Nonexistence of Complex Matrices whose Commutator equals Identity

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Theorem

Let $d \in \N_{> 0}$ be a positive natural number.

Let $\mathbf A, \mathbf B \in \C^{d \times d}$ be complex matrices.


This article, or a section of it, needs explaining.
In particular: Same applies to real matrices too, yes? I don't have the source to support that, but I don't see why not.
This theorem includes the result on the real case, because real matrices are especially complex matrices.
Obvious though it is, the real case might be worth adding as a corollary
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.
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Let $\mathbf I$ be the $d \times d$ identity matrix.


Then there is no $\mathbf A$, $\mathbf B$ such that $\mathbf A \mathbf B - \mathbf B \mathbf A = \mathbf I$.


Proof

We have that complex numbers form a commutative ring.

Aiming for a contradiction, suppose there are $\mathbf A$, $\mathbf B$ such that $\mathbf A \mathbf B - \mathbf B \mathbf A = \mathbf I$.

Then:

\(\ds \map \tr {\mathbf I}\) \(=\) \(\ds d\) Definition of Trace of Matrix
\(\ds \) \(=\) \(\ds \map \tr {\mathbf A \mathbf B - \mathbf B \mathbf A}\)
\(\ds \) \(=\) \(\ds \map \tr {\mathbf A \mathbf B} - \map \tr {\mathbf B \mathbf A}\) Trace of Sum of Matrices is Sum of Traces
\(\ds \) \(=\) \(\ds 0\) Trace of Product of Matrices

This is a contradiction.

$\blacksquare$


Sources

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