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.
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.
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]}. $
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 $\lambda\le\frac{2\|M\|\|N\|}{k}$.