# 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?

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?

• No. Why is $$\begin{bmatrix}R_{1} \\ {R_{2}}\end{bmatrix} \underset{\text{RowOp}}{\mapsto} \begin{bmatrix}R_{1} \\ {R_{2} + A R_{1}}\end{bmatrix}$$ a row operation? – user642338 Feb 7 at 6:52

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:

$$\operatorname{rk}TM = \operatorname{rk}M$$