If $(A-B)^2=O_2$ then $\det(A^2 - B^2)=(\det(A) - \det(B))^2$

Question

Let $A,B \in M_2(\mathbb{R})$ be two matrices such that $(A-B)^2=O_2$.

Prove $\det(A^2 - B^2)=(\det(A) - \det(B))^2$.

OBS. $O_2$ is the zero matrix

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Let $D=A-B$ then $D^2=O_2$ therefore, using Cayley Hamilton theorem, we get $tr(D)=\det(D)=0$. It follows $tr(A)=tr(B)=t$ and, from here, $A^2 -B^2=t(A-B) + (\det(A) - \det(B))I_2$

This is all I could get.

UPDATE

The matrices are from $M_2(\mathbb{R})$. Sorry for misleading you.

Answer 1

Let's rewrite the - correct - identity you obtained as follows: \begin{align*}& A^2 -B^2=t(A-B) + (\det(A) - \det(B))I_2\\ \iff& t(A-B)=A^2 -B^2- (\det(A) - \det(B))I_2\end{align*} Squaring both sides of your original identity gives $$(A^2 -B^2)^2=t^2\underbrace{(A-B)^2}_{=0} + 2(\det(A) - \det(B))t(A-B) + (\det(A) - \det(B))^2I_2.$$ Now substitute $t(A-B)$ given by the second equation: \begin{align*}(A^2 -B^2)^2& =2(\det(A) - \det(B))[(A^2 -B^2)- (\det(A) - \det(B))I_2] + (\det(A) - \det(B))^2I_2\\ & = 2(\det(A) - \det(B))(A^2 -B^2)- (\det(A) - \det(B))^2I_2 \end{align*}

and lastly apply Cayley Hamilton to $A^2-B^2$ and compare the coefficients.

Note that no cancellation takes place, as, using that $tr(A)=tr(B)=t$, one can prove that $$2(\det(A) - \det(B))=-(tr(A^2+B^2))$$ Indeed this is again a consequence of Cayley Hamilton, taking its trace: $$A^2-tr(A)A+\det(A)I_2=0\Rightarrow tr(A^2)-tr(A)^2+2\det(A)=0$$

Answer 2

Since $D=A-B$ is nilpotent, it is similar to the matrix $\left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right)$. So without loss of generality assume $D = \left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right)$. If $B = \left(\begin{matrix} a & b \\ c & d \end{matrix}\right)$, then $A = B+D = \left(\begin{matrix} a & b+1 \\ c & d \end{matrix}\right)$. Now $$ A^2-B^2 = (B+D)^2 - B^2 = (B^2 + BD + DB + D^2) - B^2 = BD+DB$$ since $D^2 = 0$. Notice that $$ BD+DB = \left(\begin{matrix} a & b \\ c & d \end{matrix}\right)\left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right) + \left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right)\left(\begin{matrix} a & b \\ c & d \end{matrix}\right) = \left(\begin{matrix} c & a + d \\ 0 & c\end{matrix}\right) $$ and hence $$ \det(A^2-B^2) = \det(BD+DB) = c^2. $$ On the other hand, $$ \det(A) = \det\left(\begin{matrix} a & b+1 \\ c & d \end{matrix}\right) = ad - (b+1)c = \det(B) - c$$ and hence $(\det(A) - \det(B))^2 = c^2$ as well.

Thus $\det(A^2-B^2) = c^2 = (\det(A)-\det(B))^2$, as desired.