Question about elementary row operations with block matrices Given two $n \times n$ matrices $A$ and $B$, form a new block matrix 
$$P := \begin{bmatrix}I_n&B\\-A&0\end{bmatrix}$$
Then by using only elementary row operations, show that $P$ can be transformed into
$$P' := \begin{bmatrix}I_n&B\\0&AB\end{bmatrix} $$ 

The solution to this problem is:
$$P = \begin{bmatrix}I_n&B\\-A&0\end{bmatrix} \sim \begin{bmatrix}I_n&B\\-A + AI_n &0 + AB\end{bmatrix} \sim \begin{bmatrix}I_n&B\\0&AB\end{bmatrix}$$
I don't understand this solution. Why can $A$ be multiplied from the left on the first half of the matrix and then be added to the second half of the matrix to form a sequence of elementary row operations?
 A: I guess you're trying to prove Sylvester rank inequality.
This works just like the elementary row operations. We can do this with block matrices:
$$
M = \begin{pmatrix}A&B\\C&D\end{pmatrix} \sim \begin{pmatrix}A+K\,C&B+K\,D\\C&D\end{pmatrix}
$$
What we are doing here behind the scenes is multiplying the matrix by an elementary matrix $T$:
$$
T = \begin{pmatrix}I&K\\0&I\end{pmatrix} 
$$
So we get:
$$
TM =
\begin{pmatrix}I&K\\0&I\end{pmatrix} \begin{pmatrix}A&B\\C&D\end{pmatrix} = \begin{pmatrix}A+K\,C&B+K\,D\\C&D\end{pmatrix}
$$
As you can see, $T$ has full rank and therefore invertible, which means it doesn't change rank of the matrix:
$$\operatorname{rk}TM = \operatorname{rk}M$$
A: Let 
$$
\begin{bmatrix}R_{1} \\  {R_{2}}\end{bmatrix} = \begin{bmatrix}I_n&B\\-A&0\end{bmatrix} 
$$
Then 
$$
\begin{bmatrix}R_{1} \\  {R_{2}}\end{bmatrix}  \underset{\text{RowOp}}{\mapsto} \begin{bmatrix}R_{1} \\  {R_{2} + A R_{1}}\end{bmatrix}
\implies
\begin{bmatrix}I_n&B\\-A&0\end{bmatrix} 
\underset{\text{RowOp}}{\mapsto} 
\begin{bmatrix}I_n&B\\-A+AI_{n}& 0+AB\end{bmatrix}
=
\begin{bmatrix}I_n&B\\0& AB\end{bmatrix} 
$$
Does this make it any clear?
