Prove that the rank of the block matrix is rank $A$ + rank $B$ Assume $A,B$ are $n \times n$ matrices.
Prove that the rank of the block matrix that is defined as the following
$$
\pmatrix{A& AB\\B&B+B^2}
$$
 is equal to $ \mbox{rank} A +\mbox{rank} B$.
I have absolutely no clue...  
 A: Row/column reduction of block matrices works here. First substract $B$ times the first column from the second column. Then subtract the second column from the first, giving you the matrix 
$$C = \pmatrix{A& 0\\ 0 & B}$$
This is where the matrices in the answer by Humanity come from. Note that the row operations will not change the rank of the matrix. Hence the matrix has same rank as $C$, and it is not too difficult to see that $C$ has rank equal to $\operatorname{rank} A + \operatorname{rank} B$.
A: Denoting the identity matrix by $I_n$, we have
$$M=\pmatrix{A&AB\\B&B+B^2}=\underbrace{\pmatrix{A&0\\0&B}}_{D}\underbrace{\pmatrix{I_n&0\\I_n&I_n}}_{S}\underbrace{\pmatrix{I_n&B\\0&I_n}}_{T}.$$
Now it is easy to see that $S, T$ are both invertible, while $D$ has rank equal to rank $A$ + rank $B$, hence $M$ as well. This shows also that the range of $M\in M_{2n}(K)$, seen as a linear operator on $K^n\oplus K^n$, is $\mbox{im }M=\mbox{im } D=\mbox{im }A\oplus \mbox{im }B$. 
A: Let $C,D$ be the reduced row echelon form of $A$ respectively $B$. Let $E,F$ be invertible matrices so that $EA=C$ and $FB=D$.
Then
$$\pmatrix{E& 0\\0&F}\pmatrix{A& AB\\B&B+B^2}=\pmatrix{EA& EAB\\FB&FB+FB^2}$$
Now note that all the leading ones in this matrix appear either in $EA$ or in $FB$. [If a row is $0$ in $EA$, what can you say about the same row in $EAB$?, same for $FB$].
