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

Question

Problem:

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

There is no more information

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 :)

Answer 1

$\Bbb K$ 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

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

Answer 2

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.