I was studying matrices when I came accross the following theorem:
The effect of the elementary column operations $C_{ij}$, $cC_i(c\neq 0)$ and $C_i+cC_j$, on an $m×n$ matrix $A$ is obtained by post multiplying $A$ by $(E_{ij})^t$, $(E_{i})^t$ and $(E_{ij}(c))^t$ respectively, where each elementary matrix is of order $n$ .
The notations used above have their usual meanings.(I think these notations are standard if, they need clarity please do inform me, I will edit in the details about them )
My proof for the theorem goes like this:
We consider a matrix $P$ of order $r×n$. Let the rows of $P$ are $P_1,P_2,P_3,...,P_r$ and let the columns of $A$ be $A_1,A_2,...,A_n$. Now, we consider the matrix $PA= \begin{pmatrix} P_1A_1 & P_1A_2 & ... & P_1A_n\\ P_2A_1 & P_2A_2 & ... & P_2A_n\\ ...\\ P_rA_1 & P_rA_2 & ... & P_rA_n \end{pmatrix}$. Now, if we intercange the ith column of $A$ with the jth column then the transformed matrix is $C_{ij}(A)$ and if we consider the matrix $PC_{ij}(A)$, then, $PA$, also has its ith and jth row interchanged so we can say, $PC_{ij}(A)=C_{ij}(PA)$. Now, if we multiply a scalar $c\neq 0$ to a row of $A$, then, we denote this by $C_i(c)(A)$. We consider the matrix $PC_{i}(c)(A)$ . We find that then the matrix $PA$ also has its ith column multiplied by $c$ , and hence $PC_{i}(c)(A)=C_{i}(c)(PA)$. Again, we consider the column operation where $c$ times the jth column of a matrix is added to ith column of the matrix. We denote this operation by $C_{ij}(c)(A)$. Now, we consider $PC_{ij}(c)(A)$. We find that the c times the jth column is added to the ith column in $PA$, when $A$ is tranformed under this operation and hence $PC_{ij}(c)(A)=C_{ij}(c)(PA)$. Thus we can write $C_{ij}(A)=C_{ij}(A.I)=AC_{ij}(I)=A.E_{ij}$. Similarly, $C_{i}(c)(A)=C_{i}(c)(A.I)=AC_{i}(c)(I)=AE_i(c)$ and $C_{ij}(c)(A)=C_{ij}(c)(A.I)=AC_{i}(c)(I)=AE_i(c)$.
Is the above proof correct? If so, then in the theorem above "$(E_{ij})^t$, $(E_{i})^t$ and $(E_{ij}(c))^t$" should actually be just "$(E_{ij})$, $(E_{i})$ and $(E_{ij}(c))$" (although I can't decipher what is meant by $(E_{ij})^t$, etc?) , correct?
The proof looks correct. The problem you have is with notation. Notice that the elementary matrices that interchange two $\textit{rows}$, $E_{ij}$, and multiply a row by a constant $c$, $E_{i}(c)$, are symmetric, so they also do the job for column operations. But this is not the case for the matrix giving the sum of the $i$th row and the $j$th row multiplied by a constant, $E_{ij}(c)$. If you do $AE_{ij}(c)$, then you are substituting the $i$th column, $C_i$, by $cC_i+C_j$, instead of $C_i+cC_j$. To remedy this, you just have to transpose, that is, take the product $A(E_{ij}(c))^{t}$.