit the following. The rank of A,B is in the set of integers [0,n]. Assume we know the relatively straigtfoward proofs that rank(AB)<= min(rank(A),rank(B)), -> rank(AB) = n if rank(A) and rank(B) have rank n. There are three rank possibilities. Let the rank of A = p, the rank of B = q with p,q integers. Then (p or q < n) or (p and q < n) or (p=q=n). The first two cases imply (rank(A) + rank(B) < rank(AB) + n) or (p+q < min(p,q) + n) as p+q < 2n and (rank(A) + rank(B) > rank(AB) + n) is a contradiction. This is a contrapositive proof of the strict inequality in the Sylvester theorem (rank(A) + rank(B) < rank(AB) + n). For the equality case p=q=n. Then (rank(A) + rank(B) = rank(AB)) + n) or n + n = n + n which is true because rank(A) and rank(B) have full rank, n, implies rank(AB)=n.