# Proving the group of homomorphisms is isomorphic to matrices

I'm trying to understand this theorem:

If $f, g: R^n \to R^m$ are given by the matrices $A, B \in R_{m,n}$ then $f + g$ is given by $A + B$. Thus, $\operatorname{Hom}_R (R^n, R^m)$ and $R_{m,n}$ are isomorphic as additive groups. If $R$ is commutative, they are isomorphic as $R$-modules.

How would I go about proving it? I know the last bit follows from observing that $x+x=x*2$, so we can say they are $\mathbb Z$-modules. I'm not sure about the rest though.

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Each map $f$ gives rise to a matrix and vice-versa. – Sigur Jan 2 '13 at 9:04
What does it mean to say "f is given by A"? I think if you can answer this, then you can check easily using linearity (why is f+g linear?) that (f+g)(x) and (A+B)(x) are the same for every $x\in R^n$. – Billy Jan 2 '13 at 9:50

Let $e_j = (0,\dots,1,\dots,0) \in R^n$ be the standard generators of $R^n$ as a free $R$ module, for $j = 1, \dots, n$. Similarly, let $f_i$ be the standard generators of $R^m$. Then any homomorphism of $R$-modules $f: R^n \rightarrow R^m$ gives a matrix $A_f = (a_{ij}) \in R_{m,n}$, where $f(e_j) = \Sigma_{1 \leq i \leq m} a_{ij}f_i$. If $g \in Hom_R(R^n, R^m)$ with matrix $B_g = (b_{ij})$, then for all $j = 1, \dots n$, $f + g (e_j)$ $= f(e_j) + g(e_j)$ $= \Sigma_{1 \leq i \leq m} (a_{ij} + b_{ij})f_i$. Thus, the matrix of $f + g$ is $(a_{ij} + b_{ij}) = A_f + B_g$.
Now define $\varphi: Hom_R(R^n,R^m) \rightarrow R_{m,n}$ has follows: $\varphi(f) = A_f$, where $A_f$ is as above. Then what we showed above is that $\varphi$ is a group homomorphism. Now, given a matrix in $R_{m,n}$ can you define a linear map $R^n \rightarrow R^m$. Can you then define a map $R_{m,n} \rightarrow Hom_R(R^n,R^m)$ which is an inverse of $\varphi$? This will show that $\varphi$ is an isomorphism.
Verify that if $f \in Hom_R(R^n,R^m)$, then for any $r \in R$, the matrix of $rf$ is just $rA_f$. This shows that $\varphi$ is also an $R$-linear map.
The inverse would just be $\phi^{-1}(A_f) = f$, demonstrating onto-ness. I'm not sure how to show that $ker(\phi)$ must be ${0}$ though. Where does the commutivity of R come into play in the last bit? That should all follow from $Hom_R(R^n, R^m)$ being a module homomorphism, no? – randomafk Jan 2 '13 at 22:23
Dear @randomafk: If you know that $\varphi$ has an inverse, then you know that $\varphi$ is bijective. In particular it is injective and so has trivial kernel. I am not quite sure you need commutativity of $R$ really. But then you have to differentiate between left and right modules. – Rankeya Jan 2 '13 at 22:26