NxN Matrices Commuting to Identity

Question

Does there exist for any n>2 a collection of nxn matrices that commute to the identity matrix?

Answer 1

No. If $A,B \in M_n(\mathbb{F})$ are matrices (no need for $n > 2$) then $\operatorname{trace}(AB) = \operatorname{trace}(BA)$. Working over a field $\mathbb{F}$ of characteristic zero (such as $\mathbb{R}$ or $\mathbb{C}$) we have

$$ 0 = \operatorname{trace}(AB) - \operatorname{trace}(BA) = \operatorname{trace}[A,B] \neq \operatorname{trace}[I_n] = n $$

which shows that the equation $[A,B] = I_n$ has no solution. This shows that any model in which you have two operators that satisfy such a relation (for example, the position and momentum operators in quantum mechanics, up to factors) must be an infinite dimensional one.

Answer 2

Take $n=3$.

Consider $GL_{n}(\mathbb{R})$, the set of all invertible $3\times 3$ matrices over real numbers.

Consider the subset $Z\subset GL_{n}(\mathbb{R})$ of scalar matrices.