writing proofs involving matrices Notation has always been my downfall when it comes to writing proofs. I have been given a general problem to try and write a proof based on this. This is just for research and practice. Given a matrix $e_{ij}A_{nxm}$ If the matrix is multiplied on the left by  
$$\begin{bmatrix}0&1\\0&0\end{bmatrix}$$ multiplied by $$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$  the result is $$\begin{bmatrix}c&d\\0&0\end{bmatrix}$$
Which tells me that row j replaces row i and everything else becomes zeroes
Multiplication on the left by
$$\begin{bmatrix}0&0\\1&0\end{bmatrix}$$ times $$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ produces $$\begin{bmatrix}0&0\\a&b\end{bmatrix}$$
which tells me that this kind of multiplication will put row i to row j and everything else will become zeroes. I'm just not sure how to officially write this as a proof. 
 A: Remember the basic product formula: if $A=[a_{hk}]$ is an $m\times n$ matrix and $B=[b_{hk}]$ is an $n\times p$ matrix, then the coefficient in place $(h,k)$ of $AB$ is
$$
\sum_{l=1}^n a_{hl}b_{lk}.
$$

The $n\times n$ matrix $E_{ij}$ having $1$ in place $(i,j)$ can be described by having coefficients $c_{hk}$ $(1\le h\le n$, $1\le k\le n)$ where
$$
c_{hk}=\delta_{hi}\delta_{jk}
$$
where, as usual,
$$
\delta_{rs}=\begin{cases}
1&\text{if $r=s$,}\\
0&\text{if $r\ne s$.}
\end{cases}
$$
Consider an $n\times n$ matrix $A=[a_{hk}]$; the coefficient $b_{hk}$ in place $(h,k)$ of the product $B=E_{ij}A$ can be written as
\begin{align}
b_{hk}=\sum_{l=1}^n (\delta_{hi}\delta_{lj})a_{lk}
&=\sum_{l=1}^n \delta_{hi}(\delta_{lj}a_{lk})\\
&=\sum_{l=1}^n \delta_{hi}(\delta_{lj}a_{lk})\\
&=\delta_{hi}\sum_{l=1}^n (\delta_{lj}a_{lk})\\
&=\delta_{hi}a_{jk}
\end{align}
Therefore
$$
b_{hk}=\begin{cases}
0&\text{if $h\ne i$,}\\
a_{jk}&\text{if $h=i$.}
\end{cases}
$$
This amounts exactly to saying that all coefficients of $B$ are zero except possibly for row $i$, which is the same as row $j$ of $A$.
