Proof If $AB-I$ Invertible then $BA-I$ invertible. [duplicate]

I have these problems :

• Proof If $AB-I$ invertible then $BA-I$ invertible.
• Proof If $I-AB$ invertible then $I-BA$ invertible.

I think I solve it correctly, But I'm not so sure, I'll be glad to receive feedback.

• If $AB-I$ invertible then : $$\det|AB-I| \neq 0 \implies \\ \det|A-I||B| \neq 0 \implies \\ \det|B||A-I| \neq 0 \implies\\ \det|BA-I| \neq 0$$

Therefore $BA-I$ invertible.

• If $I-AB$ invertible then :

$$\det|I-AB| \neq 0 \implies \\ \det|I-B||A| \neq 0 \implies \\ \det|I-BA| \neq 0$$

Therefore $I-BA$ invertible.

Solution 1:

There's a slick way to discover the inverse by first solving the problem for (formal) power series.

$$\begin{eqnarray} \rm (1-ab)^{-1} &=&\rm 1+ ab + a\color{#c00}{ba}b + a\color{#0a0}{baba}b +\,\cdots\\ &=&\rm 1+ a (1\, +\, \color{#c00}{ba}\ \ +\ \ \color{#0a0}{baba}\,\ +\,\cdots)b\\ &=&\rm 1+ a (1\,-\,ba)^{-1}b\end{eqnarray}\qquad\qquad$$

Simple algebra proves that this formula is universally correct (as in Kaladin's answer).

At first glance, it seems highly remarkable that such a method should work. Halmos posed the challenge of explaining why this works in one of his popular expositions in Math. Intelligencer. Some explanations are known - see here for more (see also here).

Solution 2:

Here's a way to prove this statement:

1. You want to prove that $BA-I$ is invertible if $AB-I$ is invertible. This is equivalent to proving that $AB-I$ is not invertible if $BA-I$ is not invertible.
2. Can you relate non-invertibility to some condition on the eigenvalues of $AB$ and $BA$? Hint: $0$ is an eigenvalue of $AB-I$ if and only if $1$ is an eigenvalue for $AB$.
3. Finally, what is the relation between the eigenvalues of $AB$ and those of $BA$?

If you want more hints, just ask, but please try to solve it by yourself first.

Solution 3:

If $AB-I$ is invertible with inverse $C$ then look at $-I+BCA$ as inverse of $BA-I$. \begin{align} (BA-I)(-I+BCA) &=-BA+BABCA+I-BCA\\ &= -BA+B(AB-I)CA+I\\ &=-BA+BA+I=I \end{align} and for the other side \begin{align} (-I+BCA)(BA-I)&=-BA+I+BCABA-BCA\\ &=-BA+BC(AB-I)A+I\\ &= -BA+BA+I=I \end{align} For the second part you can do something similar. Assume $I-AB$ has inverse $D$, which is $-C$, then $I-BA$ has inverse $I+BDA=-(-I+BCA)$.