Question: In this Wikipedia [page](https://en.wikipedia.org/wiki/Matrix_ring), the 5-th example claims the ring of endomorphisms of right $R$-module $M=\oplus\_{i\in I}R$ is isomorphic to the Column Finite Matrices $\mathbb{CFM}\_I(R)$; while the ring is isomorphic to Row Finite Matrices $\mathbb{RFM}\_I(R)$ when regarding $M$ as a left $R$-module.

I want to understand how the right/left module assumptions change the column/row orientation in the claim. Is it because, when $M$ is seen as a right module, we have to multiply the matrix from right side of a row "vector" in $M$? And vice versa? But if that's the case, in the first case, it doesn't seem we need the matrices to be column finite, since there are only finitely many nonzeros in the direct sum (row "vector").

That's my confusion. Thanks.

Answer: 
If you attempted to let $\mathbb{CFM}\_I(R)$ operate on the right side of row vectors from $\oplus\_I R$ you would immediately be in trouble, because the image of an element of $\oplus\_I R$ might not be in $\oplus\_I R$.

I'm taking liberties below to describe it as if it is indexed with $\omega$ so that I can talk about "first row" and whatever, but the same argument holds without a well-ordered index set (it's just more obnoxious to write.)

For example, let the "first" row of your column finite matrix be all $1$'s, and the second row be all $1$'s and the rest of the entries are all $0$. Obviously column finite, yes?

If you multiply this on the right side of the vector $[1,1,0,0,\ldots]$ (zeros on the tail forever, that is) then the result is $[2,2,2,2,\ldots]$ ($2$'s forever) which is not in $\oplus\_I R$.

But if you have these matrices operate on the left side of column vectors from $\oplus\_I R$, you will see that it does not have this problem.

> 
> You explained why "multiplying the column vectors on the left by ℂ() works", but not how it relates to $$ being a right $$-module.
> 
> 
> 

Well, technically you would be considering $\oplus\_I R$ as a left $R$ module when using RFM matrices on the right. This is to ensure the transformation stays $R$ linear. Of course, that only matters when $R$ isn't commutative.

This is the natural action of a matrix $E$ on a right of $M$ in a matrix equation: $E(xr)=(Ex)r$. If instead you wanted "$Erx=Exr$" to turn out right $R$ linear, then in general you have no way to commute the $r$ to the right of the $x$. That isn't a problem when you do it on the other side.

And likewise, the natural action of a matrix on the right is paired with the module action on the left:

$(rx)E=r(xE)$
You can roughly think of the endormorphism of a *right* $R$-module $M$ as matrices working on $M$ through multiplication on the *left*.

Perhaps this can be made clear with an example. Consider $A=k[(X\_{i})\_{i\in\mathbb{N}}]/(X\_{0}X\_{i}-X\_{0})\_{i\in\mathbb{N}},(X\_{i}X\_{j})\_{i>0,j\geq 0}$. Multiplying from the *left* with $X\_{0}$ is a morphism from $A$ to $A$ in the category of *right* $A$-modules. Indeed, for any $a,b\in A$ we have $X\_{0}(ab)=(X\_{0}a)b$. This does not work if we consider $A$ as a left module over itself, for then we would need $X\_{0}(ab)=a(X\_{0}b)$ for all $a,b$.

If we take $1,X\_{0},X\_{1},...$ as a $k$-basis for $A$, then this left multiplication is represented by the matrix $$M\_{X\_{0}}=\begin{bmatrix}0&0&0&0&0&\cdots\\1&0&1&1&1&\cdots\\
0&0&0&0&0&\cdots\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots
\end{bmatrix}$$
which is column finite but not row finite. For example, the element $1+X\_{0}+X\_{2}$ should be mapped to $2X\_{0}$. Indeed:
$$\begin{bmatrix}0&0&0&0&0&\cdots\\1&0&1&1&1&\cdots\\
0&0&0&0&0&\cdots\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots
\end{bmatrix}\begin{bmatrix}1\\1\\0\\1\\0\\\vdots\end{bmatrix}=\begin{bmatrix}0\\2\\0\\\vdots\end{bmatrix}.$$