Writing a homomorphism in multiplicative form Consider the additive group $\mathbb {Z_2}=\{0,1\}$ and consider the group $\mathbb Z_2^n$. Let $u_1,\ldots,u_d$, $(d>n)$ be not necessarily distinct , non-zero elements of $\mathbb Z_2^n$. Define the map $A:\mathbb Z_2^d\to\mathbb Z_2^n$ by sending $E_i$ to $u_i$ where $E_i=(0,\ldots,1,\ldots,0)$. 
If $u_i=(a_{1i},\ldots,a_{ni})$ then $A$ is given by the matrix $$A=\begin{pmatrix}a_{11}&\cdots&a_{1d}\\
\vdots& &\vdots\\
a_{n1}&\cdots&a_{nd}\end{pmatrix}$$ Where $a_{ij}$'s are either $0$ or $1$.
Now identify $\mathbb Z_2$ with $\{1,-1\}$ the multiplicative group of square roots of unity. In this notation how would I write the map $A$ as a matrix? Should I just replace the $0$'s by $1$ and the $1$'s by $-1$? If so then how would the $A$ operate on $E_i'=(1,\ldots,-1,\ldots1)$?
 A: In the presentation $\mathbb{Z}_2=\{-1, 1\}$ you cannot write your (group) homomorphism as a matrix! In fact in the presentation $\mathbb{Z}_2=\{0,1\}$ you cannot do so either if you are considering $\mathbb{Z}_2$ as a group not a ring!
To make this clear consider a homomorphism $\phi: \mathbb{Z}_2^2\to \mathbb{Z}_2^2$ sending $(1,0)\mapsto (a,b)$ and $(0,1)=(c,d)$. The question is whether we can write $\phi(x,y)=A(x,y)^T$ for some matrix $A$? Suppose the answer is yes, then
$$
\phi(x,y)=\begin{pmatrix}a & c\\
b& d\end{pmatrix}\begin{pmatrix}
x\\ y
\end{pmatrix}=\begin{pmatrix}
a\cdot x+c\cdot y\\
b\cdot x+d\cdot y
\end{pmatrix}
$$
There is something very wrong about the above form: What do you mean by $a\cdot x$? The only operation on $\mathbb{Z}_2$, considered only as a group, is addition! But here were using both addition and mutiplication, so we are indeed considering $\mathbb{Z}_2$ as a ring not a group! Matrix presentations for group homomorphisms are in general impossible due to the nature of matrix product needing both addition and multiplication.
Thankfully once we consider $\mathbb{Z}_2=\{0,1\}$ as a ring, then everything makes sense. But if you present $\mathbb{Z}_2=\{1,-1\}$ there is no way you can define a nice addition! The second $\mathbb{Z}_2$ is not a ring (at least in any obvious way).
Extra: There is one way of dealing with this which is not helpful at all. Define the addition on $\mathbb{Z}_2=\{-1,1\}$ as $a+b=a*b$ ($*$ is the $\mathbb{Z}$ mutiplication)and the multplication by $1\cdot a = 1$ and $-1\cdot a=a$. This is esentially renaming $0$ by $1$ and $-1$ by $1$. In this sense in $A$ you replace all 0s and 1s respectively but then you need to use this new addition and multiplication. This is however utterly useless in practice...
