If $AB-I$ is regular, then $BA-I$ is regular How do I prove that if $AB-I$ is regular $BA-I$ is regular?
I proved it only if $A$ or $B$ is regular.
if $A$ is irregular:
$$
\implies \det(AB-I) = \det(A(B-A^{-1})) = \det(A)\det(B-A^{-1}) \\
= \det(B-A^{-1})\det(A)=\det((B-A^{-1})A)=\det(BA-I) \implies
$$
because $AB-I$ is irregular $\det(AB-I) = \det(BA-I) != 0$ so $BA-I$ is regular,same if $B$ is regular.
 A: Hint. If $(AB-I)x=0$ for some $x\ne0$, what is $(BA-I)(Bx)$? Can $Bx$ be zero?
A: If I understand your question correctly, a possible answer comes from the more general fact

Let $A$ be a unital ring, and $a, b \in A$. Then $1 - a b$ is invertible if and only if $1 - b a$ is invertible.

Suppose $1 - a b$ is invertible. Then $ (1 - b a) b  = b - b a b = b (1 - ab)$. So $b$ is in the right ideal $I$ generated by $1 - ba$, and thus $b a \in I$, so that $1 = (1 - ba) + ba \in I$, and thus $1 - ba$ is right invertible. A similar argument, starting with $a (1 - ba)$, shows that $1 - ba$ is left invertible, and thus invertible.
Explicitly, $b = (1 - ba) b (1-ab)^{-1}$, so $1 = (1 - ba) + (1 - ba) b (1-ab)^{-1} a$, and thus
$$
1 = (1 - ba) (1 + b (1-ab)^{-1} a),
$$
that is,
$$
(1 - ba)^{-1} =  1 + b (1-ab)^{-1} a.
$$
A: Hint
Since you proved it for $A$ invertible so in the general case use the density of $GL_n(\Bbb R)$ in $M_n(\Bbb R)$ and the continuity of the determinant.
A: I prove this for general $A$ and $B$. We assume that $A$ and $B$ are respectively $n\times m$ and $m\times n$ matrices. $I_n$ represents the identity matrix of size $n\times n$ Consider the following matrix $M$ defined as follows:
$$
M=\left[\begin{matrix}
I_{n}&A\\
B&I_m
\end{matrix}\right]
$$
By changing the rows and columns of the matrix $M$ we can arrive at the following Matrix:
$$
M'=\left[\begin{matrix}
I_m&B\\
A&I_n
\end{matrix}\right]
$$
The only thing that matters is that by doing this, we can say the following things about the determinant of $M$ and $M'$:
$$
\mid \det(M)\mid=\mid \det(M')\mid.
$$
Now using Schur complement of $M$ and $M'$ we can see :
$$
\det(M)=\det(I_n-AB)\\
\det(M')=\det(I_m-BA).
$$
Therefore we have:
$$
\mid \det(I_n-AB)\mid=\mid \det(I_m-BA)\mid
$$
which proves what you want.
A: Preface. You have proved the equality $\det(AB-I) = \det(BA-I)$ for all cases when $A$ is regular (I'll use the term invertible instead), and you want to extend it to the case of an arbitrary $A$. There are standard ways to do such generalizations.
One way is hinted at in Sami Ben Romdhane's answer. Another way could be the algebraic analogue of it: the algebraic variety generated by $\operatorname{GL}_n(\mathbb{R})$ is $\operatorname{M}(\mathbb{R})$, therefore any polynomial identity that holds for all invertible matrices must hold for all matrices, both invertible and singular. What follows is an implementation of that idea, but in more elementary terms.
Proposed solution. Pick two arbitrary matrices $A_0$ and $B$. Also, pick some invertible matrix $A_1$. Now let us denote $A(t) = (1-t)A_0 + tA_1$. You can think of $A(t)$ as a straight line in the "space of all matrices". At $t=0$ we have $A(0) = A_0$, at $t=1$ we have $A(1) = A_1$.
Now, let's look at these two functions:
$$
\begin{align}
P(t) &= \det A(t) \\
Q(t) &= \det(A(t)B-I) - \det(BA(t) - I)
\end{align}
$$
Here are some observations about them:


*

*Both $P(t)$ and $Q(t)$ are polynomials of $t$ (do you understand why?).

*At $t=1$ we have $P(1) = \det A_1 \neq 0$. Also, $Q(1) = \det(A_1B-I) - \det(BA_1 - I) = 0$ (this follows from your work, since $A_1$ is invertible).

*At $t=0$ we have $Q(0) = \det(A_0B-I) - \det(BA_0-I)$. What we want to prove is that $Q(0)=0$.


Let us concentrate on $P(t)$ first. We know that it is a polynomial, and that it is not zero (because $P(1) \neq 0$). It is known that a nonzero polynomial on $\mathbb{R}$ can only have a finite number of roots (it is bounded by the degree of the polynomial). So, $P(t) = 0$ only for finitely many $t$. Then $P(t) \neq 0$ for all $t$ in an infinite set $T \subseteq \mathbb{R}$.
Now let us look at $Q(t)$. Whenever $t \in T$, we have $P(t) = \det A(t) \neq 0$, so the matrix $A(t)$ is invertible for each $t \in T$. But then we have $Q(t) = \det(A(t)B-I) - \det(BA(t)-I) = 0$ (you have proved this). We see that polynomial $Q(t)$ has infinitely many roots! This is only possible if $Q(t)=0$ for every $t \in \mathbb{R}$. So in fact the equality
$$
\det(A(t)B-I) = \det(BA(t)-I)
$$
holds for all $t \in \mathbb{R}$. At $t=0$ we have $\det(A_0B-I) = \det(BA_0-I)$, as required.
