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I need to prove that rings $End(\mathbb{Z}^{n})$ and $M_{n}(\mathbb{Z})$ are isomorphic.

To start with, I let $A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots& & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix} \in M_{n}(\mathbb{Z})$ (so that all the entries $a_{ij} \in \mathbb{Z}).$

Then, I defined $\Gamma(A): \mathbb{Z}^{n} \to \mathbb{Z}^{n}$ to be such that $\Gamma(A) \begin{pmatrix} z_{1} \\ z_{2} \\ \vdots \\ z_{n} \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots& & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix}\begin{pmatrix} z_{1} \\ z_{2} \\ \vdots \\ z_{n} \end{pmatrix}$, and checked that $\Gamma(A)$ is a group homomorphism by considering $\begin{pmatrix} z_{1} \\ z_{2} \\ \vdots \\ z_{n} \end{pmatrix}, \begin{pmatrix} w_{1} \\ w_{2} \\ \vdots \\ w_{n} \end{pmatrix} \in \mathbb{Z}^{n}$ and showing that $\Gamma(A)\left[ \begin{pmatrix} z_{1} \\ z_{2} \\ \vdots \\ z_{n} \end{pmatrix} + \begin{pmatrix} w_{1} \\ w_{2} \\ \vdots \\ w_{n} \end{pmatrix} \right] = \Gamma(A)\begin{pmatrix} z_{1} \\ z_{2} \\ \vdots \\ z_{n} \end{pmatrix} + \Gamma(A)\begin{pmatrix} w_{1} \\ w_{2} \\ \vdots \\ w_{n} \end{pmatrix}$

So, now, since $\Gamma(A)$ is a group homomorphism from $\mathbb{Z}^{n}$ into itself, $\Gamma(A) \in End(\mathbb{Z}^{n})$, the ring of endomorphisms on $\mathbb{Z}^{n}$.

I'm not sure how to use what I just did, though, to show that an isomorphic map exists between $End(\mathbb{Z}^{n})$ and $M_{n}(\mathbb{Z})$, or even if it helps at all. Could somebody please help me with this proof?

Thank you.

Maybe this wasn't the best first attempt, but I am honestly very confused, and constructive criticism is the best kind of criticism.

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    The endomorphisms are linear transformations are represented by the matrices explicating their action on the canonical basis. thus the two rigns are almost equal. – 2017-02-21
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    @Alephnull okay...but I was asked to show they're isomorphic. So could you please help me with that? Even if almost equal is stronger/better. – 2017-02-21
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    f(x)=Mx where M is an arbitrary matrix is an endomorphism. Moreover there is always such an M corresponding to arbitrary endomorphism f. Let i(f)=M where i is the isomorphism. – 2017-02-21
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    @Alephnull $f(x)=Mx$ is basically what I wrote in my question when I had $\Gamma(A)$ taking $\mathbb{Z}^{n}$ to $\mathbb{Z}^{n}$, right? So, now what I'm asking you is, what should $i$ be? How do I take the endomorphism (and what do elements $f$ look like?) and what kind of mapping do I put it through to have it spit me out a matrix? – 2017-02-21
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    You can do it taking $\phi:M_n(\mathbb{Z})\to\text{End}(\mathbb{Z}^n)$ given by $\phi(A)=\Gamma(A)$ and seeing if it is a ring homomorphism. If it is, can you check the kernel, and the image? – 2017-02-21
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    For surjectivity you may want to check that every endomorphism has such a matricial representation. If you have such an endomorphism $\gamma$ and $e_1=(1,0,...,0),...,e_n=(0,...,0,1)$, you can take the vectors $\gamma(e_1),...,\gamma(e_n)$ and paste them to form a matrix $A(\gamma)$. What can you say about that matrix? – 2017-02-21
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    @LeviathanTheEsper I hope you're not getting at anything like a basis, because I'm not allowed to use any of that kind of stuff yet. On the other hand, if you're saying that that matrix is the identity matrix? How does that help? – 2017-02-21
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    The idea is that of a basis, but you don't need any general knowledge from bases to use that idea. – 2017-02-21
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    And no, that matrix is not necessarily the identity. – 2017-02-21
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    For example, in $n=2$, we can take $T(u)=-u$, then $e_1=(1,0)^t,e_2=(0,1)^t$ and $T(e_1)=(-1,0)^t,T(e_2)=(0,-1)^t$ and the matrix would then be: ? ($u^t$ means "the transpose of $u$", to keep with the column notation) – 2017-02-21
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    @JessyunBourne I defined the i that way. If you want the corresponding matrix for f the matrix will be {f(e_1) |f(e_2)|....}. This is not necessary however; once you know the matrix multiplication corresponds to the application of the endomorphism this is sufficient. – 2017-02-21

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This is just a sketch. The map $$\Gamma: M_n(\mathbb{Z}^n)\rightarrow End(\mathbb{Z}^n), A\mapsto \Gamma(A) $$ is the desired isomorphism. It is easy to check that $\Gamma(A+B)=\Gamma(A)+\Gamma(B)$ as well as $\Gamma(AB)=\Gamma(A)\circ \Gamma(B)$, which means that $\Gamma$ is a ring homomorphism. To see the injectivity, let $e_i$ denote the $i$-th unit vector and note that $Ae_i$ is the $i$-th column of $A$. Thus, if $\Gamma(A)=0$, we also have $0=\Gamma(A)(e_i)=Ae_i$ for all $i$, so $A=0$.

The surjectivity remains. If $f$ is an endomorphism, let $A_f$ be the matrix with $f(e_i)$ as $i$-th column. Use the linearity of $f$ to show that $\Gamma(A_f)(x)=f(x)$ for all $x\in \mathbb{Z}^n$, that is $\Gamma(A_f)=f$.