Let $A$, $B$ be square matrices of order $2$ such that $|I_2 + AB| = 0$. Prove that $|I_2 + BA| = 0$.

In this question, I denote $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix},$$ $$B=\begin{bmatrix}e&f\\g&h\end{bmatrix}.$$ So, from $|I_2 + AB| = 0$, I'll have: $$(ae+fc+1)(gb+hd+1) = (ga+hc)(eb+fd).$$

If $|I_2 + BA| = 0$ (which we have to prove here), then $$(ae+bg+1)(cf+hd+1) = (af+bh)(ce+dg).$$ From both equations, I can see that what we're going to prove here is $cf=bg$, more precisely, $g=f$ and $b=c$.

Is my thinking path right or wrong? I still have not figured out the solution yet, I appreciate any help.

Solution 1:

We have $$ |I_2 + AB| = a e (d h + 1) - b c f g + d h + 1 $$ which is clearly invariant under swapping $a\leftrightarrow e$, $b\leftrightarrow f$, $c\leftrightarrow g$, $d\leftrightarrow h$. Hence $$ |I_2 + AB| = |I_2 + BA| $$