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If $A$ and $B$ are $n\times n$ complex matrices , then $AB-BA=I$ is impossible

I understand this by any example. But how can one explain it generally?

user190080
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Parul
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  • what do you mean by explain? Do you want some maybe some graphical intuition? – user190080 Sep 05 '16 at 15:39
  • It has an interesting generalization for (possibly infinite dimensional) linear operators. – Santiago Sep 05 '16 at 15:40
  • @Santiago I would be interested by a reference to this generalization. – Jean Marie Sep 05 '16 at 15:49
  • @JeanMarie https://en.wikipedia.org/wiki/Stone%E2%80%93von_Neumann_theorem – Santiago Sep 05 '16 at 16:04
  • I would like to draw the attention to the quantum physical applications of such relationships in particular to $QP-PQ=i\hslash$ ($\hslash$ is the normalized Plank's constant) which appeared for the first time in an article of Heisenberg in 1927(http://plato.stanford.edu/entries/qt-uncertainty/). – Jean Marie Sep 05 '16 at 16:05
  • @Santiago Thanks very much. – Jean Marie Sep 05 '16 at 16:07

2 Answers2

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Observe that $\mathrm{tr}(AB-BA)=0$, while $\mathrm{tr}(I)=n$.

carmichael561
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5

Another way could be to try the multiplicative identity

$$\det({\bf AB}) = \det({\bf A})\det({\bf B})$$

And consider the relation between determinant and eigenvalues

$$\det({\bf X})=\prod_{i=0}^{n} \lambda_i({\bf X})$$

together with the perturbation of eigenvalues by addition of identity:

$$\lambda_k({\bf X+I}) = 1+\lambda_k({\bf X})$$

mathreadler
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