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.

  • 2
    $\begingroup$ No, this is not correct. $I-AB$ has nothing to do with $\left(I-B\right) A$. $\endgroup$ – darij grinberg Dec 23 '14 at 14:20
  • $\begingroup$ @darijgrinberg Both of them are not correct? $\endgroup$ – JaVaPG Dec 23 '14 at 14:22
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    $\begingroup$ The two problems are equivalent, aren't they? Isn't $I-AB = -(AB-I)$, and $I-BA = -(BA-I)$? (And of course $X$ is invertible iff $-X$ is invertible.) $\endgroup$ – MPW Dec 23 '14 at 14:25
  • $\begingroup$ @JaVaPG: Yes, both of them. $\endgroup$ – darij grinberg Dec 23 '14 at 14:44
  • $\begingroup$ See also this question. $\endgroup$ – Bill Dubuque Mar 4 '15 at 19:08

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.

  • $\begingroup$ @M Turgeon I didn't study eigenvalues so much but I think I got the answers, If $0$ is an eigenvalue then we can conclude that the matrix is not invertible, AB and BA has the same eigenvalues, Therefore if AB invertible then BA invertible. $\endgroup$ – JaVaPG Dec 23 '14 at 14:44
  • $\begingroup$ @JaVaPG You're close, but that's not quite the answer. Recall that you are trying to prove $BA-I$ is invertible, not $BA$. $\endgroup$ – M Turgeon Dec 23 '14 at 15:12
  • $\begingroup$ @M Turgeon Therefore if $1$ is an eigenvalue for $AB$ or $BA$ then both $AB-I$, $BA-I$ are not invertible, and since we know that $AB-I$ is invertible $0$ isn't an eigenvalue which implies that $0$ isn't an eigenvalue for $BA-I$ therefore 1 is not an eigenvalue for $BA$ so $BA$ invertible. $\endgroup$ – JaVaPG Dec 23 '14 at 15:19
  • $\begingroup$ @JaVaPG Correct. $\endgroup$ – M Turgeon Dec 23 '14 at 15:21
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    $\begingroup$ good answer I really like it +1 for you $\endgroup$ – Learnmore Dec 23 '14 at 15:28

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).


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 $I+BDA=-(-I+BCA)$.

  • $\begingroup$ See my answer for a remarkable way to discover the above inverse formula. $\endgroup$ – Bill Dubuque Dec 24 '14 at 19:58

Your result will be true if $A$ or $B$ is supposed to be invertible.

Since we have $\det(ABA-A)=\det(A(BA-I))$:

$$ \det((AB-I)A)=\det (A(BA-I))\implies \det A\cdot\det (AB-I)=\det A \cdot\det(BA-I).$$

result follows if $\det A\neq0$.

Moreover $$\begin{align}\det(BA-I)\cdot\det B &= \det((BA-I)B)\\ &=\det(BAB-B)\\&=\det(B(AB-I))\\&=\det B\cdot \det(AB-I).\end{align}$$

Result follows if $\det B \neq 0$.

  • 1
    $\begingroup$ I don't think this is a counter-example. I may have made a computation mistake, but I have $$AB-I = \begin{pmatrix}1&1\\-2&-2\end{pmatrix}$$ and also $$BA-I = \begin{pmatrix}0&1\\0&-1\end{pmatrix},$$ and neither are invertible. Moreover, you give a proof for when $A$ is invertible, which is the case in your "counter-example". $\endgroup$ – M Turgeon Dec 23 '14 at 15:06

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