# Let $A,B \in M_{2x2}(\mathbb{K})$, prove that $(AB-BA)^2=-\det(AB-BA)I$

Problem:

Let $$A,B \in M_{2x2}(\mathbb{K})$$, prove that $$(AB-BA)^2=-\det(AB-BA)I$$.

Question: Is there some other way to prove this besides brute force?, and if is, could you give me some hint/solution?

Thanks so much :)

$$\Bbb K$$ is any field, right? In any event, I take $$\Bbb K$$ such.

Then for all

$$X \in M_{2 \times 2}(\Bbb K), \tag 1$$

the Cayley-Hamilton theorem holds; that is, $$X$$ satisfies it's own characteristic polynomial

$$\det(\mu I - X) = \mu^2 - \text{Tr}(X)\mu + (\det X)I ; \tag 2$$

thus,

$$X^2 - \text{Tr}(X)X + (\det X) I = 0; \tag 3$$

in particular, if

$$\text{Tr}(X) = 0, \tag 3$$

then

$$X^2 = -(\det X)I; \tag 4$$

we now use the well-known (and easily verified by direct calculation) fact that for any two

$$A, B \in M_{2 \times 2}(\Bbb K), \tag 5$$

$$\text{Tr}([A, B]) = 0; \tag 6$$

setting

$$X = [A, B], \tag 7$$

we see from (4) that

$$[A, B]^2 = -\det([A, B])I. \tag 8$$

Maybe prove that $$X^2=-\det(X)I$$ when $$X\in M_{2\times 2}(\mathbb{K})$$ with $$tr(X)=0$$. In this way you don't need to use 8 variables, just 3.