Let $A,B \in {M_2}$ and $C=AB-BA$. Why is ${C^2} = \lambda I$ true? Let $A,B \in {M_2}$ and $C=AB-BA$. Why does ${C^2} = \lambda I$?
 A: By Cayley-Hamilton we have $$C^2-tr(C)C+\det(C)I_2=0,$$ which gives $C^2=\lambda I_2$, because $tr(C)=tr(AB-BA)=0$.
A: It is a well-known fact that for any two matrices $A$ and $B$ with
$C = [A, B] = AB - BA, \tag{1}$
we have
$\text{Tr} (C) = 0, \tag{2}$
where
$\text{Tr}(X) = \text{trace}(X) = X_{11} + X_{22} \tag{3}$
for any $2 \times 2$ matrix $X = [X_{ij}]$,  $1 \le i, j \le 2$. 
It is also easy to verify (2) by direct calculation, especially in the $2 \times 2$ case (though it in fact holds for matrices of any size); with $A = [A_{ij}]$, $B = [B_{ij}]$, we have
$(AB)_{ii} = A_{i1}B_{1i} + A_{i2}B_{2i}; \tag{4}$
reversing the roles of $A$ and $B$ one writes
$(BA)_{ii} = B_{i1}A_{1i} + B_{i2}A_{2i}; \tag{5}$
if we now subtract (5) from (4) and sum over $i$, we find that all the terms of
$\text{Tr}([A, B]) = \sum_1^2 (AB - BA)_{ii} = \sum_1^2 ((AB)_{ii} - (BA)_{ii}) \tag{6}$
cancel out, leaving us with 
$\text{Tr}([A, B]) = 0; \tag{7}$
incidentally, essentially the same argument works for any two  $n \times n$ matrices $A$, $B$.
That $\text{Tr}(C) = 0$ is the first piece of the puzzle; the second is provided by the Hamilton-Cayley theorem for $2 \times 2$ matrices, which asserts that $C$ satisfies the equation
$C^2 - \text{Tr}(C) + \det(C) I = 0; \tag{8}$
see Proof that the characteristic polynomial of a $2 \times 2$ matrix is $x^2 - \text{tr}(A) x + \det (A)$ for a more detailed discussion.  Since $\text{Tr}(C) = 0$, (8) reduces to
$C^2 + \det(C) I = 0, \tag{9}$
or
$C^2 = -\det(C) I; \tag{10}$
taking $\lambda = -\det (C)$ in (10), we are done.
