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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?

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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?

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    $\begingroup$ 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? $\endgroup$ – user642338 Feb 7 at 6:52
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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:

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

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