Identity matrix over $\Bbb{Z}_2$ $\Bbb Z_2 = \{1, 0\}.$
I’ve been told there’s only one Identity matrix over all fields except $\Bbb Z_2$, in which there could be more than one.
Could anyone give an example of an Identity matrix over $\Bbb Z_2$ which isn’t the standard identity matrix?
 A: I think you've been told wrong. It's a very standard theorem of abstract algebra that should an identity exists, it is unique. Suppose there exist two different identities for your monoid/group/etc for a given operator, you can show that they are necessarily equal.
A: Given any set $S$ with a binary operation $\cdot$, there is at most one identity element. ($(S, \cdot)$ needn't be a group. The operation needn't even be associative!)
Indeed, suppose $1$ and $1'$ are identity elements, then $$1 = 1 \cdot 1' = 1',$$
where the left (resp., right) equality uses the fact that $1'$ (resp., $1$) is an identity.

So this means that there is only one identity matrix of size $n \times n$, which is the familiar one.
(By "an identity matrix", I assume you are talking about an $n \times n$ matrix $I_n$ over $\Bbb Z_2$ such that $I_n M = M = M I_n$ for all $n \times n$ matrices $M$ over $\Bbb Z_2$.)

One possible interpretation could be the fact that $-I_n$ is "also" an identity matrix but that is not saying much because the equality $1 = -1$ holds in the field $\Bbb Z_2$.
