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.
 A: 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
$$
|I_2 + AB| = |I_2 + BA|
$$
A: Note that $|I + AB| = 0$ if and only if $I + AB$ is non-invertible, which is true if and only if the equation $(I + AB) x = 0$ has a non-zero solution $x$.
Suppose that $x \neq 0$ is such that $(I + AB)x = 0$. We see that
$$
Ix + ABx = 0 \implies ABx = -x.
$$
Note that because $A(Bx) = -x \neq 0$, it must be the case that $Bx \neq 0$.
Using our observations so far, note that
$$
BA(Bx) = B(AB)x = B(-x) = -Bx.
$$
In other words, $Bx \neq 0$, and we have
$$
BA(Bx) = -Bx \implies IBx + BABx = 0 \implies (I + BA)Bx = 0.
$$
So, the equation $I + BA y = 0$ has a non-zero solution. It follows that $|I + BA| \neq 0$, which was what we wanted.
A: It is just a corollary (taking $n = m = 2$) of the result below:
For $A \in F^{m \times n}, B \in F^{n \times m}$, it follows that
\begin{equation}
\det(I_{(m)} + AB) = \det(I_{(n)} + BA). \tag{$*$}
\end{equation}
To prove identity $(*)$, consider the block matrix:
\begin{align}
\Delta = \begin{pmatrix} I_{(m)} & A \\
-B & I_{(n)} 
\end{pmatrix},
\end{align}
and verify decompositions below:
\begin{align*}
& \begin{pmatrix}
I_{(m)} & A \\
-B & I_{(n)} 
\end{pmatrix} 
\begin{pmatrix} I_{(m)} & 0 \\
B & I_{(n)} 
\end{pmatrix} =
\begin{pmatrix}
I_{(m)} + AB & A \\
0 & I_{(n)} 
\end{pmatrix}, \tag{1} \\
& \begin{pmatrix}
I_{(m)} & A \\
-B & I_{(n)} 
\end{pmatrix} 
\begin{pmatrix} I_{(m)} & -A \\
0 & I_{(n)} 
\end{pmatrix} =
\begin{pmatrix}
I_{(m)} & 0 \\
-B & I_{(n)} + BA 
\end{pmatrix} \tag{2}
\end{align*}
Taking determinants on both sides of $(1)$ and $(2)$ then yields
$$\det(\Delta) = \det(I_{(m)} + AB) = \det(I_{(n)} + BA).$$
