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Let $A$ be an $n$ by $n$ matrix over a field $K$, define $R:=\{ \sum_{i=0}^{\infty} c_i A^i$ with only finitely many $c_i \neq 0 \}$

Could anyone help me show you can turn $R$ into a commutative ring with $1$?

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    @Benjamin Lim: I assume $A$ is a set matrix?2012-05-08
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    @Benjamin Lim: Yes, unless you can see any ambiguity in the wording I used?2012-05-08
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    @Benjamin Lim: Indeed, I've been trying to use that fact to find an identity/zero (you can take the down vote off I'd you like!)2012-05-08
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    @Benjamin Lim: are you sure $A \in R$ there are only finitely many non-zero coefficients2012-05-08
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    Is there an issue? The polynomials in $A$ add and multiply in the natural way.2012-05-08
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    @LHS Suppose you can produce a matrix $B$ such that $AB = A$. Then this $B$ must also satisfy $IB = I$ because $I \in R$ no?2012-05-08
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    @Andre Nicolas: the issue is that they are power series with finitely many zero coefficients, how can you create an identity?2012-05-08
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    The $0$-th power of $A$. It is finitely many non-zero coefficients.2012-05-08
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    @Benjamin Lim: how do we know $I\in R$2012-05-08
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    @LHS What is $A^0$?2012-05-08
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    Oh dammit.. Was reading it wrong.. I'm dsylexic..2012-05-08
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    @LHS Problem solved?2012-05-08
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    Yep haha! Sorry about that..2012-05-08
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    @LHS Well I posted everything as an answer anyway.2012-05-08
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    @LHS Man you got me confused too!2012-05-08

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Let $A$ be some $n\times n$ matrix over a field $K$ and consider $R := \{ \sum_{i=0}^{\infty} c_i A_i\}$ where all but finitely many $c_i$ non-zero. Now the identity matrix is in here because it can be written as $1 \times A^0$. Therefore it follows that the identity matrix is the multiplicative identity of the ring $R$.

It is easy to see from here that $R$ can be made into a unital commmutative ring.