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This question is inspired by my previous question.

What if I removed the constraint that $M,N $ be finite dimensional? Is it possible then to find matrices $M,N$ s.t. $MN-NM=I$?

Thanks.

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Let $K$ be a commutative ring, let $$\frac{d}{dX}$$ be the endomorphism of $K[X]$ given by the differentiation with respect to $X$, and let $$M$$ be the endomorphism of $K[X]$ given by the multiplication by $X$. In formula: $$ \frac{d}{dX}\ f(X)=f'(X),\quad M\ f(X)=Xf(X). $$ Then we have $$ \frac{d}{dX}\ M-M\ \frac{d}{dX}=\text{Id}_{K[X]}. $$

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    Thanks, would you mind explaining a bit what that means? (Sorry, I am not too familiar with those words.) Thanks again!2011-11-07
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    Dear @asker: You're welcome. I hope things are clearer now. Sorry for having been cryptic...2011-11-07
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    Thanks a lot @Pierre-YvesGaillard, that is indeed a lot clearer now. I just have a small question.2011-11-07
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    Could you possibly explain why $d/dX M - M d/dX = id_{K[X]}$? I am new to operators. Presuming that $X$ is a field, what does it mean to differentiate or multiply by $X$? Thanks again! :-)2011-11-07
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    Dear @asker: Do you agree with the computation $$\frac{d}{dX}\Big(M\ f(X)\Big)-M\left(\frac{d}{dX}\ f(X)\right)=\frac{d}{dX}\Big(Xf(X)\Big)-Xf'(X)$$ $$=f(X)+Xf'(X)-Xf'(X)=f(X)?$$ Do you agree that, if $K$ is a field, then $K[X]$ (the ring of polynomials in $X$ with coefficients in $K$) is a $K$-vector space? Do you agree that $\frac{d}{dX}$ and $M$ are endomorphisms (or linear transformations) of $K[X]$?2011-11-07
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    $X$ is not a "field": just the variable $X$ and you sure know how to multiply by $X$, differentiate with respect to $X$ and compute $(xf(x))' - xf'(x)$.2011-11-07
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    Ah! Thanks again!!2011-11-07
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In infinite dimension they are not usually called matrices, but linear operators. Anyway this is still impossible in any unital normed algebra:

Assume $MN-NM=\lambda\cdot I$. Then for all $k\in\mathbb{N}^+$ we have $M^kN-NM^k=k\cdot\lambda\cdot M^{k-1}$. This can be proved by induction on $k$. Therefore $\lambda\le\frac{2\|M\|\|N\|}{k}$.