**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.2 Exercise 7.2.6**

Let $S$ be a ring with identity $1 \neq 0$. Let $n$ be a positive integer and let $A = [a_{i,j}]$ be an $n \times n$ matrix over $S$. Let $E_{i,j}$ be the element of $M_n(S)$ whose $(i,j)$ entry is 1 and whose other entries are all 0.

(1) Prove that $E_{i,j}A$ is the matrix whose $i$-th row is the $j$-th row of $A$ and all other rows are 0.

(2) Prove that $AE_{i,j}$ is the matrix whose $j$-th column is the $i$-th column of $A$ and all other columns are 0.

(3) Deduce that $E_{p,q}AE_{r,s}$ is the matrix whose $(p,s)$ entry is $a_{q,r}$ and all other entries are 0.

Solution:

(1) By definition, $E_{i,j}A = [c_{p,q}]$, where $$c_{p,q} = \sum_{k=1}^n e_{p,k}a_{k,q}.$$ Note that if $p \neq i$, then $e_{p,k} = 0$, so that $c_{p,q} = 0$. If $p = i$, then $c_{p,q} = a_{j,q}$; thus the $i$-th row of $E_{i,j}A$ is the $j$-th row of $A$, and all other entries are 0.

(2) The proof for $AE_{i,j}$ is very similar.

(3) By the above arguments, $E_{p,q}A$ is the matrix whose $p$-th row is the $q$-th row of $A$, and all other entries are $0$. Then $E_{p,q}AE_{r,s}$ is the matrix whose $s$-th column is the $r$-th column of $E_{p,q}A$, which is all zeroes except for the $p$-th row, whose entry is the $(q,r)$ entry of $A$, and all other entries are zero.