Elementary matrix proof I am supposing that $E$ is the elementary matrix obtained from $I$ (the identity matrix), by adding $\mu$ times the $m$-th row to the $l$-th row for some $\mu \in \mathbb{R}$ and $1\leq l,m\leq n$ and $l\neq m$.
I need to prove that for any $n\times n$ the matrix $B=EA$ can be obtained from $A$ by applying the same row operations.
I think I need to start by stating $E=e_{ik}$, so $e_{ik}= \left\{
  \begin{array}{2 2}
    1 \quad \text{if} \space i=k & \quad \\ \mu \quad \text{if} \space i=l \space \text{and} \space k=m & \quad \\0 \quad \text{otherwise}
  \end{array} \right.$
But I'm not sure where to go from there. I'm not very good at matrices so a bit more in depth explanation would be appreciated.
 A: Let's do it formally. I'll use the notation $E_{lm}(d)$ for the matrix obtained from the identity by adding to the $l$-th row the $m$-th row multiplied by $d$, with $l\ne m$. If $e_{ij}$ are its coefficients, then
$$
e_{ij}=\begin{cases}
1 & \text{if $i=j$}\\
d & \text{if $i=l$ and $j=m$}\\
0 & \text{otherwise}
\end{cases}
$$
Denote by $a_{ij}$ the coefficients of $A$. Then the coefficient at position $(i,j)$ in $E_{lm}(d)A$ is:


*

*for $i\ne l$,
$$
\sum_{k} e_{ik}a_{kj}=e_{ii}a_{ij}=a_{ij}
$$
because $e_{ik}=0$ for $k\ne i$ and $e_{ii}=1$.

*for $i=l$,
$$
\sum_{k} e_{lk}a_{kj}=e_{ll}a_{lj}+e_{lm}a_{mj}=a_{lj}+da_{mj}
$$
because $e_{lk}=0$ for $k\notin\{l, m\}$, while $e_{ll}=1$ and $e_{lm}=d$.
A: You can easily see that the $l^{th}$ row of $E$ will have $\mu$ at position $m$. So $b_{lj}$ (the $(l, j)$ position in $B$) is the scalar multiplication of the $l^{th}$ row of $E$ and $j^{th}$ column of $A$. But in this row everything is $0$ except position $m$ (where resides $\mu$) and position $l$ (where resides $1$), so the result will be $\mu a_{mj} + a_{lj}$. And you get just what you need : $$(b_{l1}, b_{l2}, ... b_{ln}) = \mu (a_{m1}, a_{m2}, ... a_{mn}) + (a_{l1}, a_{l2}, ... a_{ln})$$
A: $E$ is constructed by adding a single element, $\mu$, at some off-diagonal position in the identity matrix, $I$.
Let $H$ be a matrix of all zeros except for a single $\mu$ at the $(l,m)$ position: $h_{lm} = \mu$.
Then, $$E = I+H$$
so $$EA = (I+H)A = A+HA.$$
Now, it should be obvious that $HA$ is a matrix of zeros except for a single row, which is $\mu$ times some other row of $A$. Therefore, adding $HA$ to $A$ yields the result we demand.
