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$\Bbb Z_2 = \{1, 0\}.$ 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?

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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.

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  • $\begingroup$ But are there 2 equal matrixes (with different numbers) over Z2? $\endgroup$
    – aaamm
    May 22, 2021 at 18:00
  • $\begingroup$ If two matrices are equal, they don't have "different numbers". The way you write the numbers down isn't the same thing as the number represented by that symbol. The abstract object, even if it can be written down in different ways, is unique. Check out my answer here for a longer explanation: math.stackexchange.com/questions/4074774/… , and don't hesitate to ask another question if you still have doubts afterwards. $\endgroup$ May 23, 2021 at 1:11
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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 $\Bbb Z_2$.

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