Proof review/explanation: Let $\text{char}(\mathbb{K}) = 0$. It then follows that $AB-BA \ne 1 \, (A, B \in \Bbb K^{n \times n})$

Question

Let $\text{char}(\mathbb{K}) = 0$. It then follows that $AB-BA \ne 1 \, (A, B \in \Bbb K^{n \times n})$.

I first showed that $\text{trace}(AB) = \text{trace}(BA)$ for every $A, B \in \Bbb K^{n \times n}$:

Let $A=(a_{ij}), B=(b_{ij}) \in \Bbb K^{n \times n}$, so $AB=(c_{ij}) \in \Bbb K^{n \times n}$ where $c_{ij}= \sum_{k=1}^na_{ik}b_{kj}$. One now gets:

$$\text{trace}(AB) = \sum_{i=1}^n \sum_{k=1}^n a_{ik}b_{ki} = \sum_{k=1}^n \sum_{i=1}^n a_{ik}b_{ki} = \sum_{k=1}^n \sum_{i=1}^n b_{ki}a_{ik} = \text{trace}(BA)\tag{*}.$$

Now, assume $AB-BA=1$. Consider

$$AB-BA=1 \Leftrightarrow \text{trace}(AB)-\text{trace}(BA)=\text{trace}(1) \xrightarrow{(*)} \\ 0 = n,$$

which is a contradiction, so the assumption $AB-BA=1$ was wrong.

My question is: Where exactly do you need $\text{char}(\Bbb K)=0$ in this proof?

Thanks in advance!

Answer 1

The trace of the identity matrix is the sum $n1$. The trace functional is 'commutative' in the sense that $\mathrm{tr}(AB) = \mathrm{tr}(BA)$. If the characteristic is zero, then the contradiction is achieved.

Answer 2

If $\text{char}(\mathbb{K})=p$ and $p \mid n$, then you get $0=0$, which doesn't give you a contradiction.